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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. |
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Answer» Correct Answer - B 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))` |
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