Purely inductive AC circuit: what happens to current when the operating frequency is doubled? Assume constant applied voltage amplitude and ideal inductor behavior.

Introduction / Context:
For sinusoidal steady-state analysis, an ideal inductor’s opposition to AC is captured by inductive reactance X_L. Designers need intuition about how current changes with frequency at fixed voltage to predict heating, magnetic fields, and resonance behavior in filters and power systems.

Given Data / Assumptions:

• Ideal inductor L driven by a fixed-amplitude sinusoidal source.
• Inductive reactance X_L = 2 * pi * f * L.
• Linearity and no core saturation or resistance.

Concept / Approach:
Ohm’s law for AC magnitudes gives I = V / X_L for a purely inductive branch. Because X_L is proportional to frequency f, doubling f doubles X_L. Therefore, with the same applied voltage, current is inversely proportional to frequency and will be halved when f is doubled.

Step-by-Step Solution:
1) Use X_L = 2 * pi * f * L.2) Double frequency: f → 2f → X_L(new) = 2 * pi * (2f) * L = 2 * X_L.3) Current relation: I = V / X_L → I(new) = V / (2 * X_L) = I / 2.4) Conclude: current is cut in half.

Verification / Alternative check:
Numerical example: V = 10 V, L = 100 mH. At 50 Hz, X_L ≈ 31.4 ohm → I ≈ 0.318 A. At 100 Hz, X_L ≈ 62.8 ohm → I ≈ 0.159 A, approximately half, confirming the proportionality.

Why Other Options Are Wrong:
“Doubles” contradicts inverse proportionality; “no effect” ignores X_L dependence; “requires complete analysis” is unnecessary because the ideal formula fully predicts the change.

Common Pitfalls:
Including winding resistance or core effects unintentionally; while real inductors have losses, the idealized relation still guides first-order design.

Final Answer:
cuts the current through the inductors one-half