$A$ resistor of $20 \, \Omega$ and an inductor of $5 \, H$ are connected in series with a $5 \, V$ battery. What is the rate of change of current at $t = 0.25 \, s$?

An inductor $(L = 100 \, mH)$,a resistor $(R = 100 \, \Omega)$ and a battery $(E = 100 \, V)$ are initially connected in series as shown in the figure. After a long time,the battery is disconnected by short-circuiting the points $A$ and $B$. The current in the circuit $1 \, ms$ after the short circuit is

Two resistors of $10 \Omega$ and $20 \Omega$ and an ideal inductor of $10 \ H$ are connected to a $2 \ V$ battery as shown. The key $K$ is closed at time $t=0$. Find the initial $(t=0)$ and final $(t \rightarrow \infty)$ currents through the battery.

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 . . . . . . $A$.

$A$ coil carrying a steady current is short-circuited. The current in it decreases $\alpha$ times in $t_0$. The time constant of the circuit is:-

For $L-R$ circuit,the time constant is equal to