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Two single-phase transformer rated 1000 kVA and 500 kVA have per unit leakage impedance of (0.02 + j0.06) and (0.025 + j0.08) respectively. Then, the largest kVA load delivered by the parallel combination of these two transformers without overloading any one is
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- 1825.3 kVA
- 1377.44 kVA
- 1500 kVA
- 3202.74 kVA
Correct Option: B
Choosing base of 1000 kVA,
Ze1(1000 kVA) = (0.02 + j 0.06) pu
| Ze2(500 kVA) = | . (0.025 + j 0.08) p.u. = 0.05 + 0.16 j | |
| 500 |
| Now, S1 = | . S | |
| Ze1 + Ze2 |
| ⇒ S = S1 . | |
| Ze2 |
| = 1000 . | |
| 0.05 + j0.16 |
= 1377.44 kVA,
| and S2 = | . S1 | |
| Ze1 + Ze2 |
| ∴ S = 500 | = 1825.3 kVA | |
| 0.02 + j0.06 |
As, Ze1 < Ze2, transformer 1 reaches its rated kVA first,
∴ largest kVA load = 1377.44 k VA
