Frequency through series RC: A 6 kHz sinusoidal voltage is applied to a series RC circuit. What is the frequency of the voltage appearing across the resistor?

Difficulty: Easy

Correct Answer: 6 kHz

Explanation:


Introduction / Context:
Linear time-invariant (LTI) circuits do not change the input frequency; they only alter amplitude and phase of each sinusoidal component. Therefore, the frequency across any element in a simple series RC equals the source frequency.


Given Data / Assumptions:

  • Source frequency f_in = 6 kHz.
  • Series RC—linear, time-invariant.
  • No nonlinearities or modulation effects.


Concept / Approach:

An LTI circuit acts as a filter characterized by a transfer function. It scales and phase-shifts sinusoidal inputs at the same frequency; it does not generate new frequencies under linear operation.


Step-by-Step Solution:

Input: v_in(t) = Vp sin(2π * 6k * t).Output across R: v_R(t) = |H_R(jω)| * Vp * sin(2π * 6k * t + φ_R), where ω = 2π * 6k.Frequency component is unchanged at 6 kHz; only magnitude and phase are modified by H_R(jω).


Verification / Alternative check:

Measure in the lab: oscilloscope FFT will show a single line at 6 kHz for an ideal RC driven by a single-tone sinusoid.


Why Other Options Are Wrong:

12 kHz and 18 kHz imply frequency doubling or tripling, which does not occur in LTI elements. 0 Hz would require rectification or averaging, not present here.


Common Pitfalls:

Confusing time constant effects (amplitude/phase) with frequency conversion; mixing with nonlinear device behavior.


Final Answer:

6 kHz

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