Question

In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________

(a) 14.7 A

(b) 18.5 A

(c) 40 A

(d) 50 A

I have been asked this question during an interview for a job.

The doubt is from Dot Convention in Magnetically Coupled Circuits in portion Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

The correct answer is (a) 14.7 A

Explanation: Using KVL, 100 = R\(\frac{dq}{dt} + \frac{q}{C} \)

Or, 100 C = RC\(\frac{dq}{dt}\) + q

Now, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t dt\)

Or, 100C – q = (100C – qo) e^-t/RC

I = \(\frac{dq}{dt} = \frac{(100C – q_0)}{RC} e^{-1/1}\)

= 40e^-1 = 14.7 A.