Transformer example problem

Problem:

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 Ω 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

Solution:

Part A:

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:

${\frac {V_{1}}{V_{2}}}={\frac {N_{1}}{N_{2}}}$

Where $V_{1}=$Voltage across primary, $V_{2}=$Voltage across secondary, $N_{1}=$ Number of turns in primary, $N_{2}=$ Number of turns in secondary

${\frac {480\ volts}{120\ volts}}={\frac {300\ turns}{N_{2}}}$

To solve for the number of turns required for the secondary, the equation is rearranged solving for $N_{2}=$:

$N_{2}={\frac {300\cdot 120}{480}}\Rightarrow N_{2}=75\ turns$

Part B:

The voltage across the secondary is given in the problem statement as 120 volts. Using ohms law, $V=i\cdot R$, we can solve for the current in the loop ($i_{2}$ ).
$i_{2}={\frac {V_{2}}{R_{L}}}$
Where $i_{2}=$ Current through secondary, $V_{2}=$Voltage across secondary, $R_{L}=$ Load Resistor ($R_{L}=$ 100 Ω)

$i_{2}={\frac {120\ volts}{100\ \Omega }}\Rightarrow i_{2}=1.2\ A$

Part C:

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:

${\frac {i_{1}}{i_{2}}}={\frac {N_{2}}{N_{1}}}$
Where $i_{1}=$Current in primary, $i_{2}=$Current in secondary, $N_{1}=$ Number of turns in primary, $N_{2}=$ Number of turns in secondary

${\frac {i_{1}}{1.2\ A}}={\frac {75\ turns}{300\ turns}}$

Rearranging to solve for $i_{1}$:

$\ i_{1}=i_{2}\cdot {\frac {N_{2}}{N_{1}}}\Rightarrow 1.2A\cdot {\frac {75\ turns}{300\ turns}}\Rightarrow i_{1}=.3\ A$

Part D:

The induced emf of the secondary can be calculated by: $V_{2}=4.44\ \cdot {\mathit {f}}\cdot N_{2}\cdot \Phi _{m}\ \angle 0^{\circ }$<ref>Guru and Huseyin, Electric Machinery and Transformers, 3rd ed. (New York: Oxford University Press, 2001), 209.</ref> Solving for $\Phi _{m}$, we can calculate the maximum flux in the core: $\ \Phi _{m}={\frac {V_{2}}{4.44\cdot {\mathit {f}}\cdot N_{2}}}$

Where $\Phi _{m}\ =$max flux in core, $V_{2}=$Voltage across secondary, ${\mathit {f}}=$ Frequency of line, $N_{2}=$ Number of turns in secondary

$\ \Phi _{m}={\frac {120}{4.44\cdot 60\ Hz\cdot 75}}\Rightarrow \Phi _{m}=6.006\ mWb$

There is a problem here, because in an ideal transformer $N_{1}I_{1}=N_{2}I_{2}$ and the flux from $I_{2}$ opposes the flux from $I_{1}$. Magnetic circuits tells us that the flux, Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_1 = { N_1 I_1 }\over \script R } and the flux, Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_2 = { -N_2 I_2 }\over \script R} so that the total flux in any ideal transformer is zero.

References:

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Authors:

Tim Rasmussen

Reviewers:

Wesley Brown

Alex Roddy