$A$ $5\,cm$ long solenoid having $10\,\Omega$ resistance and $5\,mH$ inductance is joined to a $10\,V$ battery. At steady state,the current through the solenoid in ampere will be.....

In the circuit shown in the figure, what is the value of $I_1$ just after pressing the key $K$?

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:

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:

$A$ $20 \, H$ 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:

In an $LR$ circuit,the time constant is the time in which the current grows from zero to the value (where ${I_0}$ is the steady-state current):