Parallel R–L–C network: A 15 Ω resistor, an inductor with XL = 8 Ω, and a capacitor with XC = 12 Ω are connected in parallel across an AC source. What is the equivalent circuit impedance?

Introduction / Context:
Parallel AC networks are best handled using admittances. Each branch contributes conductance and/or susceptance, which add directly. The reciprocal of the total admittance yields the equivalent impedance. Mastery of this conversion is essential for power distribution and filter loading problems.

Given Data / Assumptions:

• Resistive branch: R = 15 Ω → Y_R = 1/R.
• Inductive branch: X_L = +8 Ω → Z_L = j8 → Y_L = 1/Z_L = −j(1/8) S.
• Capacitive branch: X_C = 12 Ω → Z_C = −j12 → Y_C = 1/Z_C = +j(1/12) S.
• Ideal components; sinusoidal steady state.

Concept / Approach:
Total admittance Y_T = Y_R + Y_L + Y_C. Then Z_eq = 1 / Y_T. The result is complex, but the question asks for the equivalent impedance magnitude (standard in such option sets).

Step-by-Step Solution:

Verification / Alternative check:
The resistive branch alone is 15 Ω. Adding reactive branches in parallel reduces the overall impedance below 15 Ω, so a value around 12–13 Ω is consistent with expectations.

Why Other Options Are Wrong:

• 127 Ω, 4,436 Ω, 6,174 Ω: These are far too large for a network that already contains a 15 Ω branch in parallel; the equivalent cannot exceed the smallest branch impedance.

Common Pitfalls:

• Treating parallel impedances like series (adding magnitudes rather than admittances).
• Forgetting the sign difference between inductive and capacitive susceptance.

Final Answer:
12.7 Ω