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):

$A$ coil of inductance $8.4 \, mH$ and resistance $6 \, \Omega$ is connected to a $12 \, V$ battery. The current in the coil is $1.0 \, A$ at approximately the time

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

An inductor with an iron core is connected in series with a resistor $R$ across an ideal $DC$ source. The circuit is in a steady state. If the iron core is removed, what happens to the current flowing through the inductor immediately after the removal of the core?

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:

$A$ $50 \, V$ potential difference is suddenly applied to a coil with $L = 5 \times 10^{-3} \, H$ and $R = 180 \, \Omega$. The rate of increase of current after $0.001 \, s$ is ....... $A/s$.