RC parallel impedance check: For a resistor R in parallel with a capacitor C at angular frequency ω, is the magnitude of the equivalent impedance given by |Z_p| = 1 / sqrt( (1/R)^2 + (ω C)^2 )?
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ACorrect
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BIncorrect
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COnly valid when ωRC << 1
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DOnly valid when ωRC >> 1
Answer
Correct Answer: Correct
Explanation
Introduction / Context:Computing the equivalent impedance of parallel networks is essential for filter and bias network design. This question examines whether a commonly used closed-form expression for the magnitude of the impedance of an R || C network is correctly stated.
Given Data / Assumptions:
- Linear components; sinusoidal steady state.
- Admittance addition for parallels: Y_total = Y_R + Y_C.
- Angular frequency ω = 2π f.
Concept / Approach:For the parallel combination, Y_R = 1/R (conductance) and Y_C = jωC (susceptance). The magnitude of total admittance is |Y| = sqrt( (1/R)^2 + (ωC)^2 ). The magnitude of impedance is |Z| = 1 / |Y|, giving |Z| = 1 / sqrt( (1/R)^2 + (ωC)^2 ). This result is general for any ω and does not require small- or large-signal approximations beyond linearity.
Step-by-Step Solution:
Form admittance: Y = 1/R + jωC. Compute magnitude: |Y| = sqrt( (1/R)^2 + (ωC)^2 ). Invert to get impedance magnitude: |Z| = 1 / |Y|. Conclude the given expression is correct.Verification / Alternative check:Check limiting cases: as ω → 0, |Z| → R; as ω → ∞, |Z| → 0, consistent with a capacitor short at high frequency.
Why Other Options Are Wrong:Incorrect: contradicts basic admittance arithmetic. Valid only when ωRC << 1 or >> 1: the expression holds for all ω under linear assumptions.
Common Pitfalls:Mixing series and parallel formulas; forgetting to invert admittance magnitude to get impedance magnitude.
Final Answer:Correct