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 time `t = 0`, switch `S` is opened, calculate
(a) current through the solenoid after the switch is opened.
(b) amount of heat generated in the solenoid.

Options

A. Current in the inductor just after removing the battery is `(E(r + R))/(rR)`
B. Total enery dissipated in the solenoid and the reistor long time after removing the battery is `(1)/(2) L(E^(2)(R + r)^(2))/(r^(2)R^(2))`
C. The amount of heat generated in the solenoid due to removing the battery is `(E^(2)L)/(2r(r + R))`
D. The amount of heat generated in the solenoid due to removing the battery is `(E^(2)L)/(2R(r + R))`

Answer

C

Explanation

Applying `KVL` in outer loop, we get `I_(0)r - E = 0`
`rArr I_(0) = (E)/(r )`
`:.` Initial energy in solenoid `= U_(0) = (1)/(2) LI^(2)underset0()= (E^(2)L)/(2r^(2))*` This
energy will be dissipared in the from of heat in `r` and `R` after opening of the switch. Since the same current flows through these resistances, therefore heat generated in each resistor is directly proprtional to its resistance.
`:.` Heat generated in solenoid
`= (r )/(r + R) U_(0)`
`= [(r)/(r + R)](E^(2) L)/(2r^(2)) = (E^(2)L)/(2r (r + R))`