Phase relationship in a purely capacitive AC circuit: how does current relate to voltage?
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ACurrent leads voltage by 90°
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BVoltage leads current by 90°
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CCurrent lags voltage by 90°
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DCurrent and voltage are in phase (0°)
Answer
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°.