In a purely capacitive AC circuit (ideal capacitor only), what is the phase relationship between the current and the applied AC voltage?

Introduction:This question targets phasor relationships in reactive circuits. In a purely capacitive AC circuit, the current responds to the rate of change of voltage, producing a fixed phase shift between the two quantities.

Given Data / Assumptions:

• Ideal capacitor, no resistance or inductance.
• Sinusoidal steady-state operation.
• Using rms phasor conventions.

Concept / Approach:Capacitor current is i(t) = C * dv/dt. For v(t) = V_m * sin(ωt), the derivative is proportional to cos(ωt) = sin(ωt + 90°), meaning the current waveform reaches its peaks a quarter-cycle earlier than the voltage.

Step-by-Step Solution:Let v(t) = V_m * sin(ωt)Then i(t) = C * dv/dt = ωC * V_m * cos(ωt)Since cos(ωt) = sin(ωt + 90°), current leads voltage by 90°Phasor form: I = j * ωC * V, argument(I) − argument(V) = +90°

Verification / Alternative check:Impedance of a capacitor is Z_C = 1 / (j * ω * C) = −j / (ωC), which has a phase of −90°. Current through Z_C thus has a phase of +90° relative to voltage, confirming the lead.

Why Other Options Are Wrong:

• 0.707 or 0.637 of voltage: These are numeric ratios unrelated to phase.
• Current lags voltage: True for inductors, not capacitors.
• In phase: Only true for pure resistors.

Common Pitfalls:

• Mixing up the capacitor and inductor phase rules.
• Confusing amplitude ratios with phase relationships.

Final Answer:Current leads voltage by 90 degrees.