Q:
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 ________
Answer & Explanation
Answer: D) 1.67 mF
Explanation: 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.
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