Problem Set 1: Difference between revisions

Revision as of 22:50, 8 February 2010

Problem

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

Solution \! Power is found only through resister

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

Both Poc and Voc are giving so solve for R

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

To calclate the core inductance we will use the formular for apparent power.

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

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

Subtituting in values and solving gives

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

Use the apparent power and the real power to find the reactive power

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

Finally

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

Solve for the inductance

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

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

@@ Line 8: / Line 8: @@
 '''Solution'''
+\!
 Power is found only through resister
-<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>
 To calclate the core inductance we will use the formular for apparent power.
-<math>S^{2}=P^{2}+Q^{2}</math>
+<math>S^{2}=P^{2}+Q^{2}\!</math>
-<math>S^{2}=|Voc*Ioc^{*}|^{2}</math>
+<math>S^{2}=|Voc*Ioc^{*}|^{2}\!</math>
 Subtituting in values and solving gives
-<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>
+<math>S^{2}=(5V*3A)^{2}=225(VA)^{2}\!</math>
 Use the apparent power and the real power to find the reactive power
-<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr</math>
+<math>Q=\sqrt{S^{2}-P^{2}}= \sqrt{225^{2}-10^{2}}= 224.8VAr\!</math>
 Finally
-<math>Q=Voc^{2}/X</math>
+<math>Q=Voc^{2}/X\!</math>
 Solve for the inductance
-<math>Q=Voc^{2}/X</math>
+<math>Q=Voc^{2}/X\!</math>
-<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H </math>
+<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H \!</math>