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 mFI had been asked this question in an interview.This intriguing question originated from Basic Concepts in portion EDC Overview of Electronic Devices & Circuits

A capacitor is charged by a constant current of 2 mA and results in a

1.	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 mFI had been asked this question in an interview.This intriguing question originated from Basic Concepts in portion EDC Overview of Electronic Devices & Circuits
Answer» 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-t1given 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.

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