A two-branch parallel AC circuit has a series R–L branch with R = 20 Ω and L = 1 H in one branch, and a pure capacitor C = 100 μF in the other branch. It is fed from a 100 V AC source. At parallel resonance, what is the input impedance of the entire circuit?

Introduction / Context:
Parallel resonance (also called anti-resonance) occurs when the net susceptance of a parallel network is zero. At this operating point the input admittance is purely real, and the input impedance peaks. This question tests recognition that the resistive part governing the input is the conductance provided by the resistive element in the R–L branch.

Given Data / Assumptions:

• Branch 1: series R–L with R = 20 Ω, L = 1 H.
• Branch 2: pure capacitor C = 100 μF.
• At resonance: imaginary parts of the total admittance cancel.
• Ideal components (no extra losses).

Concept / Approach:
Admittance of the R–L series branch is Y_RL = 1/(R + jωL). The capacitor admittance is Y_C = jωC. At the resonant frequency, imag{Y_RL} + imag{Y_C} = 0 so total susceptance is zero. The total conductance is then simply real{Y_RL} = R/(R^2 + (ωL)^2). But at the exact frequency that cancels the susceptance with the capacitor, the parallel network's input impedance reduces to Z_in = 1/real{Y_total} = 1/(1/R) = R, a standard result for a lossless C tuned against a series R–L branch.

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