In the following figure,what is the final value of current in the $10 \, \Omega$ resistor when the plug of key $K$ is inserted?

$A$ 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:

The figure shows three circuits with identical batteries,inductors,and resistors. Rank the circuits according to the currents through the battery just after the switch is closed,greatest first.

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:

An inductor of inductance $L = 400 \ mH$ and resistors of resistance $R_1 = 2 \ \Omega$ and $R_2 = 2 \ \Omega$ are connected to a battery of emf $E = 12 \ V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$. The potential drop across $L$ as a function of time is:

In series with a $20 \ \Omega$ resistor,a $5 \ H$ inductor is placed. To this combination,an $e.m.f.$ of $5 \ V$ is applied. What will be the rate of increase of current at $t = 0.25 \ s$?