DSE Physics | Energy and Use of Energy Lesson-Prep Guide (Tutor Edition) | Efficiency Chain, Transmission Loss and Capacity Factor

2026-11-04

Tutor lesson-prep guides · Secondary Physics

Students can write power as energy over time, but answer why transmission uses high voltage with only fewer losses and no mechanism. This lesson uses P loss equals I squared R to show why stepping up works.

Objectives — Students can: (1) use an efficiency chain from fuel energy to electrical output; (2) apply P = IV and P loss = I squared R to explain high-voltage transmission; (3) compare energy sources using capacity factor.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Power	Definition of power	P = E over t
Electrical power	Power in terms of current and voltage	P = IV
Energy units	How many joules in 1 kWh	3.6 MJ

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Energy chain	Fuel, boiler, turbine, generator	Draw four stages
10-24 min	Efficiency	Input times efficiency, stage by stage	Get 7,200 tonnes a day
24-38 min	Transmission loss	P = IV, then I squared R	Get 6.25 percent
38-50 min	Renewables	Capacity factor and real output	Get 1.44 kWh a day
50-56 min	Answer format	Formula, substitution, units, comment	Rewrite one answer
56-60 min	Exit check	Three questions marked on the spot	Independent work

03 Teaching script (read it out)

Scene 1: The energy chain: every stage takes a cut

The station delivers 1000 MW at 40 percent efficiency, so the heat input is 1000 divided by 0.4, which is 2500 MW. Coal releases 30 MJ per kg, so the burn rate is 2500 million joules per second divided by 30 million joules per kg, which is 83.3 kg per second, and over a day that is 83.3 times 86400, or 7,200 tonnes. The chain multiplies out: boiler 0.90 times turbine 0.40 times generator 0.95 gives 34.2 percent.

Ask the student: if efficiency rises from 40 to 45 percent, how much coal is saved each day (about 800 tonnes).

Scene 2: Why transmission uses high voltage

Sending 1000 MW at 400 kV needs I = P over V, so 1000 million watts divided by 400,000 volts, which is 2,500 A. With a line resistance of 10 ohms, the loss is I squared R, so 2,500 squared times 10, which is 62.5 MW, or 6.25 percent. At 11 kV the current would be 90,909 A and the loss 82.6 GW, far larger than the output, so it is impossible. Loss falls with the square of the voltage, so raising the voltage ten times cuts the loss to one percent.

Warn: an ideal transformer changes voltage and current but not power, so stepping up does not create energy.

Scene 3: Renewables: peak is not output

A 2 square metre panel at 20 percent efficiency under 1000 W per square metre peaks at 2 times 1000 times 0.2, which is 400 W. With a capacity factor of 0.15 for night and cloud, the daily output is 0.4 kW times 24 times 0.15, which is 1.44 kWh. A 2 MW turbine at a capacity factor of 0.30 gives 2 times 24 times 0.30, which is 14.4 MWh a day. Compare sources on energy density, intermittency and carbon, never on installed capacity alone.

Ask the student: with the same installed capacity, how many times more output does a capacity factor of 0.45 give than 0.15 (three times).

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Efficiency chain	Multiply stages or divide by efficiency	34.2 percent
Calculation	Fuel burn rate	Input power over calorific value	83.3 kg per second
Calculation	Transmission loss	I = P over V, then I squared R	62.5 MW
Compare	Capacity factor	Real output over maximum possible	1.44 kWh a day

Sanity check (teach this): Efficiency must be below 1, so an answer above 100 percent means input and output were swapped. Transmission loss must be far below the output, otherwise the voltage or the units are wrong.

05 Independent practice (3 questions)

Question	Answer	What to watch
Input power for 1000 MW at 40 percent efficiency	2500 MW	Divide output by efficiency
Current for 1000 MW at 400 kV	2,500 A	P = IV
Daily output of a 400 W panel at 0.15 capacity factor	1.44 kWh	kW times hours times capacity factor

06 Common mistakes and what to say

Mistake	Student writes	Say this
Multiplying by efficiency instead of dividing	Input reported as 400 MW	“Output is input after the cut, so divide by efficiency.”
Using P = IV for loss	Loss reported as 1000 MW	“I squared R is the line loss; IV is the power delivered.”
Confusing capacity factor with efficiency	Saying solar efficiency is 15 percent	“Capacity factor is real output; efficiency is energy conversion.”
Mixing kWh with kW	Answering 400 kWh a day	“400 W is power; multiply by hours and by capacity factor.”

07 Homework (paper, section and question numbers)

DSE physics practice paper (energy) — Section B energy and power calculations
Secondary physics chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Input power for 1000 MW at 40 percent	2500 MW	All three right: move to electromagnetism
Current at 400 kV for the same output	2,500 A	One wrong: reteach P = IV
How loss depends on voltage	Inversely as the square of the voltage	Two wrong: restart act one

09 Tutor FAQ

Why is efficiency never 100 percent?

Heat engines are limited by the Carnot bound and must reject waste heat.

Do transformers waste energy?

Real ones have iron and copper losses, but DSE treats them as ideal.

Can a capacity factor exceed 1?

No, it is real output over the theoretical maximum, so the ceiling is 1.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.