Problem Set 1: Difference between revisions

Revision as of 09:57, 5 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 15: / Line 15: @@
 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 appearent power.
+To calclate the core inductance we will use the formular for apparent power.
 <math>S^{2}=P^{2}+Q^{2}</math>
@@ Line 23: / Line 23: @@
 <math>S^{2}=|Voc*Ioc^{*}|^{2}</math>
-Subtituting in values  aand solving gives
+Subtituting in values and solving gives
 <math>S^{2}=(5V*3A)^{2}=225(VA)^{2}</math>
-<math>Q=\sqrt{P^{2}-Q^{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>
+Finally
+<math>Q=Voc^{2}/X</math>
+Solve for the inductance
+<math>Q=Voc^{2}/X</math>
-<math>{Z_{th}}=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>
+<math>X=Voc^{2}/Q=5V^{2}/224.8VAr=0.11H </math>