Understand why electrical power is sent at high voltage the easy way, with plain English intuition, a clear diagram, the I squared R loss rule, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.
Learn
Every power station sits a long way from the houses it feeds, and the wires in between are not perfect. They have a little resistance, so they warm up and waste energy as heat along the way. The clever fix is to send the power at a very high voltage so that only a tiny current flows. Because the wasted heat grows with the square of the current, a small current means almost no waste. That single idea is why the towers marching across the countryside carry electricity at hundreds of thousands of volts.
Why the wires lose energy
A transmission line is just a very long wire, and every wire has some resistance. Push a current through that resistance and it heats up, exactly like the element in a toaster. That heat is energy that left the power station but never reaches your home.
The heat wasted in the line is given by:
Ploss=I2R
The thing to notice is the little 2. The current is squared, so the loss is extremely sensitive to how much current flows. Double the current and you do not double the waste, you quadruple it. Halve the current and the waste drops to a quarter.
The trick: high voltage means low current
Now here is the whole idea. The power delivered down the line is:
P=VI
For a fixed amount of power P, voltage and current trade off against each other. If you push the voltage way up, the current drops right down to keep the product the same. And a low current, fed into I2R, means a tiny heat loss.
So the recipe is to step up the voltage to a very high value at the power station, send the power across the country at that high voltage with a small current, then step down the voltage again to a safe level near homes. Transformers do the stepping up and down.
The lines in the middle of the diagram do the long, lonely trip at high voltage and low current. That low current is the whole point: it keeps the I2R heat loss small for the entire journey.
See it for yourself
Build a circuit, give the connecting wires some resistance, and watch how much power they waste as heat. Add an ammeter and a voltmeter to read the current and the voltage lost along the line, the same losses real transmission lines beat by using high voltage.
Students often muddle two ideas, so keep them apart. The power loss is the heat wasted in the wire, and it uses the current squared. The voltage drop is how much voltage is lost along the wire, and it uses the current just once.
Ploss=I2RVdrop=IR
Both get smaller when the current is smaller, which is one more reason high voltage transmission wins.
How to actually solve one
The method is a short, repeatable recipe.
If you are given the delivered power and the transmission voltage, find the line current from P=VI, so I=VP.
Find the heat wasted in the line with Ploss=I2R. Always square the current first, then multiply by the resistance.
If asked for the voltage lost along the line, use Vdrop=IR.
To compare two cases, remember the loss follows the square of the current, so halving the current quarters the loss.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
Write the formula for power lost as heat in a transmission line.
Ploss=I2R — the loss depends on the square of the current and the line resistance.
Why is power sent across long distances at high voltage?
For a fixed power P=VI, a higher voltage means a smaller current, and since Ploss=I2R, a smaller current means far less wasted heat.
Why does halving the current cut the loss to a quarter, not a half?
Because the loss is I2R: the current is squared, so (21)2=41.
What is the difference between power loss and voltage drop?
Power loss Ploss=I2R is the heat wasted (current squared). Voltage drop Vdrop=IR is the volts lost along the line (current once).
Describe the step up then step down journey from station to home.
Step the voltage up for the long, high-voltage, low-current trip, then step it back down to a safe level (about 230 V) near homes.
Recall · Transformers
For an ideal transformer, how do input and output power relate?
V1I1=V2I2 — power in equals power out, so stepping the voltage up steps the current down by the same factor.
Recall · Magnetic Flux and Induction
What condition is needed to induce an EMF in a coil?
A changing magnetic flux: ε=−NΔtΔΦ. A steady flux induces nothing.
See the recipe in action in the Worked Examples tab, then test yourself in Try It.
Worked examples
Worked Example 1Power lost in a line
A transmission line has a total resistance of 5.0Ω and carries a current of 20 A. Find the power lost as heat in the line.
1
The heat lost in the line depends on the current through it and the line resistance, through Ploss=I2R.
Ploss=I2R
2
Put in the current and the resistance.
Ploss=(20)2×5.0
3
Square the current first, then multiply by the resistance.
Ploss=400×5.0=2000 W
Answer
Ploss=2000 W
Worked Example 2Double the voltage, halve the current
The same power is delivered down the same line, but the transmission voltage is doubled, so the current halves to 10 A. Find the new power loss in the same line.
1
The line resistance is unchanged at 5.0Ω. Only the current has changed, so use the same loss rule with the new current.
Ploss=I2R=(10)2×5.0
2
Square the smaller current first, then multiply by the resistance.
Ploss=100×5.0=500 W
3
Halving the current cut the loss to a quarter, not a half, because the loss depends on the square of the current.
2000500=41
Answer
Ploss=500 W, one quarter of before
Practice questions
Practice test
Try it yourself
6 questions, 8 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.Electrical power is sent across long distances at very high voltage. The main reason is to:
1mark
Need a hint?
For a fixed power, P=VI. If V goes up, what happens to I, and what does that do to I2R?
Show worked solution
To deliver a fixed power P=VI, a higher voltage means a smaller current. Because the heat lost in the line is Ploss=I2R, a smaller current means far less wasted heat. That is the whole reason power is stepped up to very high voltage for long distance transmission.
Q2.A line of resistance 2.0Ω carries a current of 5.0 A. The power lost as heat in the line is:
1mark
Need a hint?
Use Ploss=I2R. Square the current before you multiply.
Show worked solution
Ploss=I2R=(5.0)2×2.0=25×2.0=50 W. Forgetting to square the current gives 10 W, a very common slip.
Q3.The current in a transmission line is reduced to one third of its original value, with the line resistance unchanged. The power lost as heat in the line becomes:
1mark
Need a hint?
The loss depends on the square of the current. What is (1/3)2?
Show worked solution
Because Ploss=I2R, the loss scales with the square of the current. Reducing the current to one third multiplies the loss by (1/3)2=1/9, so the loss drops to one ninth.
Q4.A current of 4.0 A flows through a transmission line of resistance 3.0Ω. The voltage drop along the line is:
1mark
Need a hint?
The voltage drop along a line is Vdrop=IR.
Show worked solution
Vdrop=IR=4.0×3.0=12 V. This is the voltage lost along the line itself, separate from the power loss I2R.
Q5.A transmission line has a total resistance of 4.0Ω and carries a current of 10 A. Find the power lost as heat in the line, and the voltage drop along the line. Show your working.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
The power lost as heat uses the square of the current:
Ploss=I2R=(10)2×4.0=100×4.0=400 W.
The voltage drop along the line uses the current once:
Vdrop=IR=10×4.0=40 V.
Q6.A power station generates electricity at 20 kV RMS, which an ideal transformer steps up to 500 kV RMS for transmission. The transformer output is 100 MW. Which one of the following is closest to the RMS current on the 20 kV generator side?
1mark
Need a hint?
An ideal transformer conserves power, so P=VI on the generator side too.
Show worked solution
Power is conserved through an ideal transformer, so on the generator side I=VP=20×103100×106=5.0×103 A =5.0 kA (option D).
VCAA 2025 Physics Exam, Section A Q10
Frequently asked questions
Why is high voltage better for sending power a long way?
To deliver a fixed amount of power, a higher voltage lets you use a smaller current. The heat wasted in the lines depends on the square of the current, so cutting the current cuts the losses dramatically. That is why power is stepped up to very high voltage for the long trip, then stepped down again near homes.
What is the difference between power loss and voltage drop?
Power loss is the heat wasted in the line and depends on the square of the current, I squared R. Voltage drop is how much voltage is lost along the line and depends on the current once, I times R. They are related but answer different questions.
Why does halving the current cut the loss to a quarter, not a half?
Because the loss is I squared R. The current appears squared, so if you halve the current you multiply the loss by one half squared, which is one quarter. Small changes in current have a big effect on the wasted heat.
