https://fweb.wallawalla.edu/class-wiki/api.php?action=feedcontributions&feedformat=atom&user=Timothy.RasmussenClass Wiki - User contributions [en]2026-05-18T09:48:15ZUser contributionsMediaWiki 1.43.0https://fweb.wallawalla.edu/class-wiki/index.php?title=Example:_Metal_Cart&diff=8819Example: Metal Cart2010-01-29T04:41:27Z<p>Timothy.Rasmussen: </p>
<hr />
<div>==Problem==<br />
A DC generator is built using a metal cart with metallic wheels that travel around a set of perfectly conducting rails in a large circle. The rails are L m apart and there is a uniform magnetic <math>\vec B</math> field normal to the plane. The cart has a penguin,with mass m, and is driven by a rocket engine having a constant thrust <math> F_1 </math>. A wet polar bear, having stumbled out of a shack where he recently had a bad experience with a battery, lays dying across the tracks acting as a load resistance R over the rails. Find The current as a function of time. What is the current after the generator attains the steady-state condition?<br />
<br />
[[Image:Emec_cart_polarBear2.png]]<br />
<br />
<br />
<blockquote>Problem loosely based on 2.6 from Electric Machinery and Transformers, 3rd ed <br />
<ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001)</ref></blockquote><br />
==Solution==<br />
<br />
<br />
For this Problem the large circle will be represented by a pair of parallel wires and the cart as a single wire. This is illustrated below in the top and end view figures.<br />
[[Image:Emec_cart_topview.png]]<br />
<br />
<br />
[[Image:Emec_cart_endview.png]]<br />
<br />
We have two forces, <math> F_1 </math> being the force from the rocket engine and <math> F_2 </math> being the force caused by the current in the conductor and the Magnetic Field.<br />
The resulting Force <math> F_t </math> is simply the sum of <math> F_1 </math> and <math> F_2 </math><br />
<br />
<math> F_2 </math> can be found using Ampere's Law<br />
<br />
<math>\vec F=\int\limits_{c} I ~\vec dl\times \vec B~~~~~\Longrightarrow~~~~~ \vec F_2=\int\limits_{0}^{L} I ~\vec dl\times \vec B ~~~~~\Longrightarrow~~~~~ \vec F_2=- I(t) ~B~L ~~ \hat i</math><br />
<br />
We can also say that <math> I(t)=\frac{-e_m(t)}{R} </math><br />
<br />
And <math>~~ {e_m(t)}= \int\limits_{o}^{L} (\vec v \times \vec B)~\vec dl ~~~~~\Longrightarrow~~~~~ {e_m(t)}=-v~L~B </math><br />
<br />
<br />
<math> I(t)=\frac{v~L~B}{R} ~~~~~~~~~~~~~~~~~~~~~~\vec F_2=- I(t) ~B~L ~~ \hat i~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L~B}{R} ~B~L ~~ \hat i ~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L^2~B^2}{R} ~~ \hat i</math><br />
<br />
<br />
<math> \dot v=\frac{F_1}{m} ~\hat i~+\frac {F_2}{m}~ \hat i ~~~~~\Longrightarrow~~~~~ \dot v= ~\frac{F_1}{m} ~~\hat i~- \frac{v~L^2~B^2}{R m}~~\hat i ~~~~~\Longrightarrow~~~~~ \dot v~ +~ v \left(\frac{L^2~B^2}{R~m}\right) ~-~\frac{F_1}{m}~=~0 </math><br />
<br />
Now we have a lovely differential equation to work with! To attempt to find the current we will take the Laplace transform.<br />
<br />
<br />
<math> \mathcal{L}\left\{ \dot v ~+~v\left(\frac{L^2~B^2}{R~m}\right)-\frac{F_1}{m} \right\} ~~~~~\Longrightarrow~~~~~ s~V(s)~+~\frac{L^2B^2}{R~m} V(s) ~-~\frac{F_1}{m~s}</math><br />
<br />
Lets title and substitute in the variable <math>~~~~\psi=\frac{L^2~B^2}{R ~m}~~</math> to simplify things<br />
<br />
<math> s~V(s)~+~\psi~V(s)~-~\frac{F_1}{m~s}~=~0</math><br />
<br />
<math> V(s)~(s~+~\psi)~=~\frac{F_1}{m~s} ~~~~~\Longrightarrow~~~~~ V(s)~=~\frac {F_1}{m~s~(s~+~\psi)}</math><br />
<br />
Using partial fraction expansion<br />
<br />
<math>\frac {F_1}{m~s~(s~+~\psi)}~~~~~~\Longrightarrow~~~~~~\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi} </math><br />
<br />
<br />
<math>V(s)=\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi}</math><br />
<br />
<br />
<math>V(t)= \mathcal{L}\left\{ \frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi} \right\} ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0}{m~\psi}\left(u(t)~-~e^{-\psi~t} ~u(t)\right) ~~~~~\Longrightarrow~~~~~ V(t)= \frac {F_0}{m~\psi}\left(1~-~e^{-\psi~t}\right) </math><br />
<br />
<br />
<math> V(t)=\frac {F_0}{m~ \frac{L^2~B^2}{R ~m}}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right) ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0~R}{L^2~B^2}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math><br />
<br />
<br />
We know that<br />
<math>~~ e_m(t)~=~V(t)LB </math><br />
<br />
So we can substitute in V(t) to get<br />
<br />
<math> e_m(t)= \frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math><br />
<br />
And we know that <math>~~I(t)~=~\frac{e_m(t)}{R}</math> <br />
<br />
<br />
<math> I(t)=\frac{\frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)}{R}~~~~~\Longrightarrow~~~~~ I(t)~=~\frac {F_0}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math><br />
<br />
To find the steady-state current we simply look at the limit of I(t) as <math>t \rightarrow \infty</math><br />
<br />
<math>\lim_{t\rightarrow \infty} I(t)=\frac{F_0}{L~B} \left( 1- e^{-\infty}\right) ~~~~~\Longrightarrow~~~~~ I(\infty)=\frac{F_0}{L~B} </math><br />
<br />
<br />
So the Steady-State Current = <math>\frac{F_0}{L~B}</math><br />
<br />
<br />
<br />
<math> In~conclusion~it ~can ~be ~seen ~that ~a ~penguin ~driven, ~polar ~bear ~killing ~generator </math><br />
<br />
<math>~would ~be ~a ~viable ~option ~for ~alternative ~energy ~in ~Canada. </math><br />
<br />
<br />
==References==<br />
<br />
<references /><br />
<br />
<br />
==Reviewed by==<br />
<br />
[[Kirk Betz]] Read and approved 1-26-10<br />
<br />
[[Will Griffith]] Approved 1-27-10<br />
<br />
<br />
==Read By==<br />
<br />
<br />
==Comments==<br />
I completely agree with your conclusion (Tim Rasmussen)</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=8437Electromechanical Energy Conversion2010-01-20T16:42:17Z<p>Timothy.Rasmussen: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
* If it's been approved by the reviewers, move it to the articles section<br />
<br />
Does anyone know why my LaTEX stuff is changing sizes throughout my article? [[An Ideal Transformer Example]]<br />
<br />
*(John Hawkins) As I understand it, the text is made full size (larger) if there is ever a function call, i.e. something starting with a backslash, excluding some things like greek letters. I have just put "\ " (the function call for a space) at the beginning of an equation and had it work. If you don't want to change anything about your equation but just want it displayed full size, type "\,\!" (small forward space and small backward space) somewhere in your equation.<br />
*Thanks John!<br />
<br />
==Announcements==<br />
<br />
If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Example: Ampere's Law]] (Tyler Anderson)<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[AC vs. DC]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Faraday's Law]]<br />
* [[Eddy Current]]<br />
* [[Example Problems of Magnetic Circuits]]<br />
* [[Magnetic Circuit]] (John Hawkins)<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
* [[An Ideal Transformer Example]]<br />
* [[Example: Ideal Transformer Exercise]] (John Hawkins)<br />
* [[Reference Terms and Units]] (Amy Crosby)<br />
* [[Ideal Transformer Example|Example: Ideal Transformer]]<br />
* [[Problem Set 1]](Jodi Hodge)<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Gauss Meters]] (Tyler Anderson)<br />
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)<br />
* [[Magnetic Flux]] (Jason Osborne)<br />
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)<br />
* [[AC Motors]]<br />
* [[Example Problem - Toroid]] ([[Kirk Betz]])<br />
* [[Transformer_example_problem|Ideal Transformer Example]] (Tim Rasmussen)</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Tim_Rasmussen&diff=8316Tim Rasmussen2010-01-19T03:43:23Z<p>Timothy.Rasmussen: </p>
<hr />
<div>==Contact Information==<br />
Timothy.Rasmussen@WallaWalla.edu<br />
<br />
909-260-3874<br />
<br />
<br />
==Published Articles==<br />
<br />
[[AC Motors]]<br />
<br />
==Reviewed Articles==<br />
[[Ferromagnetism]]<br />
<br />
[[Nick_ENGR431_P1|Magnetostatics]]<br />
<br />
[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]]<br />
<br />
==Articles in Progress==<br />
<br />
[[Transformer example problem]]<br />
<br />
==Sandbox==<br />
<br />
[[SandBox]]</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=8285Electromechanical Energy Conversion2010-01-19T00:33:20Z<p>Timothy.Rasmussen: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
* If it's been approved by the reviewers, move it to the articles section<br />
<br />
Does anyone know why my LaTEX stuff is changing sizes throughout my article? [[An Ideal Transformer Example]]<br />
<br />
*(John Hawkins) As I understand it, the text is made full size (larger) if there is ever a function call, i.e. something starting with a backslash, excluding some things like greek letters. I have just put "\ " (the function call for a space) at the beginning of an equation and had it work. If you don't want to change anything about your equation but just want it displayed full size, type "\,\!" (small forward space and small backward space) somewhere in your equation.<br />
*Thanks John!<br />
<br />
==Announcements==<br />
<br />
If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Gauss Meters]]<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[AC vs. DC]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Faraday's Law]]<br />
* [[Eddy Current]]<br />
* [[Example Problems of Magnetic Circuits]]<br />
* [[Magnetic Circuit]] (John Hawkins)<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
* [[An Ideal Transformer Example]]<br />
* [[Example: Ideal Transformer Exercise]] (John Hawkins)<br />
* [[Reference Terms and Units]] (Amy Crosby)<br />
* [[Transformer_example_problem|Ideal Transformer Example]]<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)<br />
* [[Magnetic Flux]] (Jason Osborne)<br />
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)<br />
* [[AC Motors]]</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=AC_Motors&diff=8284AC Motors2010-01-19T00:32:42Z<p>Timothy.Rasmussen: </p>
<hr />
<div><br />
== Basic Parts and Principles ==<br />
<br />
Electric motors convert electrical energy into mechanical motion by using magnetic forces to accelerate objects. Electricity comes in two flavors: [[AC vs. DC| AC and DC]]<ref>http://fweb/class-wiki/index.php/AC_vs._DC</ref>. Therefore, electric motors need to be able to utilize at least one of these in order to operate. As a general rule, AC and [[DC Motor|DC]] motors are constructed using slightly different parts because of the different behavior of the types of electricity. Lets first look at the parts in a generic AC motor and then discuss the role they play in making motion. <br />
<br />
AC motors consist mainly of a stator and an armature<ref>http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motorac.html</ref>. The stator is fixed inside the motor. Stators are almost always made using tightly wound wire in order to yield a high magnetic flux density. The second part is the rotor, which rotates to provide movement to whatever application is desired. The rotor can also use wound wire, through which current flows or a permanent magnet. In order to get this current to the rotor without tangling wires around the rotor, metal slip rings are used to complete the circuit<ref>http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motorac.html</ref>.<br />
<br />
Magnetic flux is created when current passes through the armature wires. Since the motor uses alternating current, the magnetic field will alternate polarity. Both the stator and rotor produce magnetic fields. The basic interaction between magnetic fields indicates that opposite poles attract while like poles repel. All electric motors use this behavior to produce rotation. When the poles of the stator and rotor are the same, the force will push the two apart. Similarly, the stator and rotor will be pulled together. Motors use both of these simultaneously to impart motion to the rotor, to which the output shaft is attached. This rotation can be used to do useful mechanical work.<br />
<br />
==Synchronous AC Motors==<br />
Synchronous motors are termed "synchronous" because they inherently run at a constant velocity which is synchronized with the frequency of the AC power supply. These motors contain the same two basic components common to all motors: A rotor - the components that rotate, and a stator - the outside shell of the motor. The rotor can be made from either a permanent magnet or winding powered by a DC power source. When powered, this winding operates as a permanent magnet. The rotor has 2 poles in the simplest case, but can have many more depending on the application. The stator holds the armature winding which creates a pulsating magnetic field inside the motor. The armature winding can be either single or multi-phase depending on the configuration of the motor.<br />
<br />
Synchronous motors create a torque from the magnetic field of the rotor interacting with the alternating field created by the armature. The field created by the armature is continuously changing because the coils are powered by an AC source. As the voltage in the windings swings from positive to negative, the magnetic field also shifts<ref>http://www.allaboutcircuits.com/vol_2/chpt_13/2.html</ref>. As this field shifts from north to south, the poles on the rotor inside of the motor will be either attracted or repelled from the coils of the armature. These attraction and repulsion forces create a torque which drives the rotor.<br />
<br />
[[Image:Synchronous_Motor.JPG]]<ref>http://www.allaboutcircuits.com/vol_2/chpt_13/2.html</ref><br />
<br />
<br />
Synchronous motors are unique in that they are not self starting. This is because as soon as voltage is applied to the armature windings, the magnetic field varies at the frequency of power line. At start up, this field builds so quickly that the rotor can not get up to speed and synchronize with the varying armature winding. Several different methods can be used to start synchronous motors. First and most simply, a secondary motor can be used to start the rotor spinning at the rotational velocity corresponding to the frequency of field shift. As soon as the rotor achieves this velocity, it "snaps in" to synchronism and the secondary motor can be shut down or disconnected<ref>http://www.acsynchronousmotors.com/</ref>. Large Synchronous motors can also employ a separate starting mechanism in the rotor. A squirrel cage winding in the rotor can be fed with DC power through slip rings to bring the rotor up to speed<ref>http://www.electricmotors.machinedesign.com/guiEdits/Content/bdeee2/bdeee2_1-5.aspx</ref>. As the synchronous motor of this type starts, it essentially operates as a DC motor until it reaches the operating speed of the AC line.<br />
<br />
==Induction Motors==<br />
Induction motors are termed "induction" because there is no current supplied to the rotating coils in the rotor. The coils are closed loops which have large currents induced in them by the changing magnetic field produced in the stator coils<ref>http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/indmot.html</ref>. This is different from synchronous AC motors which can have a current supplied onto the rotors.<br />
<br />
There are two main types of induction motors. The first type is an adjustable-speed drive. These are used in the process control industry to adjust the speed of fans, compressors, pumps, blowers, etc. Also, these are used for electric traction in hybrid vehicles. The second type is a servo drive. These emulate the performance of a DC-motor drive and are used in machine tools, robotics, etc for highly precise control.<br />
<br />
====Squirrel-Cage Induction Motor====<br />
<br />
[[Image:Squirrel-cage-induction-motor.gif|thumb|right|200px|Squirrel-Cage AC Motor<ref>http://en.wikipedia.org/wiki/File:Induction-motor-3a.gif</ref>]]<br />
Squirrel cage motors are the most common forms of AC induction motors. They are commonly used in adjustable-speed applications. The cage has bars of copper or aluminum running the length of the rotor. In most working motors, the bars are skewed from following the axial direction of the motor to reduce noise. The bars are electrically shorted at each end of the rotor by end rings, and thus producing a cage like structure. <br />
<br />
The stator of an induction motor has three windings which are displaced by 120<sup>o</sup> with respect to each other. <ref>Electric Drives by Ned Mohan</ref><br />
[[Image:Stator_Windings.JPG|300px]]<br />
<br />
These stator windings are arranged around the rotor so that when energized with an alternating current they create a rotating magnetic field which sweep past the rotor. The changing magnetic field induces a current in the squirrel-cage of the rotor. The currents interact with the rotating magnetic field produced by the stator windings and produces a torque on the rotor.<ref>http://en.wikipedia.org/wiki/Induction_motor</ref>.<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Alex Roddy]]<br />
<br />
[[Tim Rasmussen]]<br />
<br />
[[Kyle Lafferty]]<br />
<br />
==Reviewers:==<br />
[[Wesley Brown]]<br />
<br />
[[Erik Biesenthal]]<br />
<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=8277Electromechanical Energy Conversion2010-01-19T00:02:38Z<p>Timothy.Rasmussen: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
* If it's been approved by the reviewers, move it to the articles section<br />
<br />
Does anyone know why my LaTEX stuff is changing sizes throughout my article? [[An Ideal Transformer Example]]<br />
<br />
*(John Hawkins) As I understand it, the text is made full size (larger) if there is ever a function call, i.e. something starting with a backslash, excluding some things like greek letters. I have just put "\ " (the function call for a space) at the beginning of an equation and had it work. If you don't want to change anything about your equation but just want it displayed full size, type "\,\!" (small forward space and small backward space) somewhere in your equation.<br />
*Thanks John!<br />
<br />
==Announcements==<br />
<br />
If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Gauss Meters]]<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[AC vs. DC]]<br />
* [[AC Motors]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Faraday's Law]]<br />
* [[Eddy Current]]<br />
* [[Example Problems of Magnetic Circuits]]<br />
* [[Magnetic Circuit]] (John Hawkins)<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
* [[An Ideal Transformer Example]]<br />
* [[Example: Ideal Transformer Exercise]] (John Hawkins)<br />
* [[Reference Terms and Units]] (Amy Crosby)<br />
* [[Transformer_example_problem|Ideal Transformer Example]]<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)<br />
* [[Magnetic Flux]] (Jason Osborne)<br />
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8250Transformer example problem2010-01-18T07:33:41Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line needs to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have to output the desired voltage. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core of the transformer<br />
<br />
[[Image:Transformer_EMEC.png]]<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \cdot \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m \ = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8249Transformer example problem2010-01-18T07:33:25Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line needs to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have to output the desired voltage. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core of the transformer<br />
<br />
[[Image:Transformer_EMEC.png]]<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \cdot \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m \ = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8248Transformer example problem2010-01-18T07:31:27Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line needs to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have to output the desired voltage. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core of the transformer<br />
<br />
[[Image:Transformer_EMEC.png]]<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \cdot \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8247Transformer example problem2010-01-18T07:26:57Z<p>Timothy.Rasmussen: /* Problem: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line needs to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have to output the desired voltage. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core of the transformer<br />
<br />
[[Image:Transformer_EMEC.png]]<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:Transformer_EMEC.png&diff=8246File:Transformer EMEC.png2010-01-18T07:22:55Z<p>Timothy.Rasmussen: Transformer Diagram for ideal transformer example problem</p>
<hr />
<div>Transformer Diagram for ideal transformer example problem</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8245Transformer example problem2010-01-18T06:06:55Z<p>Timothy.Rasmussen: /* Problem: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line needs to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have to output the desired voltage. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core of the transformer<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8244Transformer example problem2010-01-18T06:04:35Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \Rightarrow \Phi_m = 6.006 \ mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8243Transformer example problem2010-01-18T06:03:59Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math>, we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<math> \ \Phi_m = \frac{120}{4.44 \cdot 60 \ Hz \cdot 75} \rightarrow \Phi_m = 6.006 mWb </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8242Transformer example problem2010-01-18T05:52:45Z<p>Timothy.Rasmussen: /* Part C: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 \ A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8241Transformer example problem2010-01-18T05:52:26Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8240Transformer example problem2010-01-18T05:52:11Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
Where <math> \Phi_m = </math>max flux in core,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>\mathit{f} = </math> Frequency of line,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8239Transformer example problem2010-01-18T05:46:30Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_2}{4.44 \cdot \mathit{f} \cdot N_2} </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8238Transformer example problem2010-01-18T05:45:34Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_1}{4.44 \cdot \mathit{f} \cdot N_1} </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8237Transformer example problem2010-01-18T05:45:15Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_1}{4.44} \cdot \mathit{f} \cdot N_1 </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8236Transformer example problem2010-01-18T05:44:59Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<math> \ \Phi_m = \frac{V_1}{4.44 \cdot \mathit{f} \cdot N_1 </math><br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8235Transformer example problem2010-01-18T05:43:09Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \ \Phi_m </math> we can calculate the maximum flux in the core:<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8234Transformer example problem2010-01-18T05:42:10Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for <math> \Phi_m </math> we can calculate the maximum flux in the core:<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8233Transformer example problem2010-01-18T05:33:50Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8232Transformer example problem2010-01-18T05:33:29Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m \ \angle 0 45^\circ </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8231Transformer example problem2010-01-18T05:29:20Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 209.</ref><br />
Solving for<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
[[Tim Rasmussen]]<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8230Transformer example problem2010-01-18T05:28:43Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 129.</ref><br />
Solving for<br />
<br />
==References:==<br />
<references/><br />
<br />
==Authors:==<br />
==Reviewers:==<br />
==Readers:==</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8229Transformer example problem2010-01-18T05:27:57Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m </math><ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 129.</ref><br />
Solving for</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8228Transformer example problem2010-01-18T05:26:58Z<p>Timothy.Rasmussen: /* Part D: */</p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m </math><br />
Solving for</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8227Transformer example problem2010-01-18T05:26:28Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The induced emf of the secondary can be calculated by: <br />
<math> V_2 = 4.44 \ \mathit{f} \cdot N_2 \cdot \Phi_m = </math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8226Transformer example problem2010-01-18T05:22:45Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====<br />
The maximum flux in the core is given by: <br />
<math> \Phi_m = </math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8225Transformer example problem2010-01-18T05:11:33Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary. D) The maximum flux in the core<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math><br />
<br />
====Part D:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8224Transformer example problem2010-01-18T05:09:55Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns} \Rightarrow i_1 = .3 A</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8223Transformer example problem2010-01-18T05:08:40Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math><br />
<br />
Rearranging to solve for <math>i_1 </math>:<br />
<br />
<math>\ i_1 = i_2 \cdot \frac{N_2}{N_1} \Rightarrow 1.2 A \cdot \frac{75 \ turns}{300 \ turns}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8222Transformer example problem2010-01-18T05:02:00Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8221Transformer example problem2010-01-18T05:01:40Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75 \ turns}{300 \ turns}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8220Transformer example problem2010-01-18T05:00:52Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<math>\frac{i_1}{1.2 \ A} = \frac{75}{300}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8219Transformer example problem2010-01-18T05:00:01Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{75}{300}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<math>\frac{i_1}{1.2 \ A} = \frac{N_2}{N_1}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8218Transformer example problem2010-01-18T04:49:13Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math><br />
<br/><br />
Where <math> i_1 = </math>Current in primary,<br />
<math>i_2 = </math>Current in secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8217Transformer example problem2010-01-18T04:41:47Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====<br />
The ratio of primary current to secondary current is inversely proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{i_1}{i_2} = \frac{N_2}{N_1}</math></div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8216Transformer example problem2010-01-18T04:40:45Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 \ turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8214Transformer example problem2010-01-18T04:34:43Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (<math> i_2 </math> ).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8213Transformer example problem2010-01-18T04:33:57Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (i_2).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8212Transformer example problem2010-01-18T04:33:39Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (i_2).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (<math>R_L = </math> = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8211Transformer example problem2010-01-18T04:33:13Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for <math>N_2 = </math>:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (i_2).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8210Transformer example problem2010-01-18T04:31:47Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for N_2:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current in the loop (i_2).<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8209Transformer example problem2010-01-18T04:31:09Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for N_2:<br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current.<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8208Transformer example problem2010-01-18T04:30:42Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<br/><br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
To solve for the number of turns required for the secondary, the equation is rearranged solving for N_2:<br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current.<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8207Transformer example problem2010-01-18T04:28:53Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
The ratio of primary voltage to secondary voltage is directly proportional to the ratio of number of turns on the primary to number of turns on the secondary:<br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current.<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussenhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_example_problem&diff=8206Transformer example problem2010-01-18T04:27:26Z<p>Timothy.Rasmussen: </p>
<hr />
<div>===Problem:===<br />
An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply line is to output 120 V from the secondary. If a 100 &Omega; resistor is connected across the secondary, determine: A) How many turns the secondary must have. B) The current through the resistor, C)The current drawn through the primary.<br />
<br />
===Solution:===<br />
====Part A:====<br />
<br />
<math>\frac{V_1}{V_2} = \frac{N_1}{N_2}</math><br />
<br/><br />
<br/><br />
Where <math> V_1 = </math>Voltage across primary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>N_1 = </math> Number of turns in primary,<br />
<math>N_2 = </math> Number of turns in secondary<br />
<br />
<br />
<math>\frac{480 \ volts}{120 \ volts} = \frac{300 turns}{N_2}</math><br />
<br/><br />
<br/><br />
<math>N_2 = \frac{300 \cdot 120}{480} \Rightarrow N_2 = 75 \ turns </math><br />
<br />
====Part B:====<br />
The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, <math> V =i \cdot R </math>, we can solve for the current.<br />
<br/><br />
<math> i_2 = \frac{V_2}{R_L}</math><br />
<br/><br />
Where <math> i_2 = </math> Current through secondary,<br />
<math>V_2 = </math>Voltage across secondary,<br />
<math>R_L = </math> Load Resistor (R_L = 100 &Omega;)<br />
<br/><br />
<br/><br />
<math> i_2 = \frac{120 \ volts}{100 \ \Omega} \Rightarrow i_2 = 1.2 \ A </math><br />
<br />
====Part C:====</div>Timothy.Rasmussen