Question

The maximum efficiency of a 100 kVA, 1 - phase transformer is 98% and occurs at 80% of full load at 0.8 pf lagging. If leakage reactance of transformer is 5%. The voltage regulation at full load will be (Take power factor = 0.8)

• A. 2.125%
• B. 4.275%
• C. 5.673%
• D. 3.753%

Answer 1

The correct option is D 3.753%

Given maximum efficiency of transformer is 98% at 80% of full load at 0.8 p.f lagging.

$\text{At maximum efficiency,}$

$x^2{P}_{cu}=P_i$

$x=0.8$

So, $(0.8{)}^2{P}_{cu}=P_i$

$0.98=\frac{0.8\times 0.8}{0.8\times 0.8+P_i+(0.8{)}^2{P}_{cu}}$

$2P_i=0.64(\frac{1}{0.98}-1)$

$2P_i=0.01306p.u$

$P_i=(0.8{)}^2{P}_{cu}$

$=0.00653p.u$

$(0.8{)}^2{P}_{cu}=0.00653$

$P_{cu}=0.0102p.u.$

$P_{cu}=I_2^2{R}_2=0.0102p.u.$

$\%of{R}_2=1.02\%$

$\text{and leakage reactance,}$

$X_2=5\%or0.05p.u.$

$\text{Voltage regulation}=\frac{I_2{R}_{2}cos\varphi +I_2{X}_{2}sin\varphi }{V_2}$

$=\frac{0.0102\times 0.8+0.05\times 0.6}{1}=0.03816\left(or\right)3.816\%$