Parallel combination magnitude: A 100 Ω resistor is connected in parallel with a capacitor whose reactance magnitude is 100 Ω at the operating frequency. What is the magnitude of the total impedance |Z|?

Introduction / Context:
When resistive and reactive branches are in parallel, it is usually easier to work with admittances and then invert to find the impedance magnitude. This problem is a classic symmetric case where the resistor magnitude equals the reactance magnitude, simplifying the arithmetic.

Given Data / Assumptions:

• Resistor R = 100 Ω.
• Capacitive reactance magnitude |X_C| = 100 Ω (so branch impedance is −j100 Ω).
• Ideal components, single frequency.

Concept / Approach:
Admittance of the network is Y = 1/R + 1/(−jX_C) = 1/R + j(1/X_C). The magnitude is |Y| = sqrt((1/R)^2 + (1/X_C)^2). Then |Z| = 1/|Y|. With R = X_C = 100 Ω, the symmetry yields a simple numerical result.

Step-by-Step Solution:
Compute conductance: G = 1/R = 1/100 = 0.01 S.Compute susceptance: B = 1/X_C = 1/100 = 0.01 S.Admittance magnitude: |Y| = sqrt(G^2 + B^2) = sqrt(0.01^2 + 0.01^2) = 0.014142 S.Impedance magnitude: |Z| = 1/|Y| ≈ 1/0.014142 ≈ 70.7 Ω.

Verification / Alternative check:
A numeric phasor or calculator check yields the same 70.7 Ω. This is consistent with the general relation for equal R and X branches in parallel: |Z| = R/√2.

Why Other Options Are Wrong:
(a) 60 Ω is too low; (b) ~99 Ω suggests near-resistive behavior, not correct here; (c) 250 Ω is far off. The computed value matches 70.7 Ω.

Common Pitfalls:
Adding impedances directly in parallel instead of using admittance; forgetting to take magnitude after combining real and imaginary parts.

Final Answer:
70.7 Ω