A parallel plate capacitor of capacitance 10 μF and separation between the plates 5 mm, is connected to an a.c. supply of 220 V with angular frequency 314 rad/s. The displacement current in the capacitor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.69 A
No charge crosses the gap between the plates, but the electric field there keeps changing as the capacitor charges and discharges. Maxwell called the effect of this changing field the displacement current, \( I_d = \varepsilon_0\,d\phi_E/dt \); it keeps the current continuous and is equal to the conduction current in the connecting wires. So it is enough to find the current in the circuit, which is set by the capacitive reactance.
\( \displaystyle X_C = \frac{1}{\omega C} \)
\( \displaystyle I = \frac{V}{X_C} = V\omega C \)
\( \displaystyle I = 220 \times 314 \times 10\times10^{-6}\,\mathrm{A} \)
\( \displaystyle I \approx 0.69\,\mathrm{A} \)
The plate separation is not needed.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – “… 8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current?”