Problem Set 1: Difference between revisions

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Latest revision as of 23:43, 8 February 2010

PROBLEM

Find the core inductance and resistance of a transformer using measurements Voc=5V, Ioc=3A, and Poc =10W.

SOLUTION

Power is stored in resistors because of their linear behavior.

$P o c = V o c^{2} / R$

Both Poc and Voc are giving so solve for R

$R = V o c^{2} / P o c = 5 V / 10 W = 2.5 O h m s$ Answer

Apparent power relation to voltage and current is

$S^{2} = | V o c * I o c^{*} |^{2}$

The magnitude for current and voltage are given in the problem statement. Subtitute in known values and solve.

$S^{2} = (5 V * 3 A)^{2} = 225 (V A)^{2}$

Apparent power S is complex. It has a real part P, real power, and an imaginary part Q, reactive power.

$S = P + i Q$

Apparent power in the imaginary and real coordinate system is the hypotenuse of real and reactive power and can be expressed using the equation of a circle.

$S^{2} = P^{2} + Q^{2}$

Solve the above equation for reactive power using the values of apparent and real power.

$Q = \sqrt{S^{2} - P^{2}} = \sqrt{22 5^{2} - 1 0^{2}} = 224.8 V A r$

Reactive power is stored in non-linear impedance represented as X. In this case X is inductance.

$Q = V o c^{2} / X$

$X = V o c^{2} / Q = 5 V^{2} / 224.8 V A r = 0.11 H$ Answer

@@ Line 1: / Line 1: @@
-'''Problem'''
+'''PROBLEM'''
 ----
-Find the core inductance and resitence of a transformer using measurmnets Voc=5V, Ioc=3A, and Poc =10W.
+Find the core inductance and resistance of a transformer using measurements Voc=5V, Ioc=3A, and Poc =10W.
 ----
-'''Solution'''
+'''SOLUTION'''
-Power is found only through resister
+Power is stored in resistors because of their linear behavior.
-<math>Poc= Voc^{2}/R</math>
+<math>Poc= Voc^{2}/R\!</math>
 Both Poc and Voc are giving so solve for R
-<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms</math>
+<math>R=Voc^{2}/Poc=5V/10W=2.5 Ohms\!</math> ''''' Answer'''''
-To calclate the core inductance we will use the formular for appearent power.
+Apparent power relation to voltage and current is
-<math>S^{2}=P^{2}+Q^{2}</math>
+<math>S^{2}=|Voc*Ioc^{*}|^{2}\!</math>
-<math>S^{2}=|Voc*Ioc^{*}|^{2}</math>
+The magnitude for current and voltage are given in the problem statement. Subtitute in known values and solve.
-Subtituting in values  aand solving gives
+<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}\!</math>
-<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>
-<math>Q=\sqrt{P^{2}-Q^{2}}</math>
+Apparent power S is complex. It has a real part P, real power, and an imaginary part Q, reactive power.
-<math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>
+<math>S=P+iQ\!</math>
+Apparent power in the imaginary and real coordinate system is the hypotenuse of real and reactive power and can be expressed using the equation of a circle.
+<math>S^{2}=P^{2}+Q^{2}\!</math>
+Solve the above equation for reactive power using the values of apparent and real power.
+<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>
+Reactive power is stored in non-linear impedance represented as X. In this case X is inductance.
+<math>Q=Voc^{2}/X\!</math>
+<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math> '''''Answer'''''

Problem Set 1: Difference between revisions

Latest revision as of 23:43, 8 February 2010

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