Question

A solenoid of inductance $L$ with resistance $r$ is connected in parallel to a resistance $R$. A battery of emf $E$ and of negligible internal resistance is connected across the parallel combination as shown in the figure. At the time $t=0$, switch $S$ is opened, calculate amount of heat generated in the solenoid.

• A. $\frac{E^{2}L}{\pi (R+r)}$
• B. $\frac{E^{2}L}{4\pi (R+r)}$
• C. $\frac{E^{2}L}{2\pi (R+r)}$
• D. $\frac{{2E}^{2}L}{7\pi (R+r)}$

Answer 1

The correct option is C $\frac{E^{2}L}{2\pi (R+r)}$

Steady state current developed in the inductor $=\frac{R}{r}=i_0$ (say)
Now this current decrease to zero exponentially through $r$ and $R$
$i=i_0{e}^{-t/{\tau }_L}$

Where ${\tau }_L=\frac{L}{R+r}$

Energy stored in inductor

$U_0=\frac{1}{2}L{i_0}^2$

$\left(\frac{1}{2}L\right){\left(\frac{E}{r}\right)}^2$

Now this energy dissipates in $r$ and $R$ in direct ratio of resistance.

$\therefore H_r=\left(\frac{r}{R+r}\right)U_0$

$=\frac{E^{2}L}{2\pi (R+r)}$