In an AC circuit, the reactive power is 50 mW and the apparent power is 64 mW. What is the true (real) power delivered to the load?

Introduction:Power in AC circuits separates into real (true) power P, reactive power Q, and apparent power S. The power triangle relation S^2 = P^2 + Q^2 enables computation of any one when the other two are known.

Given Data / Assumptions:

• Reactive power Q = 50 mW.
• Apparent power S = 64 mW.
• Single-frequency steady state with sinusoidal quantities.

Concept / Approach:Use the power triangle identity. Solve for true power: P = sqrt( S^2 - Q^2 ). This represents the average power that performs useful work and is dissipated as heat in resistive elements.

Step-by-Step Solution:1) Compute S^2 = 64^2 = 4096 (mW)^2.2) Compute Q^2 = 50^2 = 2500 (mW)^2.3) Subtract: 4096 - 2500 = 1596.4) Take square root: P = sqrt(1596) ≈ 39.95 mW ≈ 40 mW.

Verification / Alternative check:Power factor cos(phi) = P / S ≈ 40 / 64 = 0.625. Then Q = S * sin(phi) ≈ 64 * sqrt(1 - 0.625^2) ≈ 64 * 0.781 ≈ 50 mW, which matches the given data.

Why Other Options Are Wrong:14 mW and 36 mW: too small; do not satisfy S^2 = P^2 + Q^2.114 mW: exceeds S, impossible since S ≥ P by definition and S cannot be less than P.64 mW: equals apparent power, not true power, and would imply zero reactive power.

Common Pitfalls:Confusing S with P, ignoring units, or attempting arithmetic addition P + Q to get S. The correct relationship is quadratic, not linear.

Final Answer:40 mW