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|)?
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Amust be decreased
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Bmust be increased
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Cis held constant
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Dhas no effect
Answer
Correct Answer: must be decreased
Explanation
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:
Goal: |Vc| > |Vr| ⇒ Xc > R.Adjust frequency: decrease f so that Xc = 1/(2πfC) grows and exceeds R.Therefore, to make Vc larger than Vr, the frequency must be decreased.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