Series resonance voltage across L: A 24 Ω resistor, an inductor with XL = 120 Ω, and a capacitor with Xc = 120 Ω are in series across a 60 V source at resonance. What is the inductor voltage VL?

Introduction / Context:
At series resonance, inductive and capacitive reactances are equal and opposite, cancelling in the net series impedance. Although the source voltage appears mostly across the resistor, each reactive element can have a large individual voltage due to the circulating current, a key idea called voltage magnification.

Given Data / Assumptions:

• R = 24 Ω, XL = 120 Ω, Xc = 120 Ω.
• Source voltage Vs = 60 V (rms).
• Series circuit at resonance ⇒ net reactance 0.

Concept / Approach:

At resonance, Z_total = R, so circuit current I = Vs / R. The voltage across any reactive element is I * X (with X = XL or Xc). Because X is much larger than R, the reactive voltage can exceed the source voltage.

Step-by-Step Solution:

Verification / Alternative check:

VC = I * Xc = 300 V but 180° out of phase with VL; their phasor sum is zero, leaving only the 60 V across R, consistent with resonance behavior.

Why Other Options Are Wrong:

60 V is the source/resistor voltage, not the inductor drop at resonance. 30 V is half the source and unjustified. 660 V is an inflated figure inconsistent with I and XL.

Common Pitfalls:

Assuming each element shares voltage equally, or thinking reactive drops cannot exceed the source in series resonance.

Final Answer:

300 V