Question

Two single-phase transformers, rated 1000 kVA and 500 kVA respectively, are connected in parallel on both HV and LV sides. They have equal voltage ratings of 1 kV/400 V and their per unit impedance are (0.02 +j0.07) and (0.025 +j0.0875) respectively. The largest value of the u.p.f load that can be delivered by the parallel combination at the rated voltage is

• A. 1000 kVA
• B. 1400 kVA
• C. 1751.4 kVA
• D. 500 kVA

Answer 1

The correct option is B 1400 kVA


$\text{Choose a kVA base of 1000}$
$Z_1=0.02+j0.07=0.0728\angle {74}^{\circ }$

$Z_2=(0.025+j0.0875)\times 2$

$=0.05+j0.175=0.182\angle {74}^{\circ }$

$Z_1+Z_2=0.07+j0.245=0.255\angle {74}^{\circ }$

$\text{Now,}$ $S_1=\frac{Z_2}{|Z_1+Z_2|}{S}_L...\left(i\right)$
$S_2=\frac{Z_1}{|Z_1+Z_2|}{S}_L...\left(ii\right)$

$\text{From(i),}$ $S_L=1000\times \frac{0.255}{0.182}=1400kVA$

$\text{From (ii),}$ $S_L=500\times \frac{0.255}{0.0728}$
$=1751.37kVA\simeq 1751.4kVA$

As total load is increased, the 1000 kVA transformer will be the first to reach its full load
$\therefore S_L\left(max\right)=1400kVA$