An $L-R$ circuit has a cell of $e.m.f.$ $E$,which is switched on at time $t = 0$. The current in the circuit after a long time will be

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 closed at time $t=0$. Find the initial $(t=0)$ and final $(t \rightarrow \infty)$ currents through the battery.

In the given circuit,the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of $8 \text{ A/s}$. At an instant when $R = 12 \Omega$,the value of the current in the circuit will be . . . . . . $A$.

An inductor $(L)$ - resistance $(R)$ battery circuit is switched on at $t = 0$. If the $e.m.f.$ of the battery is $E$,what is the charge passing through the battery in time constant $T$?

An inductor coil stores $U$ energy when $I$ current is passed through it and dissipates energy at the rate of $P$. The time constant of the circuit,when this coil is connected across a battery of zero internal resistance,is

In series with a $20 \ \Omega$ resistor,a $5 \ H$ inductor is placed. To this combination,an $e.m.f.$ of $5 \ V$ is applied. What will be the rate of increase of current at $t = 0.25 \ s$?