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

The Core Concept: Why a Resistor Is an Example of True Power

In alternating current (AC) theory, power is split into three categories: true power (Watts), reactive power (VAR), and apparent power (VA). The statement that a resistor is an example of true power stems from the phase relationship between voltage and current. In a purely resistive component, voltage and current are perfectly in phase (a phase angle of 0°). Because the power factor ($\cos \theta$) is exactly 1, 100% of the power delivered to the resistor is converted into work—typically heat or light. There is no energy storage and return cycle, which is the hallmark of reactive components like inductors and capacitors.

When you are calculating true power ($P$) in an AC circuit, you must isolate the resistive elements. Inductors and capacitors merely borrow and return energy to the source, contributing zero net true power over a full AC cycle. According to All About Circuits, true power is the only power that your utility meter bills you for, making it the most critical metric for both exam problems and real-world electrical design.

Practice Problem: True Power in a Series RL Circuit

Step-by-Step Algebraic Solution & Sanity Check

To solve this, we apply AC Ohm's Law and the true power formula. We must find the total impedance first, then the circuit current, and finally the power dissipated specifically by the resistive element.

Step 1: Calculate Inductive Reactance ($X_L$)
The formula for inductive reactance is $X_L = 2\pi fL$.
$X_L = 2 \times \pi \times 60 \text{ Hz} \times 0.05305 \text{ H}$
$X_L = 376.99 \times 0.05305$
$X_L \approx 20 \, \Omega$

Step 2: Calculate Total Impedance ($Z$)
In a series RL circuit, impedance is the vector sum of resistance and reactance: $Z = \sqrt{R^2 + X_L^2}$.
$Z = \sqrt{20^2 + 20^2}$
$Z = \sqrt{400 + 400}$
$Z = \sqrt{800} \approx 28.284 \, \Omega$

Step 3: Calculate RMS Circuit Current ($I$)
Using Ohm's Law for AC: $I = V / Z$.
$I = 120 \text{ V} / 28.284 \, \Omega$
$I \approx 4.2426 \text{ A}$

Step 4: Calculate True Power ($P$)
True power is only dissipated by the resistor. The formula is $P = I^2R$.
$P = (4.2426)^2 \times 20$
$P = 18.00 \times 20$
$P = 360 \text{ W}$

The Exam Trap & Independent Verification

The Trap: The most common mistake students make on this problem is using the source voltage in the DC power formula: $P = V^2 / R$. If you plug in the 120V source voltage, you get $120^2 / 20 = 720 \text{ W}$. This is entirely wrong because the inductor also drops voltage. The voltage across the resistor is not 120V; it is $I \times R = 4.2426 \times 20 = 84.85 \text{ V}$. If you use the correct resistor voltage drop: $84.85^2 / 20 = 360 \text{ W}$.

Independent Verification: You can verify your answer using the Power Triangle method. Calculate the Power Factor ($PF$) first:
$PF = \cos(\theta) = R / Z = 20 / 28.284 = 0.7071$
Next, multiply Apparent Power ($S$) by the Power Factor:
$P = S \times PF = 509.11 \text{ VA} \times 0.7071 = 360 \text{ W}$.
The math aligns perfectly.

Component Selection Decision Tree: Sizing the Resistor

Calculating 360W on paper is only half the battle. If you are building this circuit, you cannot just buy a generic 360W resistor. Power resistors must be derated for thermal management. According to Ohmite Manufacturing's derating guidelines, chassis-mount wirewound resistors should typically be derated to 50% of their nominal rating when operating in an enclosed space without forced air cooling to prevent the casing from exceeding safe touch temperatures (usually 200°C+ at full load).

Condition / Requirement	If True, Choose...	Concrete Part / Spec
$P_{dissipated} < 5\text{W}$	Standard carbon/metal film through-hole	Generic 1/4W or 1/2W axial
$5\text{W} < P_{dissipated} < 50\text{W}$	Ceramic cement or aluminum-housed wirewound	Ohmite 40F series (e.g., 43F20R0)
$P_{dissipated} > 50\text{W}$ (Our case: 360W)	Heavy-duty chassis mount, derated 50% (Need 720W total capacity)	See final pick below
FINAL PICK: 360W Load at 20Ω	Two 500W wirewounds in series for redundancy and thermal spreading	Two Ohmite 270-500-10 (10Ω, 500W each)

The Concrete Pick: To handle a 360W true power load safely, purchase two Ohmite 270-series 500W wirewound resistors (Part: 270-500-10). Wire them in series. This yields your required $20 \, \Omega$ resistance while providing a combined thermal capacity of 1000W. Derated to 50%, the array can safely dissipate 500W continuously in a stagnant air enclosure, keeping your 360W load well within the safe operating area.

FAQ: True Power Exam Gotchas

Q: Can a circuit have true power if it only contains capacitors?
A: No. Ideal capacitors only store and release energy, resulting in purely reactive power (VAR). True power requires a resistive element to convert electrical energy into heat or work. In the real world, capacitors have Equivalent Series Resistance (ESR), which dissipates a tiny amount of true power, but in exam theory, ideal capacitors dissipate zero true power.

Q: Why do we use RMS voltage for true power calculations instead of peak voltage? A: RMS (Root Mean Square) is the DC-equivalent heating value of an AC waveform. True power is fundamentally a measure of heating/work over time. If you use peak voltage ($V_p = 170\text{V}$ for a 120V RMS system) without adjusting your formulas, your power calculation will be off by a factor of 2. Always default to RMS values unless the problem explicitly states otherwise.

Q: Does the frequency of the AC source affect the true power in a pure resistor?
A: In theory, no. A pure resistor's true power dissipation ($P = I^2R$) is independent of frequency. However, frequency dictates the reactance of series/parallel inductors and capacitors, which changes the total circuit impedance, thereby altering the total current ($I$) that flows through the resistor. So, while the resistor's physics don't change, the circuit's current does.