The time constant of an inductance coil is $3 \text{ ms}$. When a $90 \Omega$ resistance is joined in series,the time constant becomes $0.5 \text{ ms}$. The inductance and the resistance of the coil are:

In the circuit shown here, the point '$C$' is kept connected to point '$A$' till the current flowing through the circuit becomes constant. Afterward, suddenly, point '$C$' is disconnected from point '$A$' and connected to point '$B$' at time $t = 0$. The ratio of the voltage across the resistance $(V_R)$ and the inductor $(V_L)$ at $t = L/R$ will be equal to:

In the circuit shown,the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $t_0$,the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time $t$ is given by:

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 ....... $H$.

Two $L-R$ circuits are shown in the figure. Based on the given current-time graph,select the correct relation between the inductances $L_1$ and $L_2$.

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):