Question

A solenoid of inductance $L$ and 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 combination. At $t=0$, battery is removed then which of the following is possible :

• A. current in the inductor just after moving the battery is $\frac{E(r+R)}{rR}$
• B. total energy dissipated in the solenoid and the resistor long time after removing the battery is $\frac{1}{2}L\frac{E^2(R+r{)}^2}{r^2{R}^2}$
• C. the amount of heat generated in the solenoid due to removing the battery is $\frac{E^{2}L}{2r(r+R)}$
• D. the amount of heat generated in the solenoid due to removing the battery is $\frac{E^{2}L}{2R(r+R)}$

Answer 1

Answer: (C)

the amount of heat generated in the solenoid due to removing the battery is $\frac{E^{2}L}{2r(r+R)}$

Current in cuircuit $=\frac{E}{Req}$
$Req=\frac{eR}{r+R}$
$I=\frac{E}{eR/(r+R)}=\frac{E(r+R)}{rR}$
Magnetic field energy is solenoid $=\frac{1}{2}L{I}^2$
Converted heat in resistance $=\frac{1}{2}{\left[\frac{E(r+R)}{rR}\right]}^2$
$=\frac{1}{2}\frac{L{E}^2(r+R{)}^2}{r^2{R}^2}$
At steady state
$I=\frac{E}{R||r}=\frac{E(R+r)}{Rr}$
$I_1=\frac{E}{r}$ (equation $1$)
After removal of battery current in conductor is
$I=I_1{e}^{-t/T}$ (equation (2))
$\tau =\frac{L}{R+r}$
so equation $2$ will be
$I=I_1{e}^{\frac{-(R+r)t}{L}}$
Heat generated in solenoid is
$dH=I^{2}dt{\int }_0^{H}dH={\int }_0^{\infty }{I}_1^2\left(e^{{\left(\frac{-(R+r)t}{T}\right)}^2}\right)d\left(t\right)r$
$H=I_1^{2}r{\left[\frac{e^{\frac{-2(R+r)t}{L}}}{-2\frac{(R+r)}{L}}\right]}_0^{\infty }=\frac{I_1^{2}r}{-2\frac{(R+r)}{L}}[0-1]=\frac{I_1^{2}rL}{-2(R+r)}[\because L_1^2=\frac{E^2}{r^2}]$
$H=\frac{E^{2}L}{2r(R+r)}$