Question

A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is ________

(a) 0.75 mF

(b) 1.33 mF

(c) 0.6 mF

(d) 1.67 mF

I had been asked this question in an interview.

This intriguing question originated from Basic Concepts in portion EDC Overview of Electronic Devices & Circuits

Answer 1

The correct answer is (d) 1.67 mF

The explanation is: The capacitor current is given as i=C*(dv/dt), where dv/dt is the derivative of voltage, dt=t2-t1    given as 10 sec and dv is the change in voltage which is given as 12V.

So, we have C=i/(dv/dt)

=> C = 2mA/(12/10) = 2mA/(1.2).

Hence C = 1.67mF.