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

The current in an $LR$ circuit builds up to $3/4$ of its steady state value in $4 \, s$. The time constant of this circuit is:

$A$ $20 \, H$ inductor coil is connected to a $10 \, \Omega$ resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is:

An ideal coil of $10 \ H$ is connected in series with a resistance of $5 \ \Omega$ and a battery of $5 \ V$. $2 \ s$ after the connection is made,the current flowing in the circuit in amperes is:

An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

The figure shows a circuit that contains three resistors with resistance $R = 2.0 \, \Omega$, two inductors with inductance $L = 2.0 \, mH$, and an ideal battery with $emf$ $E = 9 \, V$. The current $i$ just after the switch $S$ is closed will be .... $A$.