Magnitude of total impedance for a parallel RL: A 140 Ω resistor is in parallel with an inductor having 60 Ω inductive reactance. Both are across a 12 V AC source. What is |Z| of the parallel combination?

Introduction / Context:
Calculating the total impedance of a parallel RL network requires summing admittances (not impedances). This is a standard AC analysis skill, often needed for filter and load calculations.

Given Data / Assumptions:

• R = 140 Ω (resistive branch).
• XL = 60 Ω (inductive reactance of L branch).
• Parallel connection across the same AC source.
• Ideal components (no resistance in L aside from reactance).

Concept / Approach:
Total admittance Y_total = Y_R + Y_L. For the resistor, Y_R = 1/R (purely real). For the inductor, Y_L = 1/(j XL) = -j / XL (purely imaginary). Then |Z_total| = 1 / |Y_total| and |Y_total| = sqrt( G^2 + B^2 ).

Step-by-Step Solution:
G = 1 / R = 1 / 140 ≈ 0.0071429 SB = -1 / XL = -1 / 60 ≈ -0.0166667 S|Y_total| = sqrt( G^2 + B^2 )|Y_total| ≈ sqrt( 0.0071429^2 + 0.0166667^2 ) ≈ sqrt( 0.0000510 + 0.0002778 ) ≈ 0.018129 S|Z_total| = 1 / |Y_total| ≈ 1 / 0.018129 ≈ 55.15 Ω

Verification / Alternative check:
Back-calculate currents at 12 V: I_R = 12 / 140 ≈ 0.0857 A, I_L = 12 / 60 = 0.2 A at -90°. Total current magnitude ≈ sqrt(0.0857^2 + 0.2^2) ≈ 0.2178 A. Then |Z| = V / I ≈ 12 / 0.2178 ≈ 55.1 Ω, confirming the result.

Why Other Options Are Wrong:

• 5.51 Ω: Off by a factor of 10, likely from inverting at the wrong step.
• 90 Ω, 200 Ω: Do not match the computed admittance magnitude and are inconsistent with the branch values.

Common Pitfalls:
Adding impedances instead of admittances for parallel circuits, or forgetting that the inductor admittance is imaginary (negative susceptance).

Final Answer:
55.15 Ω