Series RC driven by a sine source: initially XC = R so VR = VC. If the frequency is increased, how do the resistor and capacitor voltages compare?

Introduction / Context:
Voltage division in AC circuits depends on impedances, not just resistances. In a series RC, the capacitor’s reactance varies with frequency, changing the voltage distribution. This concept underlies filters, phase shifters, and sensor conditioning.

Given Data / Assumptions:

• A series RC circuit driven by a sine wave.
• At the starting frequency, XC = R, so VR = VC.
• Frequency is then increased; R and C values are unchanged.

Concept / Approach:
Capacitive reactance is XC = 1 / (2 * π * f * C). As frequency f increases, XC decreases. In a series network, the magnitude of voltage across a component is proportional to its impedance magnitude. Therefore, with higher f, the capacitor’s impedance shrinks and takes a smaller share of the source voltage, while the resistor takes a larger share.

Step-by-Step Solution:

Verification / Alternative check:
Numerical example with Vs = 1 V, R = 1 kΩ, C chosen so XC = 1 kΩ at f0. At 2f0, XC halves to 500 Ω. Voltage division yields VR ≈ 0.894 V, VC ≈ 0.447 V, confirming VR > VC.

Why Other Options Are Wrong:

• VC > VR: Opposite of frequency effect on XC.
• VR = VC: Only true when XC = R.
• VR and VC = 0: Not in a driven sinusoidal circuit.

Common Pitfalls:

• Treating XC as constant with frequency.
• Ignoring that phase angles change but here only magnitudes are compared.

Final Answer:
VR > VC