In a series RC circuit driven by AC, how does the source voltage relate to the individual voltages across the resistor and the capacitor?

Introduction / Context:
In series AC circuits containing both resistive and capacitive elements, voltages are not in phase. Understanding how these out-of-phase voltages combine is essential for accurate circuit analysis and impedance calculations.

Given Data / Assumptions:

• Series connection of R and C across an AC source.
• Steady state sinusoidal operation.
• Ideal components with no parasitic inductance or resistance beyond R and C.

Concept / Approach:
In series circuits, the same current flows through both elements. The resistor voltage V_R is in phase with current, while the capacitor voltage V_C lags current by 90 degree (equivalently, V_C leads current by negative 90 degree). Because V_R and V_C are in quadrature, the source voltage V_S is the vector (phasor) sum, not the arithmetic sum. Therefore, |V_S| = sqrt( V_R^2 + V_C^2 ), which is strictly less than V_R + V_C for nonzero voltages.

Step-by-Step Solution:
Step 1: Note I is common in series R and C.Step 2: Compute V_R = I * R (in phase with I).Step 3: Compute V_C = I * X_C with a -90 degree phase relative to I.Step 4: Combine phasorially: |V_S| = sqrt( V_R^2 + V_C^2 ).Step 5: Conclude |V_S| is less than V_R + V_C because of the right angle between V_R and V_C.

Verification / Alternative check:
Impedance magnitude Z = sqrt( R^2 + X_C^2 ). With I = V_S / Z and V_R = I * R, V_C = I * X_C, the same relationship emerges, confirming vector addition rather than arithmetic addition.

Why Other Options Are Wrong:

• Equal to the sum: Ignores phase difference; only true if voltages were in phase.
• Same amplitude and phase everywhere: Not true in reactive circuits.
• Greater than the sum: Violates triangle inequality for vector addition.
• None of the above: Incorrect because the total is less than the arithmetic sum.

Common Pitfalls:
Adding series element voltages arithmetically in AC circuits with phase shifts leads to overestimation of source voltage. Always combine voltages as vectors when phases differ.

Final Answer:
The total voltage is less than the sum of the voltages across the resistance and capacitance.