Phase relations in capacitors — does the capacitor voltage always lead the current by 90 degrees in steady-state AC?

Difficulty: Easy

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

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