Exam Walkthrough: Why a Resistor Is an Example of True Power in AC Circuits

In AC circuit analysis, power is not a single monolithic value. It splits into true power (Watts), reactive power (VARs), and apparent power (Volt-Amps). When textbooks and exam rubrics state that a resistor is an example of true power consumption, they are highlighting a fundamental physical reality: resistors convert electrical energy irreversibly into heat, unlike inductors or capacitors which temporarily store and release energy in magnetic or electric fields.

This walkthrough breaks down a classic exam problem involving a series RL circuit. We will calculate the exact true power dissipated by the resistor, show every algebraic step, expose the most common trap students fall into, and verify the result using an independent method.

The Core Concept: Phase Angles and True Power

True power ($P$), measured in Watts, represents the actual work done or heat generated by a circuit. The general formula for true power in an AC system is:

P = V_rms × I_rms × cos(θ)

Where θ is the phase angle difference between the voltage and current waveforms. In a purely resistive component, voltage and current are perfectly in phase. Therefore, θ = 0°, and cos(0°) = 1. The equation simplifies to P = V_rms × I_rms, or equivalently, P = I_rms² × R. This unity power factor is the mathematical reason why a resistor only consumes true power and zero reactive power. For a deeper theoretical foundation, refer to the Georgia State University HyperPhysics module on AC Power.

Practice Problem: Calculating True Power in a Series RL Circuit

Method Applied: We must use AC Ohm's Law combined with the Impedance Triangle. We cannot use DC power formulas directly on the source voltage because the inductor introduces frequency-dependent reactance, meaning the source voltage is divided between the resistor and the inductor.

Step-by-Step Algebraic Solution

Here is the complete, un-skipped algebraic path to the solution.

1. Calculate Inductive Reactance ($X_L$):
   The inductor opposes AC current flow based on frequency. The formula is $X_L = 2πfL$.
   X_L = 2 × π × 60 Hz × 0.1061 H
X_L = 376.99 × 0.1061
X_L = 40.0 Ω
2. Calculate Total Circuit Impedance ($Z$):
   In a series RL circuit, resistance and reactance are 90° out of phase, so we use the Pythagorean theorem.
   Z = √(R² + X_L²)
Z = √(40² + 40²)
Z = √(1600 + 1600)
Z = √(3200)
Z ≈ 56.57 Ω
3. Calculate Total RMS Current ($I_{rms}$):
   Using AC Ohm's Law ($I = V / Z$).
   I_{rms} = 120V / 56.57 Ω
I_{rms} ≈ 2.121 A
   Note: In a series circuit, this current flows equally through both the resistor and the inductor.
4. Calculate True Power ($P$) Dissipated by the Resistor:
   Since only the resistor dissipates true power, we apply the power formula strictly to the resistive component using the series current.
   P = I_{rms}² × R
P = (2.121 A)² × 40 Ω
P = 4.4986 × 40
P ≈ 180 W

The true power dissipated by the resistor is 180 Watts.

Sanity Check and Independent Verification

Before moving to the next exam question, always verify your answer using an independent mathematical path. This catches calculator typos and conceptual errors.

The Trap: What Students Get Wrong

The most common failure point on this specific exam question is blindly applying the DC power formula P = V² / R using the source voltage. If you calculate 120² / 40, you get 360 W. This is physically impossible in this circuit because the inductor drops a significant portion of the source voltage. The 120V is not applied entirely across the resistor; the actual voltage drop across the resistor is only V_R = I × R = 2.121A × 40Ω = 84.84V. If you use 84.84V in the formula (84.84² / 40), you correctly arrive back at 180 W.

Frequently Asked Questions

Is a resistor the only component that consumes true power?

In ideal textbook theory, yes. However, in real-world bench practice, every component consumes some true power. Inductors have wire resistance (DCR), and capacitors have Equivalent Series Resistance (ESR). When you measure a real inductor warming up under load, you are measuring the true power dissipated by its internal parasitic resistance. For more on real-world component parasitics, see the Electronics Tutorials guide on AC Power and Waveforms.

How does true power differ from apparent power in a purely resistive circuit?

In a purely resistive AC circuit (no inductors or capacitors), the phase angle is zero, meaning the power factor is exactly 1. Consequently, true power (Watts) and apparent power (Volt-Amps) are numerically identical. The distinction only matters when reactive components are introduced, causing the current waveform to lead or lag the voltage waveform.

Why do we say a resistor is an example of true power rather than reactive power?

Reactive power represents energy that sloshes back and forth between the source and the load's magnetic or electric fields, doing no net work over a full AC cycle. A resistor lacks the physical geometry to store energy in fields; it immediately scatters electron kinetic energy into lattice vibrations (heat). Because this energy transfer is unidirectional and irreversible, it is classified strictly as true power.

Can a wirewound power resistor ever exhibit reactive power?

Yes, at high frequencies. A standard wirewound resistor (like a 50W aluminum-housed chassis mount) is literally a coil of nichrome wire. At 60 Hz, its parasitic inductance is negligible. But if you use that same resistor in a 100 kHz switching power supply snubber circuit, the coil's inductance creates a measurable reactance. In high-frequency RF or switching applications, engineers specifically select 'non-inductive' resistors (like carbon composition or metal film) to ensure the component remains a pure example of true power dissipation without introducing phase shift.