In a 20 Vac series RC circuit, the measured voltage across the resistor is 20 V and the measured voltage across the capacitor is 40 V. What is the magnitude of the applied source voltage (phasor sum)?

Introduction / Context:
In series RC circuits, the resistor and capacitor voltages are out of phase by 90 degrees (resistive drop in phase with current; capacitive drop lags the current by 90 degrees). Therefore, the source voltage is not the arithmetic sum of the magnitudes but the vector (phasor) sum. This question checks correct phasor addition of Vr and Vc.

Given Data / Assumptions:

• Series RC circuit with sinusoidal steady state.
• Vr = 20 V (in phase with current).
• Vc = 40 V (lags current by 90 degrees).
• We seek |Vs|, the magnitude of the applied source voltage.

Concept / Approach:
Because Vr and Vc are orthogonal in phase, |Vs| is found using the Pythagorean relationship: |Vs| = sqrt(Vr^2 + Vc^2). Numerical substitution yields the correct applied voltage magnitude.

Step-by-Step Solution:
Given Vr = 20 V and Vc = 40 V.Compute Vs magnitude: |Vs| = sqrt(Vr^2 + Vc^2).|Vs| = sqrt(20^2 + 40^2) = sqrt(400 + 1600) = sqrt(2000) ≈ 44.72 V.Rounded to the nearest whole-volt choice: 45 Vac.

Verification / Alternative check:
Phasor diagram: draw Vr on the horizontal axis and Vc downward by 90 degrees. The source vector is the hypotenuse of the right triangle formed, clearly less than Vr + Vc (60 V) and greater than max(Vr, Vc), which fits ≈ 44.7 V.

Why Other Options Are Wrong:

• 50 Vac, 60 Vac, 65 Vac: These imply arithmetic addition or incorrect geometry; the correct operation is vector addition with orthogonal components.

Common Pitfalls:

• Adding voltage magnitudes directly in series reactive circuits; phase must be considered.
• Confusing capacitor voltage phase (90 degrees out of phase with Vr) with in-phase addition.

Final Answer:
45 Vac