Difficulty: Medium
Correct Answer: More than 5%
Explanation:
Introduction / Context:Harmonic distortion in voltage sources translates differently into current depending on the load impedance versus frequency. In a series RC circuit, impedance magnitude decreases with frequency, so higher-frequency components produce comparatively larger currents than lower-frequency components, other things equal.
Given Data / Assumptions:
Concept / Approach:Current magnitude for a frequency component at angular frequency ωk is |I(ωk)| = |V(ωk)| / |Z(ωk)|. For a series RC, |Z(ω)| = sqrt(R^2 + (1/(ωC))^2). As frequency increases, the capacitive reactance 1/(ωC) decreases, reducing |Z| and increasing current. Hence, the fifth harmonic (5ω) experiences smaller impedance than the fundamental (ω), so its current proportion increases beyond the voltage’s 5%.
Step-by-Step Solution:
Given voltage fifth-harmonic ratio = 5% of fundamental.For current: I1 = V1 / |Z(ω)|, I5 = V5 / |Z(5ω)|.Since |Z(5ω)| < |Z(ω)|, the ratio I5 / I1 > V5 / V1.Therefore, current fifth-harmonic content > 5%.Verification / Alternative check:
Compute example with R and C: the impedance drop at higher harmonics always favors more current for RC loads.Why Other Options Are Wrong:
5% or less than 5%: would be true for RL loads (increasing impedance with frequency) but not RC.Equal to or more than 5%: overly vague; the correct insight is strictly more than 5% in an RC series.Approximately zero: not applicable.Common Pitfalls:
Assuming all loads behave like resistors (frequency-neutral). Reactive impedances bias harmonic currents.Final Answer:
More than 5%
Discussion & Comments