Parallel RC at AC: A resistor of 12 Ω is connected in parallel with a capacitor having reactance Xc = 5 Ω at the operating frequency. What is the magnitude of the equivalent impedance |Z| of the parallel combination?

Introduction:
For parallel AC networks, it is often easier to work with admittances. A resistor contributes a conductance G, and a capacitor contributes a susceptance B. The magnitude of the total impedance is the reciprocal of the magnitude of the total admittance.

Given Data / Assumptions:

• R = 12 Ω (resistor branch)
• Xc = 5 Ω (capacitive reactance at the operating frequency)
• Ideal components, sinusoidal steady state

Concept / Approach:

Compute admittance components: G = 1 / R and |B| = 1 / Xc. The magnitude of total admittance is |Y| = sqrt(G^2 + B^2). Then |Z| = 1 / |Y|.

Step-by-Step Solution:

Verification / Alternative check:

Intuitively, a strong capacitive branch (small Xc) in parallel reduces total impedance below the smallest branch impedance. Since 5 Ω is the smallest branch impedance, |Z| must be less than 5 Ω; 4.6 Ω is consistent.

Why Other Options Are Wrong:

• 0.20 Ω: Confuses susceptance (0.2 S) with impedance.
• 3.5 Ω and 6.0 Ω: Do not satisfy the computed 4.62 Ω result.
• 13 Ω: Greater than either branch; parallel impedance cannot exceed the smallest branch impedance.

Common Pitfalls:

• Adding impedances directly in parallel instead of adding admittances.
• Ignoring that capacitive susceptance uses magnitude when computing |Y|.

Final Answer:

4.6 Ω