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

When the switch $S$ is closed at time $t = 0$,what will be the nature of the graphs for the induced $emf$ $e$ across the inductor $L$ and the current $i$ in the circuit?

$A$ coil of inductance $1\, H$ and resistance $10\, \Omega$ is connected to a battery of $emf$ $50\, V$ and negligible internal resistance at time $t = 0.$ The ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at $t = 0.1\, s$ is:

$A$ coil of inductance $0.20 \, H$ is connected in series with a switch and a cell of $emf$ $1.6 \, V$. The total resistance of the circuit is $4.0 \, \Omega$. What is the initial rate of growth of the current when the switch is closed? (in $A \, s^{-1}$)

$A$ $10 \text{ cm}$ long perfectly conducting wire $PQ$ is moving with a velocity $1 \text{ cm/s}$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $L = 1 \text{ mH}$ and a resistance $R = 1 \ \Omega$ as shown in the figure. The horizontal rails,$L$,and $R$ lie in the same plane with a uniform magnetic field $B = 1 \text{ T}$ perpendicular to the plane. If the key $S$ is closed at a certain instant,the current in the circuit after $1 \text{ millisecond}$ is $x \times 10^{-3} \text{ A}$,where the value of $x$ is. . . . . . [Assume the velocity of wire $PQ$ remains constant $(1 \text{ cm/s})$ after key $S$ is closed. Given: $e^{-1} = 0.37$,where $e$ is the base of the natural logarithm]

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