A voltage of 5 V, 50 Hz is applied to a capacitor of capacitance 6 μF. The displacement current will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9.42 mA
Between the plates there is no conduction current; the displacement current \( i_d = \varepsilon_0\, d\phi_E/dt \) equals the conduction current in the connecting wires, which for an ac source is \( V/X_C = V\omega C \).
\( \displaystyle X_C = \frac{1}{2\pi f C} \)
\( \displaystyle i_d = 2\pi f C V = 2 \times 3.14 \times 50\,\mathrm{Hz} \times 6\times10^{-6}\,\mathrm{F} \times 5\,\mathrm{V} \)
\( \displaystyle i_d \approx 9.42\times10^{-3}\,\mathrm{A} = 9.42\,\mathrm{mA} \)