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: $R_{L}~=~0.2~\Omega ,~~X_{L}~=~0.6~\Omega ,~~R_{H}~=~0.8~\Omega ,~~X_{H}~=~1.6~\Omega ,~~R_{cH}~=~3.2~k\Omega ,~~X_{mH}~=~1.2~k\Omega$

• 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 $~~~a=2~and~s=48000~\angle \cos ^{-1}(0.7)$

$I_{2}~=~{\frac {s^{*}}{v_{2}^{*}}}~=~200\angle -46^{\circ }$

$R_{H}~+~a^{2}~R_{l}~=~1.6$ 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 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)} ${\vec {V}}_{1}~=~893\angle 10.7\circ /text{insteadof}480V$ 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 P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}} $P_{cu}=16000w$ $n~=~{\frac {Re(s)}{Re(s)~+~P_{cu}~+~P_{c}}}$