Grade by Grade
F. Electricity and Magnetism · Electricity and Magnetism

Transmitting Electrical Energy: Why the Grid Uses High-Voltage AC

F3.7 F2.6 F3.8 F1.2About 70 minutes
Learning goals
  • I can trace electrical energy from a generating station to a home, naming where transformers step the voltage up and down.
  • I can use P = VI and P = I²R to calculate power losses in transmission lines.
  • I can explain why transmitting at high voltage and low current reduces losses.
  • I can explain why AC, rather than DC, is used for most electrical energy transmission.

Words to know

electrical grid
The connected network of generating stations, transmission lines, transformer stations and distribution lines that delivers electricity.
transmission lines
High-voltage lines that carry electrical energy long distances from generating stations.
distribution lines
Lower-voltage lines that carry electrical energy through towns and neighbourhoods to customers.
transformer station (substation)
A facility with large transformers that step voltage up or down and switching equipment that routes power.
line loss
Electrical energy turned into thermal energy in the resistance of the lines: P_loss = I²R.

The longest extension cord

Ontario's biggest generating stations are often far from the cities that use the most electricity. Energy from northern hydroelectric stations and from nuclear stations on the shores of Lake Ontario and Lake Huron must travel tens or hundreds of kilometres. Every metre of wire has some resistance, so some energy is always lost as heat. The trick that keeps those losses small is to use transformers to send the energy at very high voltage.

Why high voltage? The I²R problem

A power line has a resistance R. The current I in it heats the line. The power wasted is

P_loss = I²R

The power a station sends is P = VI. For the same power, a higher voltage means a lower current: I = P ÷ V.

  • Raise the voltage 10 times and the current falls to 1/10.
  • The loss depends on I squared, so it falls to (1/10)² = 1/100.

That is why transmission lines run at hundreds of thousands of volts. Hydro One's main transmission lines in Ontario run at voltages such as 230 kV and 500 kV.

Careful: in P_loss = I²R, use the current in the line and the line's resistance. Do not use V²/R with the station voltage; the station voltage is not the voltage across the line wires.

generatingstationtens of kVstep-uptransformeruptransmissionlinese.g. 230 or 500 kVtransformerstationdowndistributionlinesa few to tens of kVlocaltransformerdownhome120 / 240 Vhigh voltage = low current = low losses on the long linesvoltages are typical examples; exact values vary
The path from a generating station to your home. Voltage is stepped up for the long trip and stepped down in stages before it reaches your outlets.
Worked example

Example 1: High voltage versus low voltage

A station sends 100 MW through transmission lines with a total resistance of 5.0 Ω. Compare the power lost if it is sent at (a) 50 kV and (b) 500 kV.

  1. (a) I = P ÷ V = (100 × 10⁶ W) ÷ (50 × 10³ V) = 2000 A. P_loss = I²R = (2000)²(5.0) = 2.0 × 10⁷ W = 20 MW. That is 20 ÷ 100 = 20% of the power.
  2. (b) I = (100 × 10⁶) ÷ (500 × 10³) = 200 A. P_loss = (200)²(5.0) = 2.0 × 10⁵ W = 0.20 MW, only 0.20%.
  3. Ten times the voltage gave one-tenth the current and one-hundredth the loss.

20 MW lost at 50 kV; 0.20 MW lost at 500 kV.

20 MW lost (20%)sent at 50 kVI = 2000 A0.2 MW lost (0.2%)sent at 500 kVI = 200 A10 × the voltage →1/10 the current →1/100 the loss
Example 1 as bars. At 500 kV the loss bar is so small it is almost invisible: one-hundredth of the loss at 50 kV.
Worked example

Example 2: Voltage drop along a line

A distribution line with a total resistance of 0.80 Ω carries 150 A. (a) What is the potential difference lost along the line? (b) What power is lost?

  1. (a) V_drop = IR = (150)(0.80) = 120 V.
  2. (b) P_loss = I²R = (150)²(0.80) = 1.8 × 10⁴ W (18 kW). Check: P = V_drop × I = (120)(150) = 18 000 W ✓.

120 V drop; 18 kW lost.

Worked example

Example 3: The cost of losses

In Example 1(b), 0.20 MW is lost continuously. How much energy is lost in one day, and what would it cost at 10¢ per kilowatt-hour?

  1. 0.20 MW = 200 kW. In 24 h: E = 200 kW × 24 h = 4800 kW·h.
  2. Cost: 4800 × $0.10 = $480 per day. At 50 kV it would be 100 times as much: $48 000 per day.
  3. (10¢/kW·h is just a round number for the example; real rates vary.)

4800 kW·h, about $480 a day.

Why AC?

High-voltage transmission only works if the voltage can be stepped up at the station and stepped down again near customers. Transformers do this simply, efficiently and with no moving parts, but they need a changing current: AC. That is the main reason the grid uses AC.

Other reasons: AC generators are simple and robust, and AC is easy to switch off safely because the current passes through zero 120 times a second.

DC does have a niche. For very long distances, or for undersea cables, high-voltage DC (HVDC) lines can lose less energy than AC lines. They need expensive converter stations at each end to change AC to DC and back, so they are used only for special links.

Stepping down to your home

At transformer stations near cities, large transformers step the voltage down for distribution lines along streets. A local transformer on a pole or in a green box on a lawn then steps it down to the 120 V / 240 V supply in your home. Lights and most outlets use 120 V; stoves, dryers and some heaters use 240 V.

Just for fun

Power lines really do droop on hot summer afternoons. Metal expands as it warms, and on the hottest days the lines are also carrying extra current for air conditioners, which heats them even more (I²R at work).

Hands-on activity

Model transmission losses

You need: paper, a calculator, and (optional) a spreadsheet.

  1. A station must deliver 50 MW through lines with a total resistance of 4.0 Ω. Make a table with columns: transmission voltage, line current, power lost, percent lost.
  2. Fill it in for 25 kV, 50 kV, 115 kV, 230 kV and 500 kV.
  3. Graph percent lost against voltage. Describe the shape.
  4. Look outside on a walk: can you spot a pole-top or lawn transformer near your home? Count how many houses seem to be connected to it. Grown-up note: look only; never touch or climb near poles, lines, green transformer boxes or fences around transformer stations.

What to look for: at 25 kV, I = 2000 A and P_loss = 16 MW (32%); at 50 kV, 1000 A and 4.0 MW (8.0%); at 115 kV, about 435 A and 0.76 MW (1.5%); at 230 kV, about 217 A and 0.19 MW (0.38%); at 500 kV, 100 A and 0.040 MW (0.080%). The loss falls very steeply as voltage rises (proportional to 1/V²). One local transformer often serves several homes.

Your turn

1

Write the formula for the power lost in a transmission line, and say what I and R stand for.

2

A station sends 20 MW at 100 kV. What is the line current?

3

If the line in the last question has a resistance of 6.0 Ω, find the power lost and the percent lost.

4

The transmission voltage is tripled with the same power sent. What happens to the current and to the power lost?

5

Why is AC, not DC, used for most transmission?

6

List the main stages that electrical energy passes through from a generating station to a home.

More practice: warm-up

1

At higher transmission voltage, is the current larger or smaller for the same power?

2

What two voltages does a typical Canadian home receive?

3

What happens to the energy lost in transmission lines?

4

Name one place the voltage is stepped down before it reaches your home.

5

What does HVDC stand for?

More practice: on your own

1

A line with resistance 2.0 Ω carries 300 A. Find the voltage drop along it and the power lost.

2

A station sends 60 MW through 3.0 Ω lines at 230 kV. Find the percent of the power lost.

3

A student calculates the loss in a 500 kV line of resistance 5.0 Ω as P = V²/R = (500 000)² ÷ 5.0 = 5.0 × 10¹⁰ W. Explain the mistake.

4

Lines that carry 400 A lose 1.2 MW. What is their resistance?

5

Give two reasons, other than transformers, why AC is used on the grid.

Challenge

1

A town needs 12 MW. It is supplied through lines of total resistance 8.0 Ω. What is the minimum transmission voltage that keeps line losses below 1.0% of the power sent? (Assume the power sent is about 12 MW.)

2

Show that, for a fixed power P and line resistance R, the power lost is P_loss = P²R ÷ V². Use it to find the factor by which losses change if the voltage goes from 115 kV to 230 kV.

3

A rural customer is at the end of a long line. The line drops 6.0 V when the customer draws 20 A. If the customer turns on a heater and draws 50 A, what is the new drop? Why might lights dim when the heater starts?

Check your understanding

Quiz

1.

Why is electrical energy transmitted at high voltage?

2.

A station sends 10 MW at 100 kV. What is the line current?

3.

If the transmission voltage is doubled for the same power, the line loss…

4.

A line of resistance 4.0 Ω carries 250 A. What power is lost?

5.

What is the main reason AC is used on the grid?

6.

Where is the voltage reduced to 120/240 V for a home?

For parents and teachers

Big idea: AC is used to transmit electrical energy because transformers can step its voltage up for low-loss transmission and down for safe use (F3.7, F3.8). Students calculate losses with P = I²R (F2.6). This also connects to the efficiency of the whole electricity system (F1.2).

How to help: On a walk, point out pole-top transformers, transmission towers and transformer stations, and ask your teen what the voltage might be at each. Always from a safe distance: never touch poles, guy wires, green boxes or fences around stations.

Common difficulties:

  • Using V²/R with the station voltage to find line losses (a very common error). Use the line current: I²R.
  • Forgetting to convert MW and kV into W and V.
  • Thinking higher voltage means more loss. For a fixed power it means less current, and much less loss.