In the given circuit,the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of $8 \text{ A/s}$. At an instant when $R = 12 \Omega$,the value of the current in the circuit will be . . . . . . $A$.

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 $t = 0.25 \ s$?

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:

The current in the $2\, \Omega$ resistance shown in the figure is:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

In the circuit shown here, the point '$C$' is kept connected to point '$A$' till the current flowing through the circuit becomes constant. Afterward, suddenly, point '$C$' is disconnected from point '$A$' and connected to point '$B$' at time $t = 0$. The ratio of the voltage across the resistance $(V_R)$ and the inductor $(V_L)$ at $t = L/R$ will be equal to:

The total heat produced in resistor $R$ in an $RL$ circuit when the current in the inductor decreases from $I_0$ to $0$ is