AC Power Fundamentals

Concept

In alternating current (AC) circuits, voltage and current vary sinusoidally and may be out of phase. Because of this phase difference caused by inductive and capacitive loads, power is categorized into three components:

1. Real Power (Active Power): The time-average power that actually performs work or is dissipated as heat.
2. Reactive Power: The power that oscillating between the source and reactive components (inductors and capacitors), doing no net work.
3. Apparent Power: The complex combination of real and reactive power, representing the total power capacity the electrical system must handle.

The ratio of real power to apparent power is known as the Power Factor (PF).

Formula & Method

Using the RMS (root-mean-square) values of voltage and current, and the phase angle difference $\theta = \theta_v - \theta_i$:

• Real Power: $P = V_{rms} I_{rms} \cos(\theta)$
• Reactive Power: $Q = V_{rms} I_{rms} \sin(\theta)$
• Apparent Power: $S = V_{rms} I_{rms}$
• Power Factor: $\text{PF} = \cos(\theta) = \frac{P}{S}$

The relationship can be visualized as a power triangle where $S = \sqrt{P^2 + Q^2}$.

Variables & Units

• $P$ = Real power, expressed in Watts (W).
• $Q$ = Reactive power, expressed in Volt-Amperes Reactive (VAR).
• $S$ = Apparent power, expressed in Volt-Amperes (VA).
• $V_{rms}, I_{rms}$ = Root-mean-square voltage (V) and current (A).
• $\theta$ = Phase angle difference between voltage and current.
• $\text{PF}$ = Power factor, a dimensionless number between 0 and 1.

Worked Example

Problem: A single-phase AC motor operates at $230 \text{ V}_{rms}$ and draws a current of $10 \text{ A}_{rms}$. The current lags the voltage by a phase angle of $30^\circ$. Calculate the apparent power, real power, reactive power, and power factor.

Calculation:

1. Identify variables: $V_{rms} = 230 \text{ V}$, $I_{rms} = 10 \text{ A}$, $\theta = 30^\circ$.
2. Apparent Power: $S = (230)(10) = 2300 \text{ VA} = 2.3 \text{ kVA}$
3. Real Power: $P = 2300 \cos(30^\circ) \approx 2300(0.866) = 1991.8 \text{ W}$
4. Reactive Power: $Q = 2300 \sin(30^\circ) = 2300(0.5) = 1150 \text{ VAR}$
5. Power Factor: $\text{PF} = \cos(30^\circ) = 0.866 \text{ (lagging)}$

Engineering Meaning

Power factor is a key indicator of energy efficiency in electrical systems. A low power factor means the system draws more current than necessary to perform a given amount of real work, which increases $I^2R$ losses in transmission lines and necessitates larger transformers. Engineers use capacitor banks to correct low lagging power factors.

Engineering Check

Ensure that calculations use RMS values for voltage and current, not peak values. Distinguish clearly between leading (capacitive) and lagging (inductive) power factors; the sign of $Q$ is traditionally positive for inductive loads and negative for capacitive loads.

Explicit Exclusions

This article covers single-phase AC power fundamentals. It explicitly excludes three-phase power systems, complex power notation ($S = P + jQ$), harmonic distortion analysis, and detailed power factor correction sizing.\n