The National Grid

The electricity generated in a power station may have to travel a hundred miles or more through cables before it reaches a home or a factory. Along the way, some of the energy is inevitably wasted as the cables warm up. The clever engineering that keeps these losses small is the reason the wires slung between pylons carry electricity at the astonishing voltage of $400\,000 \text{ V}$400000 V rather than the safe $230 \text{ V}$230 V of a household socket. The system of cables, pylons and transformers that carries electricity from power stations to consumers is called the National Grid, and this lesson explains why it transmits power at high voltage and low current, how transformers step the voltage up and down, and how this keeps transmission efficient.

By the end of this lesson you should be able to describe the structure of the National Grid, explain why electricity is transmitted at high voltage and low current to reduce power loss in the cables, describe the roles of step-up and step-down transformers, and link the efficiency of transmission to the $I^2R$I2R heating in the cables (connecting to the transformer ideas of Topic P4).

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What Is the National Grid?

The National Grid is the nationwide network of cables and transformers that connects power stations to homes, schools, shops and factories. Its job is to move electrical energy from where it is generated to where it is needed, reliably and with as little waste as possible.

The problem the grid must solve is that transmitting a large amount of power over long cables causes the cables to heat up, wasting energy. The amount of energy wasted depends on the current flowing in the cables — and, as we shall see, keeping the current low is the key to keeping the losses small. To do this, the grid transmits electricity at a very high voltage and correspondingly low current, using transformers to change the voltage at each end.

Exam Tip: The National Grid is a system of cables and transformers. Two transformers matter: a step-up transformer at the power station raises the voltage for transmission, and a step-down transformer near the consumer lowers it to a safe, usable value.

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Why Transmit at High Voltage and Low Current?

This is the central idea of the lesson. The power wasted heating the transmission cables is given by:

$P = I^2R$P=I2R

where $P$P is the power lost, $I$I is the current in the cables and $R$R is the resistance of the cables. The crucial feature is the $I^2$I2: the power wasted depends on the square of the current. Halving the current cuts the power loss to a quarter; reducing the current to a tenth cuts the loss to a hundredth. So to keep the losses small, we must keep the current as low as possible.

But the power that has to be delivered is fixed, and the power transmitted is given by:

$P = VI$P=VI

where $V$V is the voltage and $I$I is the current. To transmit a given power $P$P with a low current $I$I, the voltage $V$V must be high — because $P = VI$P=VI means that if $I$I is small, $V$V must be large to keep the product the same.

Putting the two ideas together: transmitting at high voltage allows the current to be low, and a low current means very little power is wasted as heat in the cables, because the loss goes as $I^2R$I2R. This is why the grid uses $400\,000 \text{ V}$400000 V on the transmission lines.

Worked Example 1 — Power loss depends on current squared

A cable of resistance $5 \text{ }\Omega$5 Ω carries a current of $100 \text{ A}$100 A. Calculate the power wasted heating the cable. Then find the power wasted if the current is reduced to $20 \text{ A}$20 A.

Step 1 — use $P = I^2R$P=I2R with $I = 100 \text{ A}$I=100 A: $P = 100^2 \times 5 = 10\,000 \times 5 = 50\,000 \text{ W}$P=1002×5=10000×5=50000 W.

Step 2 — repeat with $I = 20 \text{ A}$I=20 A: $P = 20^2 \times 5 = 400 \times 5 = 2000 \text{ W}$P=202×5=400×5=2000 W.

Step 3 — compare: the current fell to a fifth, and the power loss fell to $\left(\dfrac{1}{5}\right)^2 = \dfrac{1}{25}$(51​)2=251​ — from $50\,000 \text{ W}$50000 W to $2000 \text{ W}$2000 W.

Answer: reducing the current from $100 \text{ A}$100 A to $20 \text{ A}$20 A cuts the wasted power from $50\,000 \text{ W}$50000 W to $2000 \text{ W}$2000 W — a huge saving, and exactly why high-voltage, low-current transmission is used.

Worked Example 2 — High voltage means low current

A power station transmits $200 \text{ MW}$200 MW of power ($200\,000\,000 \text{ W}$200000000 W). Compare the current needed at $25\,000 \text{ V}$25000 V with the current needed at $400\,000 \text{ V}$400000 V.

Step 1 — rearrange $P = VI$P=VI to $I = \dfrac{P}{V}$I=VP​.

Step 2 — at $25\,000 \text{ V}$25000 V: $I = \dfrac{200\,000\,000}{25\,000} = 8000 \text{ A}$I=25000200000000​=8000 A.

Step 3 — at $400\,000 \text{ V}$400000 V: $I = \dfrac{200\,000\,000}{400\,000} = 500 \text{ A}$I=400000200000000​=500 A.

Answer: raising the voltage from $25\,000 \text{ V}$25000 V to $400\,000 \text{ V}$400000 V cuts the current from $8000 \text{ A}$8000 A to $500 \text{ A}$500 A. Because power loss goes as $I^2$I2, this dramatically reduces the energy wasted in the cables.

Exam Tip: The reasoning chain is: high voltage → (from $P = VI$P=VI) low current → (from $P = I^2R$P=I2R) very small power loss. Quote both equations to earn full marks — one shows the current is low, the other shows why that matters.

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Step-up and Step-down Transformers

A transformer changes the size of an alternating voltage (this links to Topic P4). The National Grid uses two kinds: