JEE Main 2021 — Alternating Current Question with Solution

Given,

AC voltage, V(t) = 20 sin $$\omega$$t volt.

Frequency, f = 50Hz

Separation between the plates, d = 2 mm = 2 $$\times$$ 10^$$-$$3 m

Area, A = 1 m²

As, $$C=\frac{{\epsilon }_{0}A}{d}$$

where, $${\epsilon }_0$$ = absolute electrical permittivity of free space = 8.854 $$\times$$ 10^$$-$$12 N^$$-$$1 kg²m^$$-$$2

$$C=\frac{{\epsilon }_0\times 1}{2\times {10}^{-3}}$$ .... (i)

Capacitive reactance $$(X_C)=\frac{1}{\omega C}$$ .... (ii)

From Eqs. (i) and (ii), we get

$$X_C=\frac{2\times {10}^{-3}}{2\times 50\pi \times {\epsilon }_0}$$ ($$\because$$ $$\omega$$ = 2$$\pi$$f)

$$=\frac{2\times {10}^{-3}}{25\times 4\pi {\epsilon }_0}$$

$$\Rightarrow X_C=\frac{2\times {10}^{-3}}{25}\times 9\times 10^9$$

$$\Rightarrow X_C=\frac{18}{25}\times 10^6\Omega$$

By using Ohm's law,

As, $$I_0=\frac{V_0}{X_C}=\frac{20\times 25}{18}\times 10^{-6}=27.78\times 10^{-6}$$

$$\Rightarrow$$ I₀ = 27.78$$\mu$$A

$$\therefore$$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $$\mu$$A.