The time constant of an $LR$ circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series with the existing circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ..... $H$.

$A$ coil of inductance $0.20 \, H$ is connected in series with a switch and a cell of $emf$ $1.6 \, V$. The total resistance of the circuit is $4.0 \, \Omega$. What is the initial rate of growth of the current when the switch is closed? (in $A \, s^{-1}$)

Consider the $LR$ circuit shown in the figure. If the switch $S$ is closed at $t = 0$, then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is

In an $L-R$ circuit,the current reaches $\frac{3}{4}$ of its maximum value in $4 \ s$. What is the time constant of the $L-R$ circuit?

$A$ coil of inductance $40 \, H$ is connected in series with a resistance of $8 \, \Omega$ and the combination is joined to the terminals of a $2 \, V$ battery. The time constant of the circuit is ...... $s$.

In the circuit shown,an inductor $L$ and a capacitor $C$ are connected as shown. $A_1$ and $A_2$ are ammeters. When the key $K$ is pressed to complete the circuit,what will be the readings of $A_1$ and $A_2$ just after closing the key $K$?