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Q. The current in an $RL$ circuits drops from $1.0 \,A$ to $10\, mA$ in the first second following removal of the battery from the circuit. If $L$ is $10\, H$, find the resistance $R$ (in ohm) in the circuit.

Electromagnetic Induction

Solution:

we have, $i = i_{0}e^{-tR/L}$
or $10^{-3} = 1 \times e^{-1R/10}$
$\Rightarrow R = 46\Omega$