In the circuits (A) and (b) switches `S_(1)` and `S_(2)` are closed at t = 0 and are kept closed for a long time.The variation of current in the two circuits for `t ge 0` are roughly shown by figure (figures are schematic and not drawn to scale):

Options

A.
B.
C.
D.

Answer

C

Explanation

For capaciotr circuit, `i = i_(0)e^(-t)/(RC)`
For inductor circuit, `i = i_(0)(1 - e^(-(Rt)/(L)))`
Hence graph (c) correctly depicts i versus t graph.