JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Current Electricity

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A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s in ______ J.

Answer: 3840

Solution

Power dissipated in a resistor is

P = I^{2} R

. Initially,

P_{1} = I_{1}^{2} R = 192

W (since energy in 1 s is 192 J). Thus,

R = \frac{192}{4 ^{2}} = \frac{192}{16} = 12 Ω

.

When current is doubled,

I_{2} = 8

A. New power

P_{2} = I_{2}^{2} R = 8^{2} \times 12 = 64 \times 12 = 768

W.

Energy dissipated in 5 s:

E = P_{2} \times t = 768 \times 5 = 3840

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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