A parallel plate capacitor of capacitance 20 μF 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 :

CONCEPT:

Capacitor- It is a two-terminal electrical device that stores energy as an electric charge.

The capacitance of the capacitor is the ratio of the charge per applied potential and it is written as;

      ----(1)

Here, Q is the charge and V is the applied potential.

CALCULATION:

Given that: Capacitance of capacitor C = 20 μF

 C = 20 × 10^-6 F

and Rate of change of potential  = 3 v/s

Now, on using the equation (1) we have;

q = CV       -----(2)

Differentiate the equation (2) with respect to time t we have;

    ----(3)

As we know the rate to change of charge per unit of time is equal to the current, therefore equation 3) is 

Now, on putting all the given values in equation (3) we have;

i_c = 20 × 10^-6 × 3

   = 60 × 10^-6 A

   = 60 μA

As we know that displacement current is equal to capacitive current we have,

i_d = i_c = 60 μA

Hence, option 2) is the correct answer.