A cable of resistance $10 Ω$ carries electric power from a generator producing 250 kW at 10000 volt. The current in the cable is

25 A

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250 A

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100 A

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1000 A

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Solution

The correct option is A 25 A
Power, P = 250 kW
$= 250 \times 10^{3} W$
Voltage, V = 10000 volt
Now, P = VI
Thus, the current in the cable can be calculated as:
$I = \frac{P}{V} = \frac{250 \times 10^{3}}{1000} A$
= 25 A
Hence, the correct answer is option (a).

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A cable of resistance 10 Ω carries electric power from a generator producing 250 kW at 10000 volt. The current in the cable is

The correct option is A 25 APower, P = 250 kW =250×103 W Voltage, V = 10000 volt Now, P = VI Thus, the current in the cable can be calculated as: I=PV=250×1031000A = 25 A Hence, the correct answer is option (a).

A cable of resistance $10 Ω$ carries electric power from a generator producing 250 kW at 10000 volt. The current in the cable is

The correct option is A 25 A
Power, P = 250 kW
$= 250 \times 10^{3} W$
Voltage, V = 10000 volt
Now, P = VI
Thus, the current in the cable can be calculated as:
$I = \frac{P}{V} = \frac{250 \times 10^{3}}{1000} A$
= 25 A
Hence, the correct answer is option (a).