In the circuit shown ${\mathrm{S}}_1 \& {\mathrm{S}}_2$ are switches. ${\mathrm{S}}_2$ remains closed for a long time and ${\mathrm{S}}_1$open. Now ${\mathrm{S}}_1$ is also closed. Just after ${\mathrm{S}}_1$is closed, The potential difference$\left(\mathrm{V}\right)$across $\mathrm{R}$  and $\frac{\mathrm{di}}{\mathrm{dt}}$(with sign) in $\mathrm{L}$.

 

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)