Every wire has some resistance, so some of the energy a circuit carries is always lost as heat along the way. One equation, $P = I^2R$, tells you exactly how much, and it explains why power companies send electricity across the country at hundreds of thousands of volts instead of household voltage.
The same amount of power has to be delivered to a town. It can be sent along the transmission wires either as a large current at low voltage, or a small current at high voltage.
Before reading: which choice wastes less energy as heat in the wires, and why? Write your prediction and a one-sentence reason.
The power-loss equation, and where it comes from
Whenever charge is pushed through a resistance, the resistance opposes the flow and that work shows up as heat, and the amount of heat depends on the square of the current.
Start from $P = VI$ and substitute the voltage across a resistor, $V = IR$, from Ohm's law:
$$P = VI = (IR)\,I = I^2R$$
This is the power dissipated as heat in a resistance $R$ carrying a current $I$. Because the current is squared, doubling the current quadruples the heat lost, while halving the current cuts the loss to a quarter. That squared dependence is exactly why engineers fight to keep transmission currents small.
The power converted to heat in a resistance is $P = I^2R$. It comes from $P = VI$ with $V = IR$. Because current is squared, power loss is very sensitive to the current, halving $I$ cuts the loss to one quarter.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: doubling the current in a wire doubles the power lost as heat.
Putting numbers into P = I²R for a real transmission line
We just saw that loss in a resistance is $P = I^2R$. That raises a question: how much energy does a real power line actually waste? This card answers it → list the current and line resistance, then substitute straight into $P = I^2R$.
The same list-rearrange-substitute routine you use for Ohm's law works here, you just need the current flowing and the resistance of the conductor.
A transmission line has a fixed resistance set by its length, material and thickness. Multiply the square of the current by that resistance to get the power wasted heating the wires.
A transmission line of resistance $4.0\ \Omega$ carries a current of $50\ \text{A}$. What power is lost as heat in the line?
To find the heat lost in a conductor, use $P = I^2R$ with the current through it and its resistance. The loss grows with the square of the current, so reducing the current is the most effective way to cut waste.
Add the highlighted method to your notes before the check below.
Quick check: A wire of resistance $2.0\ \Omega$ carries a current of $3.0\ \text{A}$. What power is lost as heat?
Pattern A, Calculate then reason
A line of resistance $5.0\ \Omega$ must deliver $100\ \text{kW}$ of power. In your book: (1) using $I = P/V$, find the line current if the power is sent at $1000\ \text{V}$, then again at $10\,000\ \text{V}$; (2) for each case calculate the power lost using $P = I^2R$; (3) write one sentence explaining why the higher transmission voltage wastes far less energy.
Match the pairs: Match each power expression to the quantities it is written in terms of.
Dot point: Solve problems relating to energy loss in circuits using P = I²R and the law of conservation of energy
Where the "lost" energy actually goes
We just learned to calculate the heat lost in a wire. That raises a question: if energy is lost from the circuit, has it been destroyed? This card answers it → no, conservation of energy means it is transformed into heat, not lost from the universe.
The "lost" energy is only lost from the useful electrical pathway; the law of conservation of energy guarantees every joule is accounted for, mostly as heat in the surroundings.
The source (battery or power station) supplies electrical energy at a certain rate, $P_{\text{supplied}}$. In a circuit this is shared between the energy delivered to the intended load and the energy turned to heat in the connecting wires:
$$P_{\text{supplied}} = P_{\text{load}} + P_{\text{lost}}$$
Nothing disappears: the heat in the wires raises their temperature and is then released to the air. This is why circuits are never $100\%$ efficient, some energy always ends up warming the wires rather than doing the job you wanted.
By conservation of energy, the power supplied equals the useful power delivered plus the power lost as heat: $P_{\text{supplied}} = P_{\text{load}} + P_{\text{lost}}$. The "lost" energy is transformed to heat in the resistance, not destroyed.
Pause, copy the highlighted energy balance into your book before the check below.
Last check: Decide if the statement is true or false.
The electrical energy "lost" in the wires of a circuit is not destroyed, it is transformed into heat, in line with the conservation of energy.
Odd one out: three of these reduce the power lost as heat in a transmission line. Which one does not?
Dot point: Solve problems relating to energy loss in circuits using P = I²R and the law of conservation of energy
A fresh set drawn from this lesson's question 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.
UnderstandBand 5(4 marks) 1. Electricity is transmitted across long distances at very high voltage rather than at the voltage used in homes. With reference to $P = I^2R$ and the conservation of energy, explain why high-voltage transmission reduces energy loss.
1 mark: to deliver a fixed power, higher voltage means lower current ($P=VI$) · 1 mark: power lost in the line is $P=I^2R$ · 1 mark: lower current means much less loss because $I$ is squared · 1 mark: the energy lost is not destroyed but transformed to heat (conservation of energy)
ApplyBand 5(4 marks) 2. A transmission line of total resistance $3.0\ \Omega$ carries a current of $40\ \text{A}$. Calculate the power lost as heat in the line. If the same energy were delivered at a current of $10\ \text{A}$, calculate the new power loss and comment on the comparison.
1 mark: $P = I^2R = 4800\ \text{W}$ · 1 mark: at $10\ \text{A}$, $P = 300\ \text{W}$ · 1 mark: the current fell by a factor of 4 · 1 mark: the loss fell by a factor of 16, because $P \propto I^2$
Dot point: Solve problems relating to energy loss in circuits using P = I²R and the law of conservation of energy
ExplainBand 5(3 marks) 3. Using the law of conservation of energy, account for what happens to the electrical energy supplied to a circuit in which a length of resistive wire becomes noticeably warm.
1 mark: the supply transfers electrical energy to the charge carriers · 1 mark: collisions with lattice ions transfer that energy to the lattice as thermal energy · 1 mark: energy is conserved, the electrical energy supplied equals the thermal energy produced plus any other output
Dot point: Solve problems relating to energy loss in circuits using P = I²R and the law of conservation of energy
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): To deliver a fixed power $P$ to consumers, the relationship $P = VI$ means that raising the transmission voltage allows the same power to be carried by a smaller current. The power wasted heating the transmission line is given by $P = I^2R$, where $R$ is the line's resistance. Because the current is squared in this expression, even a modest reduction in current produces a much larger reduction in the heat lost, for example, reducing the current to one tenth cuts the loss to one hundredth. The energy that is lost is not destroyed: by the law of conservation of energy it is transformed into heat in the resistance of the wires, which is then released to the surroundings. High-voltage, low-current transmission therefore minimises this transformation of useful electrical energy into wasted heat.
Q2 (4 marks): The heat lost in the line is $P = I^2R = (40)^2 \times 3.0 = 1600 \times 3.0 = 4800\ \text{W}$. At the lower current, $P = (10)^2 \times 3.0 = 100 \times 3.0 = 300\ \text{W}$. The current has been reduced by a factor of $4$, but the power lost has fallen by a factor of $16$, because the loss depends on the square of the current. This is the central reason electrical energy is transmitted at high voltage: for a given power delivered, $P = VI$ means a higher voltage allows a proportionally lower current, and the $I^2R$ losses then fall as the square of that reduction.
Q3 (3 marks): The supply does work on the charge carriers, giving each coulomb of charge an amount of energy equal to the potential difference, so electrical energy is delivered to the circuit at a rate $P = VI$. As the electrons drift through the resistive wire they repeatedly collide with the vibrating ions of the lattice, and at each collision they transfer some of their energy to the lattice. That energy appears as increased vibration of the ions, which is thermal energy, so the wire warms and then loses heat to the surroundings. No energy is destroyed in the process. By conservation of energy, the electrical energy supplied by the source is exactly equal to the thermal energy produced in the wire plus any energy transformed elsewhere in the circuit, which is why the total power drawn from the supply matches the total power dissipated in the components.
You can now calculate energy lost as heat with $P = I^2R$ and account for it using conservation of energy. Next lesson compares how components are connected in series and parallel circuits.
↩ Back to focus-area hubAt the start you predicted which transmission choice wastes less energy. The answer is the small current at high voltage: line loss is $P = I^2R$, and because the current is squared, lowering it dramatically reduces the heat lost. If you predicted the high-current option, revisit Card 1, power loss depends on the square of the current, not the voltage.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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