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

In the circuit shown,the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $t_0$,the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time $t$ is given by:

When a battery is connected across a series combination of self inductance $L$ and resistance $R$,the variation in the current $i$ with time $t$ is best represented by

An inductor of $10\, \text{mH}$ is connected to a $20\, \text{V}$ battery through a resistor of $10\, \text{k}\Omega$ and a switch. After a long time,when maximum current is set up in the circuit,the current is switched off. The current in the circuit after $1\, \mu\text{s}$ is $\frac{x}{100}\, \text{mA}$. Then $x$ is equal to ...... . (Take $e^{-1} = 0.37$)

In an $L-R$ circuit,for the case of increasing current,the magnitude of current can be calculated by using the formula:

An inductor of inductance $L = 400 \ mH$ and resistors of resistance $R_1 = 2 \ \Omega$ and $R_2 = 2 \ \Omega$ are connected to a battery of emf $E = 12 \ V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$. The potential drop across $L$ as a function of time is: