For the series RC voltages considered at some frequency, what change in frequency will make the capacitor’s voltage exceed the resistor’s voltage (i.e., |Vc| > |Vr|)?

Introduction / Context:
In a series RC circuit, the ratio of capacitor to resistor voltage depends on the ratio of reactance to resistance. Tuning frequency can shift which element has the larger drop, a key idea in filters and frequency-selective dividers.

Given Data / Assumptions:

• Series RC, fixed R and C.
• We compare |Vc| and |Vr| at different frequencies.
• Ideal, linear, sinusoidal operation.

Concept / Approach:

Voltage division in series gives |Vc| : |Vr| ≈ Xc : R. If Xc > R, the capacitor drop dominates; if Xc < R, the resistor drop dominates. Since Xc = 1 / (2 * π * f * C), lowering frequency increases Xc.

Step-by-Step Solution:

Verification / Alternative check:

At very low f, Xc ≫ R and almost all source voltage appears across the capacitor; at very high f, Xc → 0 and the resistor drop dominates. This confirms the frequency trend.

Why Other Options Are Wrong:

Increasing f shrinks Xc, favoring the resistor. Holding constant or 'no effect' contradicts the strong frequency dependence of Xc.

Common Pitfalls:

Confusing RC with RL behavior; ignoring that voltages in series divide in proportion to impedance magnitudes.

Final Answer:

must be decreased