Question

A capacitor of capacitance 3.00 uF is charged to a potential of 150 V. If the voltage were changed to 500 in 5.25 ms, what average current would there be ? V [Ans. 0.20 A)​

Answer 1

Answer:

Capacitance of capacitor = 3 uF

Rate of change of potential = (dV/dt) = (500-150)/5.25

= 350/5.25 = 66,666.67 V/s

Q=CV

I=dQ/dt=C*dV/dt

I = C * dV/dt

I = 3*10^-6 * 66,666.67

I = 0.2 A

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