Phase Relations In Ac Elements
When a sinusoidal voltage is applied across a pure capacitor in an AC circuit, how does the resulting current relate in phase to that applied voltage?
Select the correct option:
Solution
Current leads the voltage by 90 degrees
For a pure capacitor the instantaneous charge is (q = CV), and the current is the rate of change of charge, (i = C,dV/dt). Differentiating a sinusoidal voltage (V = V_0\sin\omega t) gives (i = \omega C V_0\cos\omega t = \omega C V_0\sin(\omega t + 90^\circ)), showing the current reaches its peak a quarter cycle before the voltage. Hence the current leads the applied voltage by ninety degrees. The option 'lags by 90 degrees' describes a pure inductor, where current builds up only after voltage is applied. The 'in phase' option applies to a pure resistor, not a capacitor. The '45 degrees' option would require a mix of resistance and reactance, not a single ideal capacitor. This quadrature relationship is exactly the NCERT result underlying why a capacitor dissipates no average power, because the average of the product of a sine and a cosine over a complete cycle is zero. The same reasoning shows that the magnitude of the current is limited by the capacitive reactance (X_C = 1/(\omega C)), so larger capacitance or higher frequency permits a bigger current. A consistency check notes that since current peaks while voltage is zero and rising fastest, the energy merely oscillates into and out of the electric field rather than being consumed, confirming the leading-current, lossless behaviour of ideal capacitive elements.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- phase relations in ac elements
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Current leads the voltage by 90 degrees
For a pure capacitor the instantaneous charge is (q = CV), and the current is the rate of change of charge, (i = C,dV/dt). Differentiating a sinusoidal voltage (V = V_0\sin\omega t) gives (i = \omega C V_0\cos\omega t = \omega C V_0\sin(\omega t + 90^\circ)), showing the current reaches its peak a quarter cycle before the voltage. Hence the current leads the applied voltage by ninety degrees. The option 'lags by 90 degrees' describes a pure inductor, where current builds up only after voltage is applied. The 'in phase' option applies to a pure resistor, not a capacitor. The '45 degrees' option would require a mix of resistance and reactance, not a single ideal capacitor. This quadrature relationship is exactly the NCERT result underlying why a capacitor dissipates no average power, because the average of the product of a sine and a cosine over a complete cycle is zero. The same reasoning shows that the magnitude of the current is limited by the capacitive reactance (X_C = 1/(\omega C)), so larger capacitance or higher frequency permits a bigger current. A consistency check notes that since current peaks while voltage is zero and rising fastest, the energy merely oscillates into and out of the electric field rather than being consumed, confirming the leading-current, lossless behaviour of ideal capacitive elements.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of phase relations in ac elements. It appeared in the 2025 exam.
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