A resistor dissipates 192   J of energy in 1   s when a current of 4   A is passed through it…

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 is _____________ J.

Solution

P=i2Rt
192=4×4×R×1
R=12Ω
P'=i'2Rt'
=8×8×12×5=3840 J

Asked in: JEE Main 2021 (31 Aug Shift 2)

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