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 ..... $H$.

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 of $2 \, H$ and a resistance of $10 \, \Omega$ are connected in series with a battery of $5 \, V$. The initial rate of change of current is ...... $A/s$.

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

In the circuit shown,$X$ is joined to $Y$ for a long time,and then $X$ is joined to $Z$. The total heat produced in $R_2$ is:

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