Power Transmission and Distribution
The eastern Australian electricity grid managed by AEMO (Australian Energy Market Operator) in 2022 transmitted power at 330–500 kV through 40,000 km of transmission lines. The transformer chain: generator (11–25 kV) → step-up (330–500 kV) → subtransmission (66–132 kV) → distribution (11–22 kV) → household (230 V). Without this step-up, resistive losses on a 500 km transmission line at 11 kV would consume 97% of generated power before it reached your home.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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:
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: 330 kV or higher; distribution: 240 V
Understand — The Role of Transformers
- Step-up transformers increase voltage at the power station
- Step-down transformers reduce voltage at substations and homes
- Transformers only work with AC, which is why the grid uses AC
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
Core Content
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, those wires would glow red. 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.
$P_{loss} = I^2 R$
Ploss = 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.
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. Australian grid: generated at 10–25 kV, stepped up to 330–500 kV for transmission, stepped down to 240 V for domestic use.
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:
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.
- 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 (240 V): Pole-top or pad-mounted transformers reduce voltage to household levels.
- Consumption: Homes, businesses, and industries use the electricity.
Transformers only work with AC. Without transformers, we could not efficiently step voltage up for transmission and down for safe use. DC cannot be transformed directly. This is why the entire electrical grid runs on AC, despite DC having some advantages for very long distances.
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.
Grid voltage chain: generation (10–25 kV) → step-up transformer (330–500 kV) → transmission → substation step-down (66/11 kV) → pole/pad transformer (240 V). AC is essential because transformers require changing flux; DC cannot be stepped without power electronics (HVDC).
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.
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.
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.
(c) Calculate the percentage power loss in each case.
$I = P/V = 500 \times 10^6 / 250 \times 10^3 = 2000$ A
$P_{loss} = I^2 R = (2000)^2 \times 2.0 = 8.0 \times 10^6$ W $= 8.0$ MW
$I = 500 \times 10^6 / 500 \times 10^3 = 1000$ A
$P_{loss} = (1000)^2 \times 2.0 = 2.0 \times 10^6$ W $= 2.0$ MW
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:
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 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.
- 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.
- AC is used because transformers require changing flux, which only AC provides.
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+
- Domestic: 240 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
- DC flux is constant → no induction
- HVDC exists but requires electronics
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.
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). AC is mandatory (not DC) because transformers — essential for stepping voltage up and down — only work with alternating current, since they require a changing magnetic flux to induce EMF in the secondary coil (1 mark).
Timed questions on power transmission. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaAt 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.