Phase relationship in a purely capacitive AC circuit: how does current relate to voltage?

Difficulty: Easy

Correct Answer: Current leads voltage by 90°

Explanation:


Introduction:
Understanding phase relationships is fundamental to AC circuit analysis. In purely reactive elements, voltage and current are out of phase. This question asks for the specific relationship in a purely capacitive circuit.


Given Data / Assumptions:

  • Ideal capacitor (no resistance).
  • Sinusoidal steady state.
  • Single-frequency excitation.


Concept / Approach:
The impedance of a capacitor is Zc = 1 / (j * 2 * pi * f * C) = −j * Xc. The negative imaginary sign indicates current leads voltage by 90 degrees. Equivalently, i(t) = C * dv/dt; the derivative emphasizes that current peaks when voltage crosses zero and vice versa, producing a +90° current lead.


Step-by-Step Solution:

Write Zc = 1 / (j * ω * C) = −j / (ω * C).Phasor current I = V / Zc = V * (j * ω * C) → +90° relative to V.Hence current leads voltage by a quarter cycle.


Verification / Alternative check:
Time-domain relation i(t) = C * dv/dt shows current is maximum at voltage zero crossings and zero at voltage peaks, consistent with a 90° lead.


Why Other Options Are Wrong:

  • Voltage leads by 90° or current lags by 90°: These describe inductive, not capacitive, circuits.
  • In phase: Requires resistive behavior; not applicable to ideal capacitors.


Common Pitfalls:
Confusing the sign convention of j or mixing inductive and capacitive phase relations. Remember ‘‘ICE’’: in a Capacitor, Current leads Voltage.


Final Answer:
Current leads voltage by 90°.

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