Why do wires get hot? (Joule loss)
A wire has a small resistance R. When a current I flows, the moving charges bump into the metal atoms and the wire heats up. This heat is wasted energy. The power lost is
Ploss = I² × R
Notice the square. Double the current and the loss becomes four times bigger. Long lines have a bigger R, so the loss is bigger too.
High-voltage transport
A power station must send power P. Power, voltage and current are linked by P = V × I. So the current is I = P ÷ V.
If we raise V ten times, I becomes ten times smaller, and the loss I²R becomes one hundred times smaller. This is why long-distance lines run at 132, 220, 400 kV or even more. Homes cannot use such a voltage, so we bring it down near the town.
Step-up and step-down transformers do this job. They work only with alternating current (AC), which is why the grid uses AC.
The grid as a graph
A grid is not one wire. It is many power stations, substations and towns joined by lines. Scientists draw it as a graph: each station or town is a node, each line is an edge. A graph with many loops gives more than one road for power.
If one line breaks in a storm, the power flows by another edge. This keeps the lights on. A grid with a single road is cheaper but less safe.
Optimising distribution
Engineers have to balance supply and demand every second. They choose the voltage, the line thickness and the routes to keep three things small: losses, cost and risk of blackout.
- A thicker wire has a smaller R, but costs more copper or aluminium.
- A higher voltage needs taller towers and better insulators.
- More lines make the grid safer, but each line costs money.
There is no perfect answer. The best design is a good balance. Smart grids use meters and computers to move power to where it is needed.
Try it: the voltage slider
Use the 3D scene. Set the voltage to 20 kV and write down the lost percentage. Now set 40 kV (double). Predict first: will the loss halve or become one quarter? Then check. Then open the network step and cut a line.
Key formulas and definitions
- P = V × I (power sent)
- I = P ÷ V
- P<sub>loss</sub> = I² × R
- Loss fraction = P × R ÷ V²
- Double V gives 1/4 of the loss
- Transformer: V<sub>s</sub> ÷ V<sub>p</sub> = N<sub>s</sub> ÷ N<sub>p</sub>
Worked examples
1. A station sends 10 MW at 100 kV. Find the current.
I = P ÷ V = 10 000 000 ÷ 100 000 = 100 A.
2. The line in example 1 has a resistance of 5 Ω. Find the power lost as heat.
P<sub>loss</sub> = I²R = 100² × 5 = 50 000 W = 50 kW. That is 0.5% of 10 MW.
3. The same 10 MW is sent at 10 kV through the same line. Find the loss.
I = 10 000 000 ÷ 10 000 = 1000 A. Loss = 1000² × 5 = 5 000 000 W = 5 MW. Half of the power is lost.
4. The voltage of a line is raised from 50 kV to 200 kV. By what factor does the loss change (same power, same wire)?
The voltage becomes 4 times, so the current is 1/4. The loss uses I², so it becomes 1/16 of the old loss.
5. A step-up transformer has 200 turns in the primary and 4000 turns in the secondary. The input is 11 kV. Find the output.
Vs = Vp × Ns ÷ Np = 11 × 4000 ÷ 200 = 220 kV.
6. A town needs 50 MW. The line has R = 4 Ω and runs at 200 kV. Find the power the station must produce (ignore other losses).
I = 50 000 000 ÷ 200 000 = 250 A. Loss = 250² × 4 = 250 000 W = 0.25 MW. The station must produce 50 + 0.25 = 50.25 MW.
Common mistakes
- Saying "higher voltage means higher current". For a fixed power, current goes down when voltage goes up.
- Forgetting the square in I²R and thinking loss only halves when the current halves. It becomes one quarter.
- Mixing up the line voltage with the small voltage drop along the wire. The loss uses the current and the line resistance.
- Thinking transformers work with batteries (DC). They need AC, because the changing current makes the magnetic field change.