# 4.6 Transmission of Electrical Energy

F.6 Physics · Topic IV Electricity and Magnetism · Lesson 1

Alternating current, r.m.s. values, how a transformer works, and why power is sent across the country at very high voltage.

## Key points

- **d.c.** flows in one direction (steady or varying); **a.c.** reverses direction periodically. Hong Kong mains: sinusoidal, **50 Hz**, **220 V r.m.s.**
- Sinusoidal a.c.: I = I~0~ sin ωt, with T = 1/f and ω = 2πf. I~0~ is the **peak** (amplitude), I the instantaneous value.
- **r.m.s. value ／ 均方根值** = the steady d.c. that gives the **same average power** (heating effect) in the same resistor. Sinusoidal only: I~rms~ = I~0~/√2.
- Average power in a resistor P̄ = I~rms~^2^R = V~rms~I~rms~. Peak instantaneous power = 2P̄ (sinusoidal).
- **Transformer ／ 變壓器** works by **electromagnetic induction**, so it needs a changing current (a.c.). V~P~/V~S~ = N~P~/N~S~; if ideal, V~P~I~P~ = V~S~I~S~.
- Power losses and cures: coil resistance → thick copper wire; **eddy currents ／ 渦電流** → laminated core; flux leakage → closed core; hysteresis → soft iron core.
- Transmission: cable current I = P/V, loss P~loss~ = I^2^R~cable~. Stepping V up n times cuts I to 1/n and the loss to 1/n^2^.

## Must know

| Quantity | Formula |
|---|---|
| Period, frequency, angular frequency | T = 1/f · ω = 2πf (rad s^−1^) |
| r.m.s. of a sinusoid | I~rms~ = I~0~/√2 ≈ 0.707 I~0~ · V~rms~ = V~0~/√2 |
| Power in a resistor | P̄ = I~rms~^2^R = V~rms~^2^/R = V~rms~I~rms~ |
| Transformer | V~P~/V~S~ = N~P~/N~S~ · ideal: V~P~I~P~ = V~S~I~S~ |
| Efficiency | η = P~out~/P~in~ = V~S~I~S~/(V~P~I~P~) |
| Cable loss | I = P/V first, then P~loss~ = I^2^R~cable~ |

Meters and labels: an a.c. voltmeter or ammeter reads the **r.m.s.** value, and the voltage marked on a supply is r.m.s.

## Common mistakes

- ✗ Using V~rms~ = V~0~/√2 for a square wave. ✓ The /√2 rule is for **sinusoidal** a.c. only. Otherwise square the values, average over one cycle, then take the square root.
- ✗ "220 V mains means the peak is 220 V." ✓ Labels and meter readings are r.m.s. Peak = 220 × √2 ≈ 311 V.
- ✗ "A step-up transformer gives more power." ✓ It changes V and I, not power: P~out~ ≤ P~in~ (equal only if ideal). And it needs a.c.: steady d.c. gives no output.
- ✗ P~loss~ = V^2^/R with V = 132 kV. ✓ The transmission voltage is not the p.d. across the cable. Find I = P/V first, then P~loss~ = I^2^R.
- ✗ "High voltage cuts the loss because the cable resistance drops." ✓ R is unchanged. The **current** drops, and the loss is I^2^R.

## Quick check

1. A sinusoidal supply has a peak voltage of 12 V. What does an a.c. voltmeter across it read?
2. An ideal transformer has 1000 primary turns and 50 secondary turns and is connected to the 220 V mains. Find the secondary voltage.
3. 100 kW is sent along cables of total resistance 2 Ω. Compare the power lost at 1 kV and at 10 kV.

## Answers

1. 12/√2 ≈ 8.49 V (the r.m.s. value).
2. V~S~ = 220 × 50/1000 = 11 V.
3. At 1 kV: I = 100 A, loss = 100^2^ × 2 = 20 kW. At 10 kV: I = 10 A, loss = 200 W, one hundred times smaller.

## Did you know?

The low hum near a large transformer is not the current you hear. The alternating magnetic field makes the iron core expand and contract very slightly twice in every cycle, so a 50 Hz transformer hums at 100 Hz.