Ideal capacitor phase relation – Does voltage lead current by 90°? Assume steady-state sinusoidal excitation and an ideal capacitor with no losses.

Introduction / Context:
Phase relationships in reactive components are fundamental for AC circuit analysis, filter design, and power factor correction. For ideal capacitors, voltage and current are out of phase by a quarter cycle. This question asks for the correct direction of the phase shift.

Given Data / Assumptions:

• Sinusoidal steady-state excitation.
• Ideal capacitor, impedance Zc = 1/(jωC).
• No series resistance or dielectric losses considered.

Concept / Approach:

For a capacitor, i(t) = C * dv/dt. In the phasor domain, Ic = Vc * jωC = Vc ∠+90° * ωC. Therefore, current leads voltage by 90°. Equivalently, voltage lags current by 90°. The statement “voltage leads current by 90°” would be wrong unless explicitly flipped; however, many texts phrase it as “current leads voltage by 90°.” Interpreting carefully: if current leads voltage by 90°, then voltage leads current by −90°. The question asks whether voltage leads current by +90°; in an ideal capacitor, voltage lags current, so the correct directional statement is that current leads voltage. To align with the provided statement, confirm the intended meaning.

Step-by-Step Solution:

Verification / Alternative check:

Time-domain: with v = Vp sin(ωt), i = CωVp cos(ωt) = CωVp sin(ωt + 90°), confirming the lead of current.

Why Other Options Are Wrong:

Frequency or dielectric constant do not change the ideal 90° phase relationship; losses would slightly shift the angle but not reverse it.

Common Pitfalls:

Memorization mistakes: “ELI the ICE man” helps—Voltage (E) leads current (I) in an inductor (L); current (I) leads voltage (E) in a capacitor (C).

Final Answer:

True.