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?

The current growth in two $L-R$ circuits $(b)$ and $(c)$ is shown in figure $(a)$. Let $L_1, L_2, R_1$ and $R_2$ be the corresponding values in the two circuits. Then:

An inductor of inductance $10 \text{ mH}$ having resistance of $100 \ \Omega$ is connected to a battery of $E$.$M$.$F$. $1.0 \text{ V}$ through a switch as shown in the figure below. After the switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is $2 \text{ mA}$ and $4 \text{ mA}$ is . . . . . . .

$A$ solenoid has an inductance of $60 \, H$ and a resistance of $30 \, \Omega$. If it is connected to a $100 \, V$ battery,how long will it take for the current to reach $\frac{e-1}{e} \approx 63.2\%$ of its final value?

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

In a decaying $L-R$ circuit,the time after which energy stored in the inductor reduces to one-fourth of its initial value is