The circuit shown consists of a switch $S$,a battery $B$ of emf $E$,a resistance $R$,and an inductor $L$. What is the current in the circuit at the instant the switch $S$ is closed?

$A$ bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\, V$ battery through a switch. On pressing the switch,the

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 $L-R$ 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

Switch $S$ is closed at $t = 0$ in the circuit shown. The change in magnetic flux in the inductor $(L = 500 \, mH)$ from $t = 0$ to an instant when it reaches steady state is ...... $Wb$.

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