Chapter 10 — AC Power Analysis
Chapter Overview (click to expand)
AC power analysis distinguishes between three types of power — real, reactive, and apparent — that together characterize how efficiently energy is delivered to a load. This chapter develops the mathematics of complex power and the power triangle, and explains power factor correction, an essential technique for reducing wasted energy in electrical systems. **Key Takeaways** 1. Real (average) power P, measured in watts, represents energy actually consumed; reactive power Q, measured in VARs, represents energy oscillating back and forth between source and reactive elements. 2. Apparent power S is the product of RMS voltage and current, and equals the magnitude of complex power S = P + jQ. 3. Power factor (cos θ) measures how effectively a load uses apparent power; capacitor banks are added to correct a lagging power factor and reduce transmission losses.10.1 Instantaneous Power
Instantaneous power is the power at any specific instant:
For sinusoidal signals: (p(t) = \dfrac{V_m I_m}{2}\cos(\theta_v - \theta_i) + \dfrac{V_m I_m}{2}\cos(2\omega t + \theta_v + \theta_i)]
This reveals two components: a constant term (the average power) and an oscillating term at twice the signal frequency. The oscillating component means power flows back and forth between the source and reactive elements every quarter cycle — it can actually go negative, meaning energy flows back to the source.
10.2 Real Power (Average Power)
Real power (also called average power or active power) is the time-averaged power over one complete cycle. It represents the power that actually does useful work — heating elements, turning motors, producing light and sound.
Measured in watts (W). \(\theta = \theta_v - \theta_i\) is the phase angle between voltage and current.
The factor \(\cos\theta\) is the power factor — it determines how much of the apparent power does useful work. Your electric bill is based on real power (kilowatt-hours).
| Phase Angle θ | cos θ | Interpretation |
|---|---|---|
| 0° | 1.00 | Pure resistive — all power is real |
| 30° | 0.866 | Moderate reactive component |
| 60° | 0.50 | Mostly reactive |
| 90° | 0.00 | Pure reactive — zero real power |
Diagram: RMS and Average Power Visualizer
10.3 Reactive Power
Reactive power (Q) represents energy oscillating between the source and reactive elements without being consumed. It's measured in volt-amperes reactive (VAR).
Inductive loads: \(Q > 0\) (lagging). Capacitive loads: \(Q < 0\) (leading).
Energy Storage vs. Dissipation
The distinction between real and reactive power comes down to what happens to the energy each half-cycle.
In a resistor, energy flows in one direction only — from source to resistor — and is permanently converted to heat. The instantaneous power \(p_R(t) = v^2(t)/R\) is always non-negative, and its average (real power P) is positive.
In an inductor, energy is stored in the magnetic field during the first quarter-cycle (when current is increasing), then returned to the source during the second quarter-cycle (when current decreases). The net energy delivered over a full cycle is zero. This "back-and-forth" energy is reactive power.
In a capacitor, energy is stored in the electric field while the capacitor charges, then returned as it discharges. Again, the net energy per cycle is zero.
Sign Convention
The sign of Q tells you the direction of reactive energy flow:
- \(Q > 0\) (positive, inductive): The load absorbs VARs from the source. Motors, transformers, and inductors all have positive Q. The current lags the voltage.
- \(Q < 0\) (negative, capacitive): The load supplies VARs to the circuit. Capacitors generate reactive power. The current leads the voltage.
This sign convention is why capacitors are used for power factor correction — they supply the VARs that inductive loads demand, so the utility doesn't have to.
Why Reactive Power Matters Practically
Reactive power doesn't do useful work, but it still requires current to flow. Consider a motor with P = 800 W and Q = 600 VAR:
The utility must supply 8.33 A even though the motor only needs \(800/120 = 6.67\) A worth of useful current. The extra current causes real losses in transmission lines (\(I^2 R_{line}\)) — losses that the utility pays for but the customer's watt-hour meter never registers. This is why utilities charge large industrial customers a power factor penalty and why power factor correction saves money.
10.4 Apparent Power
Apparent power (S) is the product of RMS voltage and RMS current — the total "power handling" demanded from the source. Measured in volt-amperes (VA).
Apparent power determines wire sizing, transformer ratings, generator capacity, and circuit breaker ratings — equipment must handle the full current regardless of power factor.
Example: A motor draws 10 A from a 120 V source at PF = 0.8: - \(S = 120 \times 10 = 1{,}200\) VA - \(P = 1{,}200 \times 0.8 = 960\) W - \(Q = 1{,}200 \times 0.6 = 720\) VAR
The motor does only 960 W of work, but the supply must have capacity for 1,200 VA.
10.5 Complex Power
Complex power combines real and reactive power into a single complex number:
where \(\mathbf{I}^*\) is the complex conjugate of the current phasor.
Phasor Derivation of \(\mathbf{S} = \mathbf{V}\mathbf{I}^*\)
Let the voltage and current phasors be:
The complex conjugate of the current phasor simply negates the angle:
Multiplying:
Let \(\theta = \theta_v - \theta_i\) (the impedance angle). Using RMS values (\(V_{rms} = V_m/\sqrt{2}\), \(I_{rms} = I_m/\sqrt{2}\)):
This confirms: \(P = V_{rms}I_{rms}\cos\theta\) and \(Q = V_{rms}I_{rms}\sin\theta\).
Why the Conjugate?
We use \(\mathbf{I}^*\) (not \(\mathbf{I}\)) so that P and Q have the correct physical signs. Using \(\mathbf{V}\mathbf{I}\) instead would swap inductive and capacitive signs.
Worked Example: Computing Complex Power from Phasors
Problem: A load has voltage phasor \(\mathbf{V} = 120\angle 0°\) V (rms) and draws current phasor \(\mathbf{I} = 6\angle -36.87°\) A (rms). Find \(\mathbf{S}\), P, Q, and the power factor.
Step 1 — Complex conjugate of current:
Step 2 — Complex power:
Step 3 — Convert to rectangular form:
Step 4 — Read off results:
Verification: \(\sqrt{576^2 + 432^2} = \sqrt{331776 + 186624} = \sqrt{518400} = 720\) VA ✓
| Quantity | Symbol | Units | Formula |
|---|---|---|---|
| Real Power | \(P\) | W | \(V I \cos\theta\) |
| Reactive Power | \(Q\) | VAR | \(V I \sin\theta\) |
| Apparent Power | \(S\) | VA | \(VI = \|\mathbf{S}\|\) |
| Complex Power | \(\mathbf{S}\) | VA | \(P + jQ = \mathbf{V}\mathbf{I}^*\) |
Use the phasor circuit solver to compute complex power S = VI* for different load impedances.
10.6 The Power Triangle
The power triangle shows the geometric relationship between P, Q, and S — they form a right triangle:
- Hypotenuse: Apparent power S (VA)
- Horizontal leg: Real power P (W)
- Vertical leg: Reactive power Q (VAR)
- Angle θ: Phase angle between voltage and current
The power triangle is identical in shape to the impedance triangle (\(|Z|\), R, X).
Diagram: Power Triangle Explorer
10.7 Power Factor: Leading and Lagging
Power factor (PF) = \(\cos\theta = P/S\). It ranges from 0 to 1 and measures how efficiently apparent power converts to real work.
Lagging power factor — current lags voltage (inductive loads: motors, transformers, ballasts): - \(\theta > 0\), \(Q > 0\), load absorbs VARs from source
Leading power factor — current leads voltage (capacitive loads: correction capacitors, some power supplies): - \(\theta < 0\), \(Q < 0\), load supplies VARs to circuit
| Power Factor | Description | Typical Load |
|---|---|---|
| 1.0 | Unity | Resistive heaters, incandescent bulbs |
| 0.95 lagging | Excellent | Well-corrected industrial motors |
| 0.85 lagging | Good | Typical industrial mix |
| 0.70 lagging | Poor | Uncompensated motors |
| 0.50 lagging | Very poor | Lightly loaded motors |
The impedance triangle below shows the geometric relationship between R, X, Z, and power factor angle θ.
10.8 Power Factor Correction
Power factor correction improves PF by adding capacitors in parallel with inductive loads. The capacitors supply reactive power locally, reducing the reactive current drawn from the source.
Benefits: reduced line current, less I²R loss, increased infrastructure capacity, avoided utility penalties.
Derivation of Required Capacitor Size
The goal is to add a capacitor whose reactive power \(Q_C\) exactly cancels the excess reactive power of the load.
Starting from the power triangle: the load has real power P and reactive power \(Q_1 = P\tan\theta_1\). The corrected system should have the same P but reduced reactive power \(Q_2 = P\tan\theta_2\):
This \(\Delta Q\) must be supplied by the correction capacitor. The reactive power of a capacitor connected across voltage \(V_{rms}\) at angular frequency \(\omega\) is:
Setting \(|Q_C| = \Delta Q\) and solving for C:
Worked Example: Full Power Factor Correction Design
Problem: A factory load draws P = 10 kW at PF = 0.70 lagging from a 240 V, 60 Hz supply. The utility requires PF ≥ 0.95. Find the capacitor needed and the resulting line current reduction.
Step 1 — Characterize the existing load:
Step 2 — Target reactive power at corrected PF:
Step 3 — Required capacitor reactive power:
Step 4 — Capacitor value:
Use a standard 320 μF capacitor.
Step 5 — Verify reduced line current:
Current reduction: \((59.5 - 43.9)/59.5 = 26\%\) less current from the utility.
Line loss reduction: \(I^2 R\) losses drop by \(1 - (43.9/59.5)^2 = 1 - 0.544 = 45.6\%\) — nearly half the wasted power in the cables is eliminated.
10.9 Power in Resistors, Capacitors, and Inductors
Each passive component has distinct power behavior:
Resistors — voltage and current are in phase (\(\theta = 0\)):
Capacitors — current leads voltage by 90°, no real power consumed:
Inductors — current lags voltage by 90°, no real power consumed:
| Element | Real Power P | Reactive Power Q | PF |
|---|---|---|---|
| Resistor | \(I^2 R\) | 0 | 1 |
| Capacitor | 0 | \(-V^2/X_C\) | 0 leading |
| Inductor | 0 | \(+V^2/X_L\) | 0 lagging |
Use the simulation below to explore how real and reactive power change as you vary the component type, resistance, reactance, and supply voltage:
10.10 Maximum Power Transfer in AC
For maximum power transfer from a source with internal impedance \(Z_s = R_s + jX_s\) to load \(Z_L\):
The load impedance must be the complex conjugate of the source impedance.
At conjugate matching: \(R_L = R_s\) (equal resistances) and \(X_L = -X_s\) (reactances cancel). The result is resonance — maximum current and maximum power transfer — but efficiency is only 50%. This trade-off is acceptable in RF and audio signal systems; power delivery systems optimize for efficiency instead.
10.11 Efficiency and Power Gain
Efficiency:
| Device | Typical Efficiency |
|---|---|
| Power transformer | 95–99% |
| AC induction motor | 80–95% |
| LED bulb | 35–50% |
| Incandescent bulb | 2–5% |
| Switching power supply | 85–95% |
Power gain measures amplification:
10.12 Worked Example: Complete Power Analysis
Problem: A series RL load with R = 30 Ω and \(X_L = 40\) Ω is connected to a 120 V RMS, 60 Hz source. Find all power quantities and the capacitor needed for unity PF correction.
Step 1 — Impedance and current:
\(Z = 30 + j40 = 50\angle 53.13°\) Ω \(\quad I_{rms} = 120/50 = 2.4\) A
Step 2 — Powers:
[P = I^2 R = (2.4)^2(30) = 172.8 \text{ W}] [Q = I^2 X_L = (2.4)^2(40) = 230.4 \text{ VAR}] [S = I^2|Z| = (2.4)^2(50) = 288 \text{ VA} \qquad PF = P/S = 0.6 \text{ lagging}]
Step 3 — Power factor correction to unity:
\(Q_C = -230.4\) VAR needed. \(C = 230.4 / (2\pi \times 60 \times 120^2) = 42.5\) μF
After correction: current drops from 2.4 A to \(172.8/120 = 1.44\) A — a 40% reduction!
10.13 Chapter Summary
Complete AC power quantity reference:
| Quantity | Symbol | Units | Formula | Notes |
|---|---|---|---|---|
| Instantaneous power | \(p(t)\) | W | \(v(t)\cdot i(t)\) | Varies at 2ω |
| Real (active) power | \(P\) | W | \(V_{rms} I_{rms}\cos\theta\) | Useful work; billed by utility |
| Reactive power | \(Q\) | VAR | \(V_{rms} I_{rms}\sin\theta\) | \(Q>0\) inductive; \(Q<0\) capacitive |
| Apparent power | \(S\) | VA | (V_{rms} I_{rms} = | \mathbf{S} |
| Complex power | \(\mathbf{S}\) | VA | \(P + jQ = \mathbf{V}\mathbf{I}^*\) | Phasor computation |
| Power factor | PF | — | \(\cos\theta = P/S\) | 0 to 1; lagging or leading |
Power triangle relationship:
Per-element power behavior:
| Element | P | Q | PF |
|---|---|---|---|
| Resistor | \(I^2 R > 0\) | 0 | 1 (unity) |
| Inductor | 0 | \(I^2 X_L > 0\) | 0 lagging |
| Capacitor | 0 | \(-V^2/X_C < 0\) | 0 leading |
Power factor correction: - Add capacitor in parallel with inductive load - Required VAR: \(Q_C = P(\tan\theta_1 - \tan\theta_2)\) - Required capacitance: \(C = Q_C / (\omega V_{rms}^2)\) - Result: reduced line current, lower \(I^2R\) losses, no change in real power delivered
AC maximum power transfer: \(Z_L = Z_s^*\) (conjugate matching) - \(R_L = R_s\), \(X_L = -X_s\) (reactances cancel, creating resonance) - Maximum real power: \(P_{max} = |V_s|^2/(4R_s)\) - Efficiency: 50% — acceptable for signal applications, not for power delivery
Key sign rules: - \(\theta = \theta_v - \theta_i\): positive when current lags (inductive/lagging) - Complex power uses \(\mathbf{I}^*\) (conjugate), not \(\mathbf{I}\), to get correct P and Q signs - Utilities penalize PF below ~0.90–0.95 for large industrial customers