Question

For the circuit shown in the figure, initially the switch is closed for a long time so that steady state has been reached. Then at $t=0$, the switch is opened, due to which current in the circuit decays to zero. The heat generated in the inductor is [$L$ = self inductance of inductor, $r$ = resistance of inductor] :

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

Answer 1

The correct option is C

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

Let heat generated in inductor be $H_1$ and in $R$ be $H_2$ then :
Total Heat generated

$H_1+H_2=\frac{1}{2}L\times {\left(\frac{E}{r}\right)}^2=\frac{E^{2}L}{2r^2}..........\left(1\right)$
Ratio of Heat generated

$\frac{H_1}{H_2}=\frac{r}{R}...............\left(2\right)$

from (1) & (2)

$H_1+\frac{H_{1}R}{r}=\frac{E^{2}L}{2r^2}$

$H_1=\frac{E^{2}L}{2r(r+R)}$