True power in series RC: A 2 kΩ resistor is in series with a 0.002 µF capacitor across an AC source. If the circuit current is 6.50 mA, what is the true (real) power dissipated?

Introduction / Context:
In AC circuits, true power (in watts) is dissipated only by resistive components. Reactive elements like ideal capacitors and inductors exchange energy with the source but do not consume average power. Therefore, real power in a series RC is determined solely by the resistor using P = I_rms^2 * R.

Given Data / Assumptions:

• R = 2 kΩ = 2000 Ω.
• I_rms = 6.50 mA = 0.0065 A.
• Ideal capacitor (no ESR), sinusoidal steady state.

Concept / Approach:
Apply the resistor power formula directly: P = I^2 * R. The capacitor contributes zero average power, although it affects current and phase. Accurate unit conversion is key to avoid magnitude errors.

Step-by-Step Solution:

Verification / Alternative check:
Cross-check using voltage across the resistor: V_R = I * R = 0.0065 * 2000 = 13 V; then P = V_R * I = 13 * 0.0065 ≈ 0.0845 W = 84.5 mW, confirming the result independently.

Why Other Options Are Wrong:

• 845 mW and 130 mW: Overestimates from factor-of-10 mistakes.
• 13 mW: Underestimate; likely confusion between V_R and P or unit mishandling.

Common Pitfalls:

• Treating reactive components as dissipative; ideal capacitors do not add to true power.

Final Answer:
84.5 mW