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A parallel plate capacitor of capacitance 20 μF20 \text{ }\mu\text{F} is being charged by a voltage source whose potential is changing at the rate of 3 V/s3 \text{ V/s}. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively

A

Zero, 60 μA60 \text{ } \mu\text{A}

B

60 μA60 \text{ } \mu\text{A}, 60 μA60 \text{ } \mu\text{A}

C

60 μA60 \text{ } \mu\text{A}, zero

D

Zero, zero

Step-by-Step Solution

Capacitance of capacitor C=20 μFC = 20 \text{ } \mu\text{F} =20×10−6 F= 20 \times 10^{-6} \text{ F} Rate of change of potential dVdt=3 V/s\frac{dV}{dt} = 3 \text{ V/s} q=CVq = CV dqdt=CdVdt\frac{dq}{dt} = C \frac{dV}{dt} ic=20×10−6×3i_c = 20 \times 10^{-6} \times 3 =60×10−6 A= 60 \times 10^{-6} \text{ A} =60 μA= 60 \text{ } \mu\text{A} As we know that id=ic=60 μAi_d = i_c = 60 \text{ } \mu\text{A}

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