Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

The current growth in two $L-R$ circuits $(b)$ and $(c)$ is shown in figure $(a)$. Let $L_1, L_2, R_1$ and $R_2$ be the corresponding values in the two circuits. Then:

The figure shows an $L-R$ circuit. When the switch $S$ is closed,the currents through resistors $R_1, R_2$ and $R_3$ are $I_1, I_2$ and $I_3$ respectively. The values of $I_1, I_2$ and $I_3$ at $t=0 \, s$ are:

An inductor of inductance $10 \text{ mH}$ having resistance of $100 \ \Omega$ is connected to a battery of $E$.$M$.$F$. $1.0 \text{ V}$ through a switch as shown in the figure below. After the switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is $2 \text{ mA}$ and $4 \text{ mA}$ 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:

$A$ 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