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;

\(C=\frac{q}{V}\)      ----(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 \(\left( {\frac{{dV}}{{dt}}} \right) \) = 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;

\(\frac{{dq}}{{dt}} = C\frac{{dV}}{{dt}}\)    ----(3)

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

\(i_c= C\frac{{dV}}{{dt}}\)

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.