$A$ coil of resistance $50 \Omega$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after the time $t=0.1 s$ of the battery being connected, then the inductance of the coil is:

Two figures are shown as Fig. $A$ and Fig. $B$. The time constant of Fig. $A$ is $\tau_{A}$ and the time constant of Fig. $B$ is $\tau_{B}$. Then:

$A$ bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\, V$ battery through a switch. On pressing the switch,the

$A$ coil of inductance $8.4 \, mH$ and resistance $6 \, \Omega$ is connected to a $12 \, V$ battery. The current in the coil is $1.0 \, A$ at approximately the time

For the given circuit,the current $i$ through the battery when the key is closed and the steady state has been reached is .....$A$.

$A$ $10 \text{ cm}$ 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]