Difficulty: Easy
Correct Answer: increases
Explanation:
Introduction / Context:Understanding how impedance varies with frequency is essential in filter design and AC network analysis. In a parallel RC circuit, the capacitor branch admittance depends directly on frequency; therefore, the overall impedance is frequency-sensitive in a predictable way.
Given Data / Assumptions:
Concept / Approach:Admittance of the capacitor is Y_C = jωC (magnitude ωC), which decreases as frequency decreases (since ω = 2πf). The total admittance of the parallel network is Y_total = 1/R + jωC. As ω decreases, |Y_total| decreases toward 1/R, causing the total impedance Z_total = 1 / |Y_total| to increase toward R.
Step-by-Step Solution:
Lower f → smaller ωC → capacitor branch contributes less admittance.Total admittance drops toward 1/R.Since Z_total = 1 / |Y_total|, a decrease in admittance means an increase in impedance.Verification / Alternative check:Limiting cases: At very low f, capacitor behaves like open circuit → Z_total ≈ R (highest). At very high f, capacitor admittance dominates → Z_total becomes small (lowest). This trend confirms impedance increases as frequency decreases.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:increases
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