For AC circuit analysis, what is the equivalent impedance magnitude of a 100 Ω resistor connected in parallel with a pure inductor whose reactance X_L is 200 Ω?

Introduction / Context:
Combining resistive and reactive elements in parallel is a common task in AC circuit design, filter synthesis, and impedance matching. Unlike series combinations that add directly, parallel combinations interact through admittances. Computing the equivalent impedance helps predict current sharing, phase angle, and power factor. This problem asks for the magnitude of the impedance of a resistor in parallel with a purely inductive reactance.

Given Data / Assumptions:

• R = 100 Ω (purely resistive).
• X_L = 200 Ω (purely inductive, reactance +j200 Ω).
• Elements are connected in parallel at a single frequency.
• We seek the magnitude |Z_eq|.

Concept / Approach:
Use complex impedance algebra. For R in parallel with jX: Z_eq = (R * jX) / (R + jX). The magnitude simplifies to |Z_eq| = (R * X) / sqrt(R^2 + X^2) because |j| = 1 and magnitude of the denominator is sqrt(R^2 + X^2). Plug in R = 100 and X = 200. This approach avoids converting to admittances explicitly while remaining exact for a pure reactance in parallel with a resistor.

Step-by-Step Solution:
Write Z_R = 100, Z_L = j200.Compute product magnitude: |Z_R * Z_L| = |100 * j200| = 20000.Compute denominator magnitude: |Z_R + Z_L| = |100 + j200| = sqrt(100^2 + 200^2) = sqrt(50000) ≈ 223.606.Compute |Z_eq| = 20000 / 223.606 ≈ 89.44 Ω.Match the closest listed value → 89.3 Ω.

Verification / Alternative check:
Admittance method: Y = 1/100 + 1/(j200) = 0.01 − j0.005 S. |Y| = sqrt(0.01^2 + (−0.005)^2) ≈ 0.01118 S → |Z| = 1/|Y| ≈ 89.44 Ω. Same result confirms correctness.

Why Other Options Are Wrong:
77.7 Ω and 88.8 Ω are off the precise computation. 224 Ω is roughly the denominator magnitude, not the final impedance. “None” is incorrect because 89.3 Ω matches the calculation within rounding.

Common Pitfalls:
Adding impedances in parallel as if they were series; forgetting to take magnitudes; mixing up admittance and impedance or losing the j sign; rounding too early, which introduces noticeable error.

Final Answer:
89.3 ohm