Question

A parallel plate capacitor of capacitance $20\mu F$ is being charged by a voltage source whose potential is changing at the rate of $3V{s}^{-1}$. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :

• A. $60\mu A,0$
• B. $0,0$
• C. $0,60\mu A$
• D. $60\mu A,60\mu A$

Answer 1

The correct option is D $60\mu A,60\mu A$

Given, capacitance of capacitor
$C=20\mu F=20\times {10}^{-6}F$
Rate of change of potential $\left(\frac{dV}{dt}\right)=3\frac{V}{s}$
$\because q=CV$
or, $\frac{dq}{dt}=C\frac{dV}{dt}$
or, $i_C=20\times {10}^{-6}\times 3=60\times {10}^{-6}A=60\mu A$
As we know that $i_d=i_c=60\mu A$
[Alternate solution : we know that $i_d=i_c$ and the displacement current has finite value when potential difference across capacitor is changing]