Power transmission

Lighting a room that is far from the power station

The electricity that lights your room may have been made hundreds of kilometres away. It travels along wires. Wires have resistance, so they heat up. Every joule of heat in the line is a joule that never reaches your home. How do we send a lot of power over a long way and lose only a little?

The heat in the line

Let the power station send the power $P$ at a voltage $V$. The line has a total resistance $r$, and the current in it is $I$. If the power factor is $1$, the power sent is $P = VI$. We want $I$, so:

The line has resistance, so it heats at the rate $I^2r$. Put in the current:

The share of the power that is lost is:

The voltage $V$ is squared in the bottom. If the voltage is ten times larger, the current is ten times smaller, and the loss is a hundred times smaller. This is why the voltage is raised for the journey.

A worked example

A power station sends $500$ kW along a line of total resistance $12\ \Omega$. Compare the loss when it sends at $5.0$ kV with the loss when it first steps the voltage up to $50$ kV. The power factor is $1$.

1. At $5.0$ kV. $I = \dfrac{P}{V} = \dfrac{500 \times 10^3\ \text{W}}{5.0 \times 10^3\ \text{V}} = 100$ A. The heat is $I^2r = (100\ \text{A})^2 \times 12\ \Omega = 120\,000\ \text{W} = 120$ kW. This is $\dfrac{120}{500} = 24\ \%$ of the power.
2. At $50$ kV. $I = \dfrac{500 \times 10^3\ \text{W}}{50 \times 10^3\ \text{V}} = 10$ A. The heat is $(10\ \text{A})^2 \times 12\ \Omega = 1200\ \text{W} = 1.2$ kW. This is $\dfrac{1.2}{500} = 0.24\ \%$ of the power.

The voltage went up ten times. The current went down ten times. The loss went down a hundred times, from $120$ kW to $1.2$ kW.

How much voltage is lost along the line? At $50$ kV the current is $10$ A. The voltage lost along the line is $Ir = 10\ \text{A} \times 12\ \Omega = 120$ V. This is only $0.24\ \%$ of $50\,000$ V. At $5.0$ kV the current is $100$ A, and the voltage lost is $1200$ V, which is $24\ \%$ of $5000$ V.

Take care: the voltage of the line is not across the line

A very common slip is to write the loss as $\dfrac{V^2}{r}$, with $V$ the transmission voltage. Test it with the numbers above. At $50$ kV it gives $\dfrac{(50\,000\ \text{V})^2}{12\ \Omega} = 208$ MW. This is far more than the $500$ kW that was sent, which cannot be right.

The mistake is that $V$ is the voltage between the two wires of the line, as measured at the station. It is not the voltage across the resistance of the line. Only the small voltage lost along the line, $Ir$, is across it. So the heat is $I^2r$, or $\dfrac{(Ir)^2}{r}$, and never $\dfrac{V^2}{r}$.

From the power station to your socket

The electricity goes through several transformers on its way.

1. The alternator. The generator at the station makes an alternating voltage of about $11$ kV.
2. The step-up transformer. It raises the voltage to a very high value for the long lines. In India the lines on tall towers carry $132$, $220$ or $400$ kV.
3. The substation. A step-down transformer lowers the voltage to about $33$ kV and then $11$ kV for a town.
4. The pole transformer. Near your home a last step-down transformer gives the $230$ V of the socket.

All of this works because the current is alternating. A transformer can raise or lower an alternating voltage with an efficiency of more than $95\ \%$. For a steady voltage there is no such simple device. Some very long lines now use direct current, with electronic converters at each end, but the loss $I^2r$ follows the same rule.

Why not thicker wire, or a still higher voltage?

We could also cut the loss by making $r$ small, with thicker wires. But a thick copper or aluminium cable costs more and weighs more, and the towers would have to be stronger. A higher voltage costs too. The wires need larger insulators and must be kept farther from the tower and from the ground, and at the highest voltages the air starts to leak. Engineers choose a voltage and a wire so that the sum of all these costs is the least.

Power factor and the line

You have met the power factor of a circuit before. A load with a power factor $\cos\phi$ takes a current $I = \dfrac{P}{V\cos\phi}$ for the power $P$. A low power factor makes the current in the line larger, so the loss is larger by $\dfrac{1}{\cos^2\phi}$. A power factor of $0.8$ makes the loss $1.56$ times larger. So the supply companies ask big users to keep their power factor close to $1$.

Start at $5$ kV. Make the transmission voltage $10$ kV. By what factor does the share lost fall? Now go back to $5$ kV and set the power factor to $0.7$. What happens to the loss?

Practise with new numbers

Try it in Python

The program sends $800$ kW through a line of $10\ \Omega$ at five different voltages. For each voltage it finds the current, the loss and the share lost.

Common mistakes

• Writing the loss as $V^2/r$. The voltage $V$ is not across the line. Use $I^2r$.
• Forgetting that the current is $P/V$. Do not take the current as a fixed number. It changes when the transmission voltage changes.
• Keeping kW or kV in a formula for W and V. Change them first: $1\ \text{kW} = 1000\ \text{W}$ and $1\ \text{kV} = 1000\ \text{V}$.
• Thinking that a higher voltage means a larger current. The same power at a higher voltage needs a smaller current.
• Thinking the transformer makes power. It changes voltage and current. The power does not grow.
• Mixing up the share lost and the loss. The loss is in watt. The share is the loss divided by the power sent.

Remember these

• The line current is $I = \dfrac{P}{V}$ (power factor $1$), or $\dfrac{P}{V\cos\phi}$.
• The heat in the line is $I^2r = \dfrac{P^2r}{V^2}$, and the share lost is $\dfrac{Pr}{V^2}$.
• A transformer raises the voltage at the station and lowers it near the user. Ten times the voltage gives a hundred times less loss.
• The voltage lost along the line is $Ir$. It is not the whole voltage of the line.
• A low power factor raises the line current and the loss.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A power station sends $2.0$ MW at $40$ kV. What current is in the line?

2. Why is electric power sent at a very high voltage, and not at the $230$ V that a house uses?

3. The voltage at which the power is sent is made three times larger. By what factor does the loss in the line change?

4. A line carries $20$ A and has a resistance of $5.0\ \Omega$. What is the heat in the line, and what voltage is lost along it?