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):

$A$ constant voltage of $25 \, V$ is applied to a series $L-R$ circuit at $t=0$ by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

An inductor coil carries a steady-state current of $2.0 \ A$ when connected across an ideal battery of $emf$ $4.0 \ V$. If its inductance is $1.0 \ H$,find the time constant of the circuit in $sec$.

In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

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 ..... $H$.

In the circuit shown,the cell is ideal. The coil has an inductance of $4H$ and zero resistance. $F$ is a fuse of zero resistance and will blow when the current through it reaches $5A$. The switch is closed at $t = 0$. The fuse will blow after how much time?