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A single phase transmisson line of impedance j 0.8 ohm supplies a resistive load of 500 A at 300 V. The sending end power factor is
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- unity
- 0.8 lagging
- 0.8 leading
- 0.6 lagging
Correct Option: D
VS = Vr + IZs
= 300 + j 0.8 × [500 × 0.8 – 500 × 0.6]
= 300 + j 0.8 [400 – j 300]
= 300 + j 320 + 240 = 540 + j 320
| = | = 0.6 lag | |
| 500 |
