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

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Solution

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Capacitance of capacitor $$C = 20 \mu F$$

$$= 20 \times 10^{-6} F$$

Rate of change of potential is,$$\left(\dfrac{dV}{dt} \right) = 3 v/s$$

$$\implies q = CV$$

$$\implies \dfrac{dq}{dt} = C \dfrac{dV}{dt}$$

$$i_c = 20 \times 10^{-6} \times 3$$

$$= 60 \times 10^{-6} A$$

$$= 60 \mu A$$

As we know that $$i_d = i_C = 60 \mu A$$

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