The shape of the lesson
By the end of the lesson, learners can
- Set the turns: Change the turns ratio and watch the secondary voltage and current follow the equations.
- Why high voltage wins: A higher transmission voltage means a smaller current, and the cable loss falls steeply.
- Power: VₚIₚ = VₛIₛ for an ideal transformer.
- Current: rearrange to find the missing current.
- Loss: power wasted in a cable is I²R; a small current means a tiny loss.
One picture every idea returns to
For an ideal transformer the voltage ratio equals the turns ratio: Vp divided by Vs equals Np divided by Ns. If power is conserved, Vp times Ip equals Vs times Is. High-voltage transmission uses a small current, so the cables waste far less power as heat.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Turns, volts, and loss.. The ratio of turns on the two coils fixes the ratio of the voltages exactly. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Set the turns | Change the turns ratio and watch the secondary voltage and current follow the equations. Project tab 6 of the lab (Transformer equations) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Why high voltage wins | A higher transmission voltage means a smaller current, and the cable loss falls steeply. Model the worked example Worked example on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why high voltage wins, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| In the National Grid the voltage is stepped up for transmission so the current is small, because the power wasted in the cables is the current squared times the resistance. | A smaller current means far less heat lost. |
Transformer equations and transmission: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Write the equation linking the voltages and currents of an ideal (100% efficient) transformer, and state what it tells you about the power.
A transformer steps 230 V up to 1150 V. The primary current is 5.0 A. Calculate the secondary current, assuming the transformer is 100% efficient.
A current of 8.0 A flows in a transmission cable of resistance 3.0 Ω. Calculate the power wasted as heat in the cable.
In a transmission cable the current is reduced to half its original value. State what happens to the power wasted in the cable, and explain.
Explain why electrical power is sent through the grid at a very high voltage rather than a low one. Refer to the current and the power lost in the cables.
A power of 20 kW is sent along a cable of resistance 2.0 Ω. Calculate the power wasted in the cable when it is sent at 200 V, and again when it is sent at 20 000 V. Comment on the difference.
Answer key and marking notes
- Vₚ × Iₚ = Vₛ × Iₛ
The power put into the primary equals the power taken from the secondary. ✓✓
- Iₛ = VₚIₚ ÷ Vₛ = (230 × 5.0) ÷ 1150
Iₛ = 1.0 A
voltage up 5 times, so current down 5 times
- P = I²R = 8.0² × 3.0 = 64 × 3.0
P = 192 W
- The power wasted falls to one quarter. ✓
- The loss is I²R, and halving the current gives (1/2)² = 1/4 of the loss. ✓
- For a fixed power, a higher voltage means a smaller current. ✓
- The power wasted heating the cables is I²R. ✓
- A smaller current wastes much less energy, so transmission is more efficient. ✓
At 200 V:
I = P ÷ V = 20000 ÷ 200 = 100 A, loss = 100² × 2.0 = 20 000 WAt 20 000 V:
I = 20000 ÷ 20000 = 1.0 A, loss = 1.0² × 2.0 = 2.0 WThe low-voltage cable wastes almost all the power; the high-voltage cable wastes almost none. ✓
7 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A transformer has 1000 turns on the primary coil and 50 turns on the secondary. The primary voltage is 240 V. What is the secondary voltage? (Vₚ ÷ Vₛ = Nₚ ÷ Nₛ)
- A12 V
- B4800 V
- C19 V
- D480 V
A step-down transformer changes 240 V to 12 V. The secondary coil has 100 turns. How many turns are on the primary coil?
- A2000
- B5
- C480
- D1200
A transformer is 100% efficient. Its primary takes 0.50 A at 240 V. If the secondary voltage is 12 V, what is the secondary current? (VₚIₚ = VₛIₛ)
- A10 A
- B0.025 A
- C0.50 A
- D120 A
An ideal transformer is 100% efficient. What does this mean for the power in the primary and secondary coils?
- AThe power delivered by the secondary equals the power supplied to the primary.
- BThe secondary power is always greater than the primary power.
- CThe voltage is the same on both sides.
- DNo power is transferred to the secondary.
Electricity is transmitted across the country at very high voltage. Why is high voltage used?
- AHigh voltage means a low current, so less power is lost as heat in the cables (P = I²R).
- BHigh voltage makes the electricity travel faster.
- CHigh voltage reduces the resistance of the cables.
- DHigh voltage stores more energy in the wires.
A transmission cable of resistance 2.0 Ω carries a current of 10 A. How much power is wasted as heat in the cable? (P = I²R)
- A200 W
- B20 W
- C40 W
- D400 W
A step-up transformer at a power station increases the voltage from 25 kV to 400 kV before transmission. Assuming it is ideal, what happens to the current?
- AThe current decreases by the same factor that the voltage increases.
- BThe current increases by the same factor as the voltage.
- CThe current stays the same.
- DThe current falls to zero.
Quiz key
Vₛ = 240 × 50 ÷ 1000 = 12 V. Fewer secondary turns give a lower voltage (step-down). Option B uses the ratio upside down, which would step the voltage up.
Nₚ = Nₛ × Vₚ ÷ Vₛ = 100 × 240 ÷ 12 = 2000 turns. The primary must have more turns than the secondary for a step-down transformer, which rules out the smaller options.
VₚIₚ = VₛIₛ, so Iₛ = (240 × 0.50) ÷ 12 = 120 ÷ 12 = 10 A. Stepping voltage down steps current up, so the secondary current is larger than the primary.
In an ideal transformer the output power equals the input power (VₛIₛ = VₚIₚ), since no energy is lost. So if the voltage is stepped up, the current is stepped down by the same factor.
For a fixed power P = VI, a higher voltage means a smaller current. Since the heat lost in the cables is P = I²R, a smaller current greatly reduces these losses. The voltage does not change the cables' resistance or the speed of transmission.
P = I²R = 10² × 2.0 = 100 × 2.0 = 200 W. Option B forgets to square the current (10 × 2); D squares the resistance instead.
Since power is conserved in an ideal transformer (VₚIₚ = VₛIₛ), stepping the voltage up by a factor of 16 steps the current down by the same factor of 16. This lower current is what reduces heat losses in the transmission cables.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.