Power wasted in transmission cables

1. Transmission cables are made of metal and therefore have resistance, so the current in them does work against that resistance and heats the cables and the surrounding air.
2. The power dissipated in a cable is $P=I^{2}R$, where $I$ is the current in the cable in amperes and $R$ is the resistance of the cable in ohms.
3. Because the current is squared in that expression, the wasted power is very sensitive to the current: halving the current cuts the waste to a quarter.
4. The resistance of a cable increases with its length, so a long transmission route wastes more power than a short one carrying the same current.
5. Thick cables of low-resistance metal such as aluminium keep $R$ small, but the resistance can never be reduced to zero.
6. Energy dissipated in this way heats the surroundings and is no longer available to the consumer, so it is paid for by the generator but delivered to nobody.

Raising the potential difference

1. The power that must be delivered is fixed by what consumers are using, and it is linked to the transmission potential difference and current by $P=VI$.
2. Rearranging that relationship gives $I=\frac{P}{V}$, so transmitting the same power at a higher potential difference needs a smaller current.
3. A smaller current in the cables means much less power is dissipated, because the loss depends on $I^2$ rather than on $I$.
4. Raising the transmission potential difference by a factor of ten cuts the current to a tenth and cuts the wasted power to a hundredth of its previous value.
5. A step-up transformer at the power station provides the high potential difference for transmission, and a step-down transformer near the consumer returns it to a safe level.
6. The overall effect is that a far larger fraction of the generated energy arrives at the consumer, so the grid is much more efficient than it would be at a low potential difference.
7. Transmitting at $230V$ over long distances would mean currents of thousands of amperes and losses larger than the power delivered, which is why the grid cannot work that way.

Working through the numbers

1. Identify the fixed quantity first, which is almost always the power that has to be delivered.
2. Find the current from $I=\frac{P}{V}$ using the transmission potential difference, not the domestic one.
3. Put that current into $P=I^{2}R$ with the resistance of the whole route to find the power dissipated.
4. Subtract the dissipated power from the transmitted power to find what reaches the consumer, and divide to find a percentage where one is asked for.
5. Watch for prefixes, since a mixture of megawatts, kilovolts and ohms in one question is the commonest source of a wrong power of ten.

Costs and practical limits

1. A smaller current allows thinner and lighter cables, so pylons can be spaced further apart and the network is cheaper to build.
2. Cables that carry less current run cooler, so they expand and sag less and suffer less wear over time.
3. A very high potential difference is dangerous, so cables are suspended high above the ground on insulators and the pylons carry warning signs.
4. Very high potential differences need thicker insulation, taller pylons and larger, more expensive transformers, so the cost rises with the potential difference chosen.
5. At extremely high potential differences some energy is also lost to the air around the cables, which sets a practical ceiling on how far the potential difference can usefully be raised.
6. The potential difference used on each part of the grid is therefore a compromise between the energy saved and the cost and danger of the equipment needed.