The correct option is (a) 14.7 A
To elaborate: Using KVL, 100 = \(R\frac{dq}{dt} + \frac{q}{C}\)
100 C = RC\(\frac{dq}{dt}\) + q
Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} \int_0^t \,dt\)
100C – q = (100C – qo)e^-t/RC
I = \(\frac{dq}{dt} = \frac{100C – q_o}{RC} e^{-1/1}\)
∴ e^-t/RC = 40e^-1 = 14.7 A.