Phase relations in capacitors — does the capacitor voltage always lead the current by 90 degrees in steady-state AC?
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ATrue
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BFalse
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CTrue only at resonance
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DTrue only when series resistance is zero
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ETrue for DC but not AC
Answer
Correct Answer: False
Explanation
Introduction / Context:Understanding phase relationships between voltage and current for reactive components is essential in AC circuit design, power factor correction, and filter development. The question checks the specific lead-lag behavior of an ideal capacitor.
Given Data / Assumptions:
- Ideal capacitor in sinusoidal steady state.
- Angular frequency is omega and current is nonzero.
- No nonideal parasitics unless discussed as a caveat.
Concept / Approach:
For an ideal capacitor, current leads voltage by 90 degrees. Equivalently, voltage lags current by 90 degrees. This comes directly from i = C * dv/dt, which shows current proportional to the derivative of voltage.
Step-by-Step Solution:
For v(t) = V_m * sin(omega t), dv/dt = omega * V_m * cos(omega t).Thus i(t) = C * dv/dt = C * omega * V_m * cos(omega t) = C * omega * V_m * sin(omega t + 90 degrees).Therefore current leads voltage by 90 degrees, not the other way around.So the statement that “capacitor voltage leads current” is incorrect.Verification / Alternative check:
Phasor form: I = j * omega * C * V. The operator j indicates that the current phasor is rotated +90 degrees relative to the voltage phasor, confirming current leads voltage in a capacitor.
Why Other Options Are Wrong:
- Resonance does not reverse the fundamental capacitor relation.
- Series resistance slightly alters phase from exactly 90 degrees but does not cause voltage to lead current.
- The DC option is irrelevant because a capacitor blocks steady DC after transients.
Common Pitfalls:
Memorization errors between inductors and capacitors. A helpful mnemonic: “ICE” and “ELI.” In a capacitor (C), current (I) leads voltage (E). In an inductor (L), voltage (E) leads current (I).
Final Answer:
False