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If the potential difference between the end of a wire of fixed resistance is doubled. How much does the electric power increase?

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4 times

2 times

Remains same

None

answer is A.

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Detailed Solution

The electric power increases by four times.
We know that,

P = \frac{V^{2}}{R}

where

V

is the potential difference,

R

is resistance, and

P

is power.

P = \frac{V^{2}}{R}

Given that:
Resistance

= R

(which is constant in this case)
Potential difference

V^{'} = 2 V

(as the potential difference is doubled)
Therefore,

P^{'} = \frac{{(2 V)}^{2}}{R}

P^{'} = \frac{{4 V}^{2}}{R}

P^{'} = 4 P

(as

P = \frac{V^{2}}{R}

)

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