1. I_d = V₀ω Ccosω t 2. I_d = V₀/ω Ccosω t 3. I_d = V₀/ω Csinω t 4. I_d = V₀ω Csinω t

Answer: 1. I_d = V₀ω Ccosω t Key idea: displacement current equals the conduction current that charges the capacitor. q = CV = CV₀sinω t I_d = I_c = dq/dt = CV₀ωcosω t This matches option 1. Options 2 and 3 divide by ω C instead of multiplying (the capacitive reactance is X_C=1/ω C , so the amplitude is V₀/X_C=V₀ω C ), and option 4 has sin instead of cos ; the current leads the voltage by 90°.

Topic: Displacement current· Electromagnetic Waves

A capacitor of capacitance 'C', is connected across an ac source of voltage V, given byV = V_0 ω tThe displacement current between the plates of the capacitor, would then be given by :

1. \(I_d = V_0\omega C\cos\omega t\)

2. \(I_d = \dfrac{V_0}{\omega C}\cos\omega t\)

3. \(I_d = \dfrac{V_0}{\omega C}\sin\omega t\)

4. \(I_d = V_0\omega C\sin\omega t\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

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Answer: 1. \(I_d = V_0\omega C\cos\omega t\)

Key idea: displacement current equals the conduction current that charges the capacitor.

\(q = CV = CV_0\sin\omega t\)

\(I_d = I_c = \dfrac{dq}{dt} = CV_0\omega\cos\omega t\)

This matches option 1. Options 2 and 3 divide by \(\omega C\) instead of multiplying (the capacitive reactance is \(X_C=1/\omega C\), so the amplitude is \(V_0/X_C=V_0\omega C\)), and option 4 has \(\sin\) instead of \(\cos\); the current leads the voltage by 90°.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A capacitor of capacitance 'C', is connected across an ac source of voltage V, given byV = V_0 ω tThe displacement current between the plates of the capacitor, would then be given by :

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