JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Current Electricity

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When a potential difference V is applied across a wire of resistance R , it dissipates energy at a rate W . If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

(A) $1/4 W$
(B) $1/2 W$
(C)2 W
(D)4 W✓ Correct

Answer: (D)

Solution

Initially, the power dissipated in the wire of resistance

R

at potential difference

V

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

Cutting the wire into two equal halves gives each half a resistance

\frac{R}{2}

. Connecting them in parallel across the same supply gives an equivalent resistance

R_{e q} = \frac{1}{2} \times \frac{R}{2} = \frac{R}{4}

The new power dissipated is

W^{'} = \frac{V ^{2}}{R /4} = \frac{4 V ^{2}}{R} = 4 W

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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