Source voltage at series resonance: In a series resonant circuit with measured voltages VC = 125 V, VL = 125 V, and VR = 40 V, what is the magnitude of the source voltage?

Introduction / Context:
At series resonance, the inductor and capacitor voltages can be large but are equal and opposite in phase, effectively canceling in the phasor sum. The source voltage therefore equals the resistive drop, a key property of resonant circuits that explains their high circulating currents and voltage magnification across reactive elements.

Given Data / Assumptions:

• Series resonant circuit.
• Measured: V_L = 125 V, V_C = 125 V, V_R = 40 V (rms values).

Concept / Approach:
Phasor addition in series circuits: V_S = V_R + j(V_L − V_C). At resonance, V_L and V_C are 180° out of phase and equal in magnitude, so V_L − V_C = 0. Thus V_S reduces to the purely real resistive drop.

Step-by-Step Solution:

Verification / Alternative check:
Observing that current at resonance is I = V_R / R and Z_total = R confirms that the source voltage must equal the resistive component only.

Why Other Options Are Wrong:

• 125 V, 250 V, 290 V: These incorrectly add reactive drops which cancel vectorially at resonance.

Common Pitfalls:

• Adding voltages arithmetically instead of vectorially, ignoring phase relationships.

Final Answer:
40 V