Purely inductive AC branch: In an ideal purely inductive circuit driven by a sinusoidal source, what is the phase relationship between voltage and current?

Introduction / Context:Understanding phase relationships in ideal reactive elements is essential for AC circuit analysis and power calculations. Inductors and capacitors shift current and voltage by 90°, but in opposite directions. Remembering which quantity leads in which element is a key conceptual skill.

Given Data / Assumptions:

• Purely inductive branch (no resistance).
• Sinusoidal steady-state excitation.
• Ideal inductor behavior (no core losses or winding resistance).

Concept / Approach:For an inductor, v_L = L * di/dt. If current is sinusoidal, its derivative (and therefore voltage) leads the current by 90°. Equivalently, in phasor form, the inductor’s impedance is jXL, which has a phase of +90°, indicating that voltage across the inductor leads the current through it by 90°.

Step-by-Step Solution:Assume i(t) = I_peak * sin(ωt).Then v(t) = L * di/dt = L * I_peak * ω * cos(ωt) = L * I_peak * ω * sin(ωt + 90°).Thus, the voltage waveform is shifted +90° relative to current.Therefore, in a purely inductive circuit, voltage leads current by 90°.

Verification / Alternative check:Phasor impedance Z_L = jXL; multiplying current phasor by j rotates it +90°, producing the voltage phasor—graphically confirming the lead.

Why Other Options Are Wrong:Current leads by 90° or in-phase: describe capacitive or resistive cases, not inductive.Voltage lags current by 90°: opposite of inductor behavior; that is for capacitors (current leads in capacitors).

Common Pitfalls:Mixing up inductor and capacitor phase rules. A popular mnemonic: “ELI the ICE man” (E leads I in L; I leads E in C).

Final Answer:voltage leads current by 90°