For a 24 V(rms) parallel RL circuit with R = 45 Ω and XL = 1100 Ω, what is the true (real) power drawn from the source?

Introduction:
True power (real power) in AC circuits is the portion of power actually dissipated as heat in resistive elements. In parallel RL circuits, voltage is the same across each branch, so the real power is determined entirely by the resistor branch, while the inductor branch stores and returns energy without net real power consumption.

Given Data / Assumptions:

• Parallel RL circuit with ideal elements.
• R = 45 Ω (resistive branch).
• XL = 1100 Ω (inductive branch).
• Source voltage: V = 24 V(rms).

Concept / Approach:
In a parallel circuit, true power P is P = V^2 / R for the resistive branch. The inductor branch has reactive power only (Q), contributing no average real power. Therefore, total real power equals the resistor's real power.

Step-by-Step Solution:
1) Use P = V^2 / R for the resistive branch.2) Substitute V = 24 V, R = 45 Ω: P = 24^2 / 45.3) Compute 24^2 = 576.4) Compute P = 576 / 45 = 12.8 W.

Verification / Alternative check:
Admittance method: G = 1 / R = 0.02222 S, so I_R = V * G = 24 * 0.02222 ≈ 0.5333 A. Then P = I_R^2 * R ≈ (0.5333)^2 * 45 ≈ 12.8 W, matching the V^2 / R result.

Why Other Options Are Wrong:
313.45 W: far exceeds V^2 / R; impossible with a 24 V source and 45 Ω resistor.44.96 W and 22.3 W: inconsistent with P = V^2 / R and with the computed branch current.

Common Pitfalls:
Including reactive power in real power calculations or mistakenly using series formulas. In parallel circuits, compute real power per branch and sum; inductors contribute zero average real power.

Final Answer:
12.8 W