Year 12 Physics Module 6 ⏱ ~20 min 5 MC · 2 Short Answer Lesson 18 of 21

Power Transmission and Distribution

You'll be able to calculate power loss for given transmission parameters.

Power networks step voltage up for efficient transmission, then down for distribution and safe use. For fixed transmitted power, higher voltage means lower current and much smaller $I^2R$ line loss.

Today's hook: Send the same power at 100 kV or 500 kV through the same resistive lines. Which case heats the lines more, and why do transformers make the voltage change practical?

0/5TASKS

Learn0 of 6 steps complete

Retrieve power, voltage and current relationships before applying them to transmission.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

A power station generates 100 MW of electrical power.

If this power is transmitted at 100 kV, what is the current in the transmission lines?
If the transmission lines have total resistance 5.0 ohms, how much power is lost as heat?
If the same power is transmitted at 500 kV instead, what is the new power loss? What does this tell you about high-voltage transmission?

Warm-up, the formula for power loss in a transmission line is:

Learning Intentions

goals

Know, Power Transmission

Power loss in transmission lines: $P_{loss} = I^2 R$
For fixed power, higher voltage means lower current and lower loss
Typical transmission: hundreds of kilovolts; Australian low-voltage distribution is nominally 230/400 V

Understand, The Role of Transformers

Step-up transformers increase voltage at the power station
Step-down transformers reduce voltage at substations and homes
Conventional AC grids use transformers efficiently; HVDC links use power-electronic converters

Can Do, Calculate and Compare

Calculate power loss for given transmission parameters
Compare power loss at different transmission voltages
Explain why the grid uses AC and high-voltage transmission

Scan these before reading

vocab

Transmission linesHigh-voltage power lines that carry electricity over long distances from power stations to cities.

Power lossEnergy lost as heat in transmission wires: $P_{loss} = I^2 R$.

SubstationA facility containing transformers that step voltage up or down for transmission or distribution.

GridThe interconnected network of power stations, transmission lines, and distribution systems.

Misconceptions to fix

✗ Wrong: High-voltage transmission wastes more energy because voltage is higher.

✓ Right: Higher voltage reduces current for the same power ($I = P/V$). Since losses depend on $I^2 R$, lower current means dramatically less heat loss, even though voltage is higher.

✗ Wrong: One current type is always superior for every grid link.

✓ Right: Conventional grids developed around AC because transformers change voltage efficiently. HVDC can be advantageous for some long-distance or submarine links, but requires converter stations to connect with AC networks.

Cross-lesson links: L19 examined eddy current applications. L20 shows how the complete M6 toolkit, Faraday's law, transformers, AC generation, combines in a national electricity grid. L20 is M6's most integrating application and directly answers the HSC syllabus question 'why is AC used for power transmission?'

Core Content

Derive how fixed power and higher voltage reduce line current and $I^2R$ loss.

Why High Voltage?

+5 XP

The physics of efficient power transmission

Touch the metal case of a power point adaptor after it has been running for a few hours: it is warm. That warmth is $I^2R$ loss, electrical energy converted to heat in the wiring inside. Now scale this up: a 500 km transmission line carrying enough current to power a city. If the current is too high, the conductors overheat: they sag as they expand, voltage drop along the line grows, and protection equipment trips the circuit before the line is damaged. AEMO solves this by running the eastern Australian grid at 330–500 kV, ten to forty times the voltage a generator produces, which reduces the current by the same factor and $I^2R$ losses by the square of that factor.

Power loss in transmission

$P_{loss} = I^2 R$

P_loss = power lost as heat (W) · I = current in lines (A) · R = total line resistance (Ω)

The power transmitted is $P = VI$. For a fixed power $P$, if we increase voltage $V$, the current $I$ must decrease. Since power loss depends on $I^2$, doubling the voltage halves the current and quarters the power loss.

Worked example, comparing transmission voltages

Transmitting 100 MW with line resistance R = 10 Ω:

At 100 kV: $I = P/V = 100 \times 10^6 / 100 \times 10^3 = 1000$ A
$P_{loss} = (1000)^2 \times 10 = 10$ MW (10% loss)
At 500 kV: $I = 100 \times 10^6 / 500 \times 10^3 = 200$ A
$P_{loss} = (200)^2 \times 10 = 0.4$ MW (0.4% loss)

Increasing transmission voltage by 5× reduces power loss by a factor of 25. This is why the grid uses 330 kV or higher for long-distance transmission.

Transmission loss: $P_\text{loss} = I^2R$ (W). Since $I = P/V$, doubling voltage halves current and quarters power loss. A representative Australian chain is generation at about 10–25 kV, step-up for transmission, then step-down to nominal 230/400 V low-voltage distribution.

Pause, copy the $P_\text{loss} = I^2R$ formula and the doubling-voltage effect into your book before moving on.

400 MW is transmitted at 400 kV through lines with $R = 5.0$ Ω. The power loss is:

Map generation, transmission, substations and nominal 230/400 V distribution.

The Grid: From Power Station to Home

+5 XP

The journey of electricity through the distribution system

We just saw why high voltage reduces transmission losses. That raises a question: how is this actually implemented across hundreds of kilometres? This card answers it → multiple transformer stages step voltage up for transmission and back down for safe household use.

Electricity follows a well-defined path from generation to consumption, with transformers at each stage stepping voltage up or down as required.

A representative Australian chain. Transformers work because the supply is AC. For a fixed power, raising the voltage lowers the current, and the heating loss in the lines, P_loss = I²R, falls with the square of the current.

Text description

Generator: produces AC at about 10–25 kV.
Step-up transformer at the power station (fewer primary turns than secondary turns): raises the voltage and lowers the current by the same factor.
Transmission lines at 330 kV or more carry a low current. Energy is lost as heat in the lines, P_loss = I²R, which is small because I is low.
Substation step-down transformer (more primary turns than secondary): lowers the voltage to 66 kV or 11 kV for regional distribution.
Local pole-top or pad-mounted step-down transformer: lowers the voltage to nominal 230/400 V.
Homes and businesses use the energy.
Key relationships: I = P/V, so doubling V halves I; P_loss = I²R, so the loss falls to a quarter.

Generation (10–25 kV): Power stations generate AC at relatively low voltages.
Step-up (330 kV+): A transformer at the power station increases voltage for long-distance transmission.
Transmission: High-voltage lines carry power efficiently across hundreds of kilometres.
Substation step-down (66 kV or 11 kV): Near cities, transformers reduce voltage for regional distribution.
Local step-down (nominal 230/400 V): Pole-top or pad-mounted transformers reduce voltage to low-voltage distribution levels.
Consumption: Homes, businesses, and industries use the electricity.

Why AC?

Transformers require changing flux, so they operate directly with AC rather than steady DC. This made voltage conversion a major historical and practical reason for conventional AC networks. Modern grids are not exclusively AC: HVDC links use rectifiers and inverters where their system advantages justify converter cost and complexity.

Modern HVDC (high-voltage direct current) links are sometimes used for very long submarine cables or inter-system connections, they use power electronics (rectifiers and inverters) to convert AC to DC and back, avoiding the reactive power losses AC suffers over long cable runs.

Representative grid chain: generation (about 10–25 kV) → step-up transformer → high-voltage transmission → substations → nominal 230/400 V low-voltage distribution. AC enables direct transformer voltage conversion; HVDC instead uses power-electronic converter stations and can suit selected long links.

Pause, write the voltage chain from generation to home (with voltage values) into your book before moving on.

The electrical grid uses DC because it is safer than AC.

For a fixed transmitted power, doubling the transmission voltage reduces the current by a factor of 2.

Domestic voltage in Australia is 330 kV.

Calculate current, line loss and percentage loss with units and a sending/receiving power ledger.

Worked Example, Power Transmission Calculation

+5 XP

Compare losses at different transmission voltages step-by-step

We just saw the voltage chain from generator to home. That raises a question: how do we calculate the actual numerical improvement when voltage is changed? This card answers it → use $I = P/V$ then $P_\text{loss} = I^2R$; doubling voltage always quarters the loss.

Problem

A power station generates 500 MW of electrical power. The transmission lines have total resistance 2.0 Ω.

(a) Calculate the power loss if transmitted at 250 kV.

(b) Calculate the power loss if transmitted at 500 kV.

Step 1, Part (a): At 250 kV

$I = P/V = 500 \times 10^6 / 250 \times 10^3 = 2000$ A

$P_{loss} = I^2 R = (2000)^2 \times 2.0 = 8.0 \times 10^6$ W $= 8.0$ MW

Step 2, Part (b): At 500 kV

$I = 500 \times 10^6 / 500 \times 10^3 = 1000$ A

$P_{loss} = (1000)^2 \times 2.0 = 2.0 \times 10^6$ W $= 2.0$ MW

Step 3, Part (c): Percentage loss

At 250 kV: % loss $= (8.0 / 500) \times 100 = 1.6\%$

At 500 kV: % loss $= (2.0 / 500) \times 100 = 0.4\%$

Conclusion: Doubling the voltage halves the current and quarters the power loss.

Transmission calculation steps: (1) $I = P/V$ (convert MW→W, kV→V); (2) $P_\text{loss} = I^2R$; (3) % loss $= (P_\text{loss}/P_\text{total})\times100$. Worked result: 250 kV → 1.6% loss; 500 kV → 0.4% loss. Doubling voltage quarters the loss.

Pause, write the three calculation steps and the unit-conversion reminder into your book before moving on.

If transmission voltage is doubled while power stays constant, power loss:

Power ledger and trade-offs

Keep the locations separate: sending-end power $P_s$, line loss $I^2R_{\text{line}}$, and receiving-end power $P_r=P_s-I^2R_{\text{line}}$ in the simplified model. Percentage loss is $100(P_s-P_r)/P_s$. Raising voltage reduces current for a target power, but requires more insulation and clearance and can increase corona effects; HVDC additionally requires expensive converters.

Compare high-voltage benefits with insulation, clearance, corona and converter trade-offs.

Activity 1, Calculate Transmission Losses

ApplyBand 4

Practise the power loss calculation at different voltages

A power station generates 200 MW of power. The transmission lines have total resistance 8.0 Ω.

Calculate the current if power is transmitted at 200 kV.
Calculate the power loss in the transmission lines.
Calculate the percentage of generated power that is lost.
Repeat calculations for transmission at 400 kV. By what factor did the power loss decrease?
Explain why power companies use very high voltages for transmission despite the cost of transformers and insulation.

Activity check, for 200 MW transmitted at 200 kV through 8.0 Ω lines, the power loss (in MW) is _____.

Explain why conventional grids use AC transformers while selected links use HVDC converters.

Activity 2, Concept Check

UnderstandBand 4

Explain why the grid uses AC and high-voltage transmission

Using your knowledge of transformers and power loss, explain two reasons why the Australian electricity grid transmits power at very high voltages as alternating current, rather than at low voltage as direct current.

Three of these statements about power transmission are correct. Pick the odd one out.

Synthesis, Connect the Ideas

Power loss in transmission lines is $P_{loss} = I^2 R$.
For fixed power, increasing transmission voltage reduces current and dramatically reduces losses.
Transformers step voltage up for transmission and down for distribution and use.
Conventional AC networks use transformers for efficient voltage conversion; selected HVDC links use power electronics instead.

Key Formulae

$P_{loss} = I^2 R$
$I = P / V$
$V_s / V_p = N_s / N_p$

Grid Voltages

Generation: 10–25 kV
Transmission: 330 kV+
Low-voltage distribution: nominal 230/400 V

Key Principle

Double V → half I → quarter $P_{loss}$
$P_{loss} \propto 1/V^2$ for fixed $P$

Why AC?

Transformers need changing flux
Steady DC does not provide transformer-changing flux
HVDC exists but requires converter stations

Analyse6 marks

Electrical energy is transmitted from a power station to a distant city at hundreds of thousands of volts rather than at the voltage used in homes. Analyse the role of transformers in this system and the reason for the high transmission voltage. (6 marks)

Unpack the verb

Analyse is this dot point's own verb. The components are the two transformers and the line; the relationship is between current and power loss.

Plan: write the two power expressions that are easy to confuse, show which one governs the loss, then place each transformer in the system.

The entire question turns on there being two different powers in play — the power delivered and the power wasted — and they depend on different things.

Assess5 marks

A student writes: "High-voltage transmission wastes more energy, because the voltage is higher and the power lost is $P = VI$." Assess this statement. (5 marks)

Unpack the verb

Assess means judge it against criteria, which here means testing the claim against the physics rather than declaring it wrong.

Plan: identify what the student has got right, identify the substitution error, give the correct relationship, then state the verdict with the actual direction of the effect.

Finding the half that is correct makes the diagnosis specific. The student has not invented a formula; they have applied a real one to the wrong quantity.

Practice1 retrieval set + 2 SAQs

Check transmission calculations, grid reasoning and trade-offs with mixed practice.

Quick recall, power transmission

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks

+5 XP

ApplyBand 4(4 marks) 1. A power station generates 800 MW at 20 kV. This is stepped up to 400 kV for transmission through lines with total resistance 4.0 Ω.

Calculate the current in the transmission lines. (1 mark)
Calculate the power loss in the transmission lines. (2 marks)
Calculate the percentage of power lost during transmission. (1 mark)

1 mark: correct current using $I=P/V$ at transmission voltage · 2 marks: correct $P_{loss} = I^2R$ with working · 1 mark: correct percentage

EvaluateBand 6(3 marks) 2. Evaluate the claim: "Power companies use high-voltage AC transmission purely to save money on cable costs." Assess whether this claim is correct, partially correct, or incorrect, and justify your answer using physics principles.

1 mark: identifies claim as partially correct or incorrect with justification · 1 mark: correctly explains $P_{loss} = I^2R$ relationship with voltage · 1 mark: identifies AC requirement for transformer operation

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (4 marks): (a) $I = P/V = 800 \times 10^6 / 400 \times 10^3 = 2000$ A (1 mark). (b) $P_{loss} = I^2 R = (2000)^2 \times 4.0 = 1.6 \times 10^7$ W $= 16$ MW (2 marks, method + correct answer). (c) % loss $= (16/800) \times 100 = 2.0\%$ (1 mark).

Q2 (3 marks): The claim is partially correct (1 mark). While reducing cable costs through less material is a benefit, the primary reason for high-voltage transmission is to minimise power losses. From $P_{loss} = I^2 R$, and $I = P/V$, transmitting at higher voltage reduces current and dramatically reduces resistive heating losses, power loss scales as $1/V^2$ for fixed power (1 mark). Conventional grids developed around AC because transformers step voltage up and down efficiently and need a changing magnetic flux to induce an emf in the secondary, which steady DC does not provide (1 mark). This is a historical and practical reason rather than a physical necessity: modern HVDC links carry DC at high voltage using power-electronic converter stations.

Reviewrevisit + module checkpoint

Revisit the power ledger and the AC/HVDC trade-off before completion.

Check what actually stuck

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →

How did your thinking change?

At the start you were asked about AEMO's eastern Australian grid 2022: why did they choose 500 kV rather than 100 kV for transmission, and to verify the 25× factor on a 100 MW line with R = 5 Ω.

The answers: at 100 kV, $I = P/V = 10^8/10^5 = 1000$ A and $P_{loss} = I^2 R = 10^6 \times 5 = 5$ MW (5%). At 500 kV, $I = 200$ A and $P_{loss} = 40000 \times 5 = 0.2$ MW (0.2%). Ratio = $(1000/200)^2 = 25$. AEMO chose 500 kV specifically because of this $I^2 \propto 1/V^2$ relationship, the loss decreased 25-fold by raising voltage 5-fold.

Extend: Some modern long-distance power lines use HVDC (high-voltage direct current) instead of AC. Research why this might be advantageous for very long distances, and explain the trade-off involved.

I have completed this lesson

← Previous: Transformers, Theory and Operation Next: AC Generators →

Module overview · Physics · Checkpoint 4 · Module quiz