Given $L_1 = 1\, mH, R_1 = 1\,\Omega, L_2 = 2\,mH, R_2 = 2\,\Omega$. Neglecting mutual inductance,find the time constant (in $ms$) for the circuit shown.

$A$ 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 $t=0$?

Two resistors of $10\, \Omega$ and $20\, \Omega$ and an ideal inductor of $10\, H$ are connected to a $2\, V$ battery as shown. The key $K$ is inserted at time $t = 0$. The initial $(t = 0)$ and final $(t \rightarrow \infty)$ currents through the battery are

An inductance $L$ and a resistance $R$ are first connected to a battery. After some time,the battery is disconnected,but $L$ and $R$ remain connected in a closed circuit. Then the current reduces to $37\%$ of its initial value in

$A$ resistor of $20 \, \Omega$ and an inductor of $5 \, H$ are connected in series with a $5 \, V$ battery. What is the rate of change of current at $t = 0.25 \, s$?

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 ..... $H$.