Current Electricity

JEE Main level, numerical answer, medium

245 of 380

Question 245 of 380

A resistor dissipates 192J192J of energy in 1s1s when a current of 4A4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s5s is _____________ J.J.

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Answer: 3840

P=i2RtP={i}^{2}Rt
192=4×4×R×1192=4\times 4\times R\times 1
R=12ΩR=12\Omega
P′=(i′)2RtP'={\left(i'\right)}^{2}Rt'
=8×8×12×5=3840J=8\times 8\times 12\times 5=3840J

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