View source for Example: Step Down Transformer

==Problem Statement==
Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.

Given: <math> R_L~=~0.2~\Omega,~~X_L~=~0.6~\Omega,~~ R_H~=~0.8~\Omega,~~X_H~=~1.6~\Omega,~~ R_{cH}~=~3.2~k\Omega,~~X_{mH}~=~1.2~k\Omega</math>

*this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross

==Solution==

From the given data of the step down transformer we can see that
<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math>

<math>I_2~=~\frac{s^*}{v_2^*}~=~200\angle -46^\circ </math>

<math> R_H~+~a^2~R_l~=~1.6</math>
<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\vecV_2\right)</math>
<math>\vec V_1~=~893\angle 10.7\circ /text{instead of} 480 V </math>
<math>P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}</math>
<math>P_{cu}=16000 w </math>
<math> n~=~\frac{Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>