An inductor coil stores $64 \, J$ of magnetic field energy and dissipates energy at the rate of $640 \, W$ when a current of $8 \, A$ is passed through it. If this coil is joined across an ideal battery,find the time constant of the circuit in seconds.

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. The magnetic field $\vec{B}$ is constant and directed out of the paper. The parallel wires have no resistance,and the velocity $v$ is constant. The switch $S$ is closed at time $t = 0$.

Consider the $LR$ circuit shown in the figure. If the switch $S$ is closed at $t = 0$, then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is

$A$ coil of inductance $5\,H$ is joined to a cell of $emf$ $6\,V$ through a resistance $10\,\Omega$ at time $t = 0$. The $emf$ across the coil at time $t = \ln \sqrt{2}\,s$ is:....$V$

Regarding the given circuit,the correct statement is:

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is: