A resistor dissipates 192 J of energy in 1 s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s in______J.
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Solution
$\begin{aligned} & E=i^2 R t \\ & 192=16(R)(1) \\ & R=12 \Omega \\ & E^1=(8)^2(12)(5) \\ & =3840 \mathrm{~J}\end{aligned}$
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