The time taken for the magnetic energy to reach $25\%$ of its maximum value,when a solenoid of resistance $R$ and inductance $L$ is connected to a battery,is:

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

$A$ constant voltage of $25 \, V$ is applied to a series $L-R$ circuit at $t=0$ by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

$A$ $10 \text{ cm}$ long perfectly conducting wire $PQ$ is moving with a velocity $1 \text{ cm/s}$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $L = 1 \text{ mH}$ and a resistance $R = 1 \ \Omega$ as shown in the figure. The horizontal rails,$L$,and $R$ lie in the same plane with a uniform magnetic field $B = 1 \text{ T}$ perpendicular to the plane. If the key $S$ is closed at a certain instant,the current in the circuit after $1 \text{ millisecond}$ is $x \times 10^{-3} \text{ A}$,where the value of $x$ is. . . . . . [Assume the velocity of wire $PQ$ remains constant $(1 \text{ cm/s})$ after key $S$ is closed. Given: $e^{-1} = 0.37$,where $e$ is the base of the natural logarithm]

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

An inductor with an iron core is connected in series with a resistor $R$ across an ideal $DC$ source. The circuit is in a steady state. If the iron core is removed, what happens to the current flowing through the inductor immediately after the removal of the core?