A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400V. If the plate area is 60 cm2, then the value of displacement current for 10–6 sec. will be:
Answer & explanation
Correct answer: option 4
$I_{D}=\varepsilon_0 \frac{\partial \phi_{E}}{\partial t}=\varepsilon_0 \frac{EA}{t}=\frac{\varepsilon_0 VA}{dt}$
or $I_{D}=\frac{8.85 \times 10^{-12} \times 400 \times 60 \times 10^{-4}}{2 \times 10^{-3} \times 10^{-6}}$
= 1.062 × 10-4 A