A cable of resistance $10 Ω$ carries electric power from a generator producing $250 k W$ at $10, 000 V$ . The current in the cable is

1000 A

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250 A

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100 A

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25 A

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Solution

The correct option is D $25 A$
Given $P = 250 k W = 250 \times 1000 W$ and $V = 10000 V$
As $P = V I$
so current in the cable $I = \frac{P}{V} = \frac{250 \times 1000}{10000} = 25 A$

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A cable of resistance 10Ω carries electric power from a generator producing 250kW at 10,000V. The current in the cable is

The correct option is D 25AGiven P=250kW=250×1000W and V=10000VAs P=VIso current in the cable I=PV=250×100010000=25A

A cable of resistance $10 Ω$ carries electric power from a generator producing $250 k W$ at $10, 000 V$ . The current in the cable is

The correct option is D $25 A$
Given $P = 250 k W = 250 \times 1000 W$ and $V = 10000 V$
As $P = V I$
so current in the cable $I = \frac{P}{V} = \frac{250 \times 1000}{10000} = 25 A$