# 🌊 AC Phasors and Power in Real Systems

## Learning Objectives

By the end of this lesson you can:

- Convert a sinusoidal voltage or current into a phasor using RMS magnitude and phase angle.
- Write the impedance of a resistor, inductor, and capacitor and combine them with complex Ohm's law.
- Compute real power $P$, reactive power $Q$, apparent power $S$, and power factor for a single load.
- Explain why utilities transmit at high voltage and correct power factor close to unity.

## Quick Review

Last lesson we followed the chain: generation at 11 to 25 kV, step-up transformer, transmission at 115 to 765 kV, substation step-down, distribution at 4 to 35 kV, then the customer. Transformers only change voltage and current levels; the power handed through stays roughly the same.

Two habits from Modules 1 and 2 still apply. Ohm's law and Kirchhoff's laws are unchanged, but every quantity becomes a complex number. And a load that stores energy in a magnetic or electric field behaves very differently from a resistor, which is exactly what the phasor notation lets us track.

## Theory

### From a sine wave to a phasor

In a steady-state AC system every voltage and current has the same frequency, usually 50 or 60 Hz. The frequency therefore carries no information for a given circuit, so we keep only two numbers: magnitude and phase.

$$
v(t) = V_m \cos(\omega t + \phi) \quad \longleftrightarrow \quad \mathbf{V} = V_{rms}\angle \phi
$$

The phasor magnitude is normally the RMS value, because power formulas then come out clean: $V_{rms} = V_m/\sqrt{2}$. A 325 V peak sine is the familiar 230 V RMS mains.

Note the direction of the conversion: a phasor is a static complex number. The rotating $\omega t$ term is understood, not written.

### Impedance: complex Ohm's law

Impedance $Z = R + jX$ plays the role of resistance in AC, and the same algebra you already know works, with $\mathbf{V} = \mathbf{I}Z$.

| Element | Time domain | Impedance $Z$ | Current relative to voltage |
| --- | --- | --- | --- |
| Resistor | $v = Ri$ | $R$ | In phase |
| Inductor | $v = L\,di/dt$ | $j\omega L$ | Lags by 90° |
| Capacitor | $i = C\,dv/dt$ | $1/(j\omega C) = -j/(\omega C)$ | Leads by 90° |

Inductors oppose change in current, so current cannot jump ahead of the driving voltage. Capacitors do the opposite, which is why the two reactances have opposite signs and can cancel in a series branch.

### The three powers

Instantaneous power $p(t) = v(t)i(t)$ pulses at twice the line frequency. For design work we care about its average and its oscillating part:

- Real power $P = V I \cos\varphi$ in watts. This is the part that does useful work.
- Reactive power $Q = V I \sin\varphi$ in VAR. This is energy sloshing between the source and the load's fields.
- Apparent power $S = V I$ in VA, with $S = P + jQ$ and $|S| = VI$.

The angle $\varphi$ is the angle of the impedance, so $\varphi = \angle Z$. A lagging power factor means inductive (current lags voltage, $Q > 0$); a leading power factor means capacitive ($Q < 0$).

> [!NOTE]
> ==Real power P does work; reactive power Q only sloshes back and forth.== Utilities bill for $P$ but must still size cable, transformers, and generators for $S$, which is why they push customers toward power factor near 1.

Two consequences follow directly. Losing cable capacity to reactive current is expensive, so capacitor banks are installed near inductive loads. And for a fixed delivered power, raising the voltage lowers the current, which cuts $I^2R$ line loss quadratically. That single relation is the whole economic argument for the transmission voltages from the previous lesson.

```mermaid
flowchart LR
    A["Time domain: v(t), i(t)"] --> B["Phasors: V, I at RMS"]
    B --> C["Impedance: Z = R + jX"]
    C --> D["Solve V = I Z with KVL and KCL"]
    D --> E["Power: S = P + jQ"]
```

## Worked Examples

**Example 1: series RL load.** A 230 V RMS, 50 Hz source feeds $R = 30\ \Omega$ and $X_L = 40\ \Omega$.

1. $Z = 30 + j40 = 50\angle 53.1^\circ\ \Omega$.
2. $\mathbf{I} = 230\angle 0^\circ / 50\angle 53.1^\circ = 4.6\angle -53.1^\circ$ A.
3. $P = I^2 R = 4.6^2 \times 30 = 635$ W.
4. $Q = I^2 X_L = 4.6^2 \times 40 = 846$ VAR.
5. $S = 230 \times 4.6 = 1058$ VA, power factor $= \cos 53.1^\circ = 0.6$ lagging.

Check: $635^2 + 846^2 = 1.12\times10^6$, and $\sqrt{}$ of that is 1058. The triangle closes.

**Example 2: power factor correction.** A plant draws 10 kW at 0.8 lagging from 400 V, 50 Hz. Target: 0.95 lagging.

1. Original $Q = P \tan\varphi = 10 \times \tan(36.87^\circ) = 7.5$ kVAR.
2. New $Q = 10 \times \tan(18.19^\circ) = 3.29$ kVAR.
3. The capacitor supplies $7.5 - 3.29 = 4.21$ kVAR.
4. $C = Q_C / (\omega V^2) = 4210 / (314 \times 400^2) \approx 84\ \mu\text{F}$.

The real power did not change. Only the current drawn from the utility dropped, from 31.3 A to 26.3 A.

## Guided Practice

Work each one on paper before opening the answer. Keep track of RMS versus peak, and write the angle of every phasor.

**1.** $R = 40\ \Omega$ in series with $X_C = 30\ \Omega$, driven by 100 V RMS. Find $Z$, $\mathbf{I}$, power factor, $P$, and $Q$.

**2.** A motor draws 5 kW at 0.7 lagging from 230 V, 50 Hz. Find $S$, the line current, and $Q$.

**3.** Using the motor from exercise 2, size a capacitor to raise the power factor to 0.9 lagging.

**4.** A single-phase equivalent line has $R = 5\ \Omega$ and must deliver 50 MW at 0.9 power factor. Compare the line loss at 20 kV against 400 kV.

<details><summary>Click to reveal the answers</summary>

**1.** $Z = 40 - j30 = 50\angle -36.9^\circ\ \Omega$, so $\mathbf{I} = 2\angle +36.9^\circ$ A. Current leads. $P = I^2R = 160$ W, $Q = I^2X_C = 120$ VAR delivered by the capacitor (written as $-120$ VAR for the load). Power factor $= 0.8$ leading.

**2.** $S = 5000/0.7 = 7.14$ kVA, so $I = 7140/230 = 31.1$ A. $\varphi = 45.6^\circ$, $Q = 5 \tan 45.6^\circ = 5.10$ kVAR.

**3.** New $Q = 5 \tan(\cos^{-1} 0.9) = 5 \times 0.484 = 2.42$ kVAR. Capacitor supplies $5.10 - 2.42 = 2.68$ kVAR. $C = 2680/(314 \times 230^2) \approx 161\ \mu\text{F}$.

**4.** At 20 kV: $S = 55.6$ MVA, $I = 2778$ A, loss $= I^2R = 38.6$ MW. At 400 kV: $I = 139$ A, loss $= 96$ kW. Same power, roughly 400 times less loss. This is why stepping up matters.

</details>

## Expert Corner

- **Peak versus RMS is the mistake I see most.** Mixing a peak voltage into $P = VI\cos\varphi$ gives you twice the correct power. Pick RMS at the source and stay there.
- **Keep the $j$ in sight.** Reactances add as signed numbers, and forgetting the negative sign on $X_C$ flips your leading/lagging conclusion.
- **Lagging means inductive.** Current lags voltage in a motor or transformer. Lead with capacitors to correct it, never with inductors.
- **Rule of thumb for correction:** each 0.01 of power factor improvement near unity costs little capacitance but saves real cable capacity. Below 0.85 the gains get large and cheap.
- **Check with the triangle.** $S^2 = P^2 + Q^2$ catches almost every arithmetic slip in a power problem.

## Summary and Next Steps

A phasor keeps magnitude and phase and discards a frequency you already know. Impedance extends Ohm's law into AC, reactance signs tell you lag or lead, and the power triangle separates the work you sell from the current you must carry.

Cloze check on the essentials:

- A phasor stores the {{c1::magnitude and phase}} of a sinusoid, not its frequency.
- For a peak value $V_m$, the RMS value is {{c2::V_m/\sqrt{2}}}.
- Real power is $P = $ {{c3::VI \cos\varphi}}, and reactive power is $Q = $ {{c4::VI \sin\varphi}}.
- The impedance of a capacitor is {{c5::1/(j\omega C)}}, which is negative imaginary.
- Power factor is defined as {{c6::cos\varphi}}, the cosine of the impedance angle.

```flashcard
[
  { "front": "What two quantities does a phasor keep?", "back": "Magnitude (usually RMS) and phase angle; the frequency is implicit." },
  { "front": "Relationship between peak and RMS for a sine wave?", "back": "V_rms = V_m / sqrt(2), about 0.707 times the peak." },
  { "front": "Impedance of an inductor?", "back": "Z = j times omega times L, positive imaginary, current lags by 90 degrees." },
  { "front": "Impedance of a capacitor?", "back": "Z = 1/(j omega C) = -j/(omega C), negative imaginary, current leads by 90 degrees." },
  { "front": "Complex form of Ohm's law for AC?", "back": "V phasor = I phasor times Z, with Z = R + jX." },
  { "front": "Formula for real power P?", "back": "P = V I cos(phi) in watts, where phi is the impedance angle." },
  { "front": "Formula for reactive power Q?", "back": "Q = V I sin(phi) in VAR." },
  { "front": "Relationship between S, P, and Q?", "back": "S = P + jQ, so |S| = sqrt(P^2 + Q^2) and |S| = V I." },
  { "front": "What does a lagging power factor indicate?", "back": "An inductive load: current lags voltage, so Q is positive." },
  { "front": "How is power factor defined?", "back": "Power factor = cos(phi) = P / S." },
  { "front": "How do you correct a lagging power factor?", "back": "Add a capacitor bank that supplies the excess reactive power locally." },
  { "front": "Why does high-voltage transmission reduce losses?", "back": "For the same power, higher voltage means lower current, and loss is I^2 R." },
  { "front": "What does the power triangle check?", "back": "That S^2 = P^2 + Q^2 holds after your calculation." },
  { "front": "How does instantaneous power vary in an AC circuit?", "back": "It pulses at twice the line frequency; P and Q describe its average and oscillating parts." }
]
```

Next up is the module checkpoint, where you will combine the generation-to-distribution chain with these phasor and power calculations in exam-style problems. Rework exercises 1 and 3 once without notes before you start.