Question

A solenoid of inductance $L$ having internal resistance $r$ is connected in parallel to a resistance $R$ . A battery of emf $E$ and of negligible internal resistance is connected across this parallel combinations. At $t=0$ switch $S$ opened, then

• A. Current in the inductor just before opening of the switch is$=\frac{E(r+R)}{rR}$
• B. Total energy dissipated in the solenoid and the resistor long time after opening of the switch is $\frac{1}{2}\frac{L{E}^2}{r^2}$
• C. The amount of heat generated in the solenoid after opening of switch is $\frac{1}{2}\frac{L{E}^2}{(R+r{)}^2}$
• D. The amount of heat generated in the solenoid after opening of the switch is $\frac{1}{2}\frac{L{E}^2}{(R+r)r}$

Answer 1

Answer: (B), (D)

The amount of heat generated in the solenoid after opening of the switch is $\frac{1}{2}\frac{L{E}^2}{(R+r)r}$

Steady state current developed in the inductor
$i_0=\frac{E}{r}$
Now this current decreases to zero exponentially through $r$ and $R$
$\therefore I=i_0{e}^{\frac{t}{\tau }}$
Where $\tau =\frac{L}{R+r}$
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 $r\text{(selenoid)}$ and $R$ in direct ratio if resistance.
Energy dissipated in inductor is
$\therefore H_r=\left(\frac{r}{R+r}\right)U_0=\frac{E^{2}L}{2r(R+r)}$