In a 30 V series RLC circuit, the total impedance is Z = 20 Ω and the resistance is R = 10 Ω. What is the true (real) power consumed by the circuit under these conditions?

Introduction:
True power in AC circuits refers to the real power dissipated as heat in resistive elements. In a series RLC circuit, reactive components (inductor and capacitor) exchange energy with the source but do not consume net real power over a cycle. This problem tests using impedance, current, and power relationships to compute real power.

Given Data / Assumptions:

• Source RMS voltage: V = 30 V.
• Total series impedance magnitude: Z = 20 Ω.
• Resistive part: R = 10 Ω.
• Ideal sinusoidal steady state; reactive elements consume no average real power.

Concept / Approach:
Key formulas: current magnitude I = V / Z. Real (true) power P is given by P = I^2 * R = V * I * cos(phi). In a series RLC, cos(phi) = R / Z, so an equivalent single-step form is P = V^2 * (R / Z^2). Any of these consistent formulas will yield the same result.

Step-by-Step Solution:
1) Compute current: I = V / Z = 30 / 20 = 1.5 A.2) Use P = I^2 * R = (1.5)^2 * 10 = 2.25 * 10 = 22.5 W.3) Cross-check with P = V^2 * (R / Z^2) = 30^2 * (10 / 20^2) = 900 * (10 / 400) = 900 * 0.025 = 22.5 W.

Verification / Alternative check:
Power factor cos(phi) = R / Z = 10 / 20 = 0.5. Then P = V * I * cos(phi) = 30 * 1.5 * 0.5 = 22.5 W, confirming consistency among methods.

Why Other Options Are Wrong:
15.0 watts: would require a smaller current or resistance; not supported by the given numbers.30.0 watts: equals V * I without power factor; ignores that cos(phi) < 1.45.0 watts: equals V * I with cos(phi) = 1; impossible because reactive components make cos(phi) < 1.

Common Pitfalls:
Forgetting to include power factor or mixing instantaneous values with RMS. Another frequent error is using Z instead of R in P = I^2 * R; only the resistive part dissipates real power.

Final Answer:
22.5 watts