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.

Answer

B

Explanation

Steady state current developed in the inductor
`=E/r =i_0(say)`
a. Now this current decreases to zero exponentialy through `r` and `R`.
`:. i=i_0r^(-t//tau_L)`
where `tau_L=L/(R+r)`
Energy stored in inductor,
`U_0=1/2Li_0^2=(1/2L)(E/r)^2`
Now, this energy dissipates in `r` and `R` in direct ratio of resistance.
`:. H_r=(r/(R+r))U_0=(E^2l)/(2r(R+r))`