Show that if two inductors with equal inductance $L$ are connected in parallel then the equivalent inductance of the combination is $\frac{L}{2}.$ The inductors are separated by a large distance.

Two inductances $L_{1 }\& L_2$ are connected in series & are separated by a large distance.

$\left(\mathrm{a}\right)$ Show that their equivalent inductance is$L_1+L_2.$

$\left(\mathrm{b}\right)$ Why must their separation be large?

The circuit shown in figure is in the steady state with switch ${\mathrm{S}}_1$ closed and ${\mathrm{S}}_2$ is opened. At $\mathrm{t} = 0, {\mathrm{S}}_1$ is opened and switch ${\mathrm{S}}_2$ is closed.

Find the first instant $\mathrm{t}$, when energy in inductor becomes one third of that in capacitor.

The radius of the circular conducting loop shown in figure is $\mathrm{R}$. Magnetic field is decreasing at a constant rate $\mathrm{\alpha }$. Resistance per unit length of the loop is $\mathrm{\rho }$. Then current in wire $\mathrm{AB}$ is ($\mathrm{AB}$ is one of the diameters)

Consider the conducting square loop shown in the figure. If the switch is closed and after some time it is opened again, the closed loop will show