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

$A$ coil of inductance $1\,H$ and resistance $100\,\Omega$ is connected to a battery of $6\,V$. Determine approximately:
$(a)$ The time elapsed before the current acquires half of its steady-state value.
$(b)$ The energy stored in the magnetic field associated with the coil at an instant $15\,ms$ after the circuit is switched on. (Given $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

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 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 . . . . . . $A$.

In the circuit shown,an inductor $L$ and a capacitor $C$ are connected as shown. $A_1$ and $A_2$ are ammeters. When the key $K$ is pressed to complete the circuit,what will be the readings of $A_1$ and $A_2$ just after closing the key $K$?

$A$ blackbox $(BB)$ which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown in figure $(A)$. After the switch $S$ is closed at time $t=0$, the current $I$ as a function of time $t$ is shown in figure $(B)$. From this, we can infer that the blackbox contains: