Question

In the circuit given below, the capacitor is initially having a charge of 10 C.  1 second after the switch is closed, the current in the circuit is ________

(a) 14.7 A

(b) 18.5 A

(c) 40.0 A

(d) 50.0 A

I got this question by my school teacher while I was bunking the class.

My question comes from Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor topic in division Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Right answer is (a) 14.7 A

The best explanation: 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.