Q 12-06-138JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
In the circuit shown, the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $(t_0)$, the switch $S_1$ is opened and $S_2$ is closed. The behaviour of the current $I$ as a function of time $t$ is given by:
Answer: (A) see figure
With $S_1$ closed the current grows exponentially towards $\varepsilon/R$:
$$I = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)$$
At $t_0$ the cell is cut off and $S_2$ closes the R–L loop, so the current decays exponentially from its value at $t_0$, without any jump:
$$I = I(t_0)\,e^{-R(t - t_0)/L}$$
So the graph is an exponential rise up to $t_0$ followed by an exponential decay.
Solution by Sreeraj P, M.Sc Physics