Energy dissipated in the cables

Definition

Dissipation

The spreading of energy into the thermal energy stores of the surroundings, where it is stored in a less useful way.

Every cable in the national grid has resistance, even though the cables are thick and made of good conductors such as aluminium.
When a current passes through a resistance, the cable heats up, so energy is transferred to the thermal energy store of the cable.
This energy then spreads into the air around the cable, so it is dissipated to the surroundings and never reaches a consumer.
The rate at which energy is dissipated in a cable is given by $P = I^2 R$ , where $P$ is power in watts ( $W\text{W}$ ), $I$ is current in amperes ( $A\text{A}$ ) and $R$ is resistance in ohms ( $Ω\Omega$ ).
The current appears squared in this equation, so doubling the current makes the heating four times greater.
Halving the current cuts the heating to a quarter, and cutting the current to a tenth reduces the heating to one hundredth.
The current in the cables is therefore the single biggest factor that decides how much energy the grid wastes.

Why high voltage improves efficiency

Definition

Efficiency

The useful output energy transfer divided by the total input energy transfer, which gives the proportion of energy that is transferred usefully.

The power transferred along a cable is linked to the potential difference and the current by $P = V I$ .
For a fixed amount of power, a larger potential difference means a smaller current, because $V$ and $I$ multiply together to give the same $P$ .
Raising the potential difference by a factor of $10$ therefore lowers the current by a factor of $10$ .
A current ten times smaller makes the heating in the cables $10^2 = 100$ times smaller.
Less energy is dissipated to the surroundings, so a greater proportion of the input energy reaches homes and businesses.
A greater useful share of the input energy is what a higher efficiency means, since $transfer\text{efficiency} = \dfrac{\text{useful output energy transfer}}{\text{input energy transfer}}$ .
This is why the national grid transmits power at hundreds of thousands of volts, and only a few per cent of the energy is dissipated in the transmission cables.

Example

Comparing the losses at two potential differences

A power station sends $W2\,000\,000\ \text{W}$ along cables with a total resistance of $Ω5\ \Omega$ .
At a potential difference of $V10\,000\ \text{V}$ , rearrange $P = V I$ to find the current.

\dfrac{P}{V} = \dfrac{2\,000\,000}{10\,000} = 200\ \text{A}

The power dissipated in the cables at this current is found with $P = I^2 R$ .

200^2 \times 5 = 200\,000\ \text{W}

This wastes $10%10\%$ of the power sent.
At a potential difference of $V100\,000\ \text{V}$ , the current is ten times smaller.

\dfrac{2\,000\,000}{100\,000} = 20\ \text{A}

20^2 \times 5 = 2000\ \text{W}

This wastes only $0.1%0.1\%$ of the power sent, so raising the potential difference ten times has cut the wasted power a hundred times.

Why the voltage is lowered again

A potential difference of $kV400\ \text{kV}$ is lethal, and it can even make a spark jump through the air to a nearby object.
Step-down transformers therefore reduce the potential difference before the power reaches each locality, so it is safe to bring into streets and homes.
The cables after a step-down transformer carry a larger current, but they are short, so the extra energy dissipated in them stays small.
Transformers waste a small amount of energy themselves, but far less than would be wasted by sending a large current over hundreds of kilometres.
Using high voltage for long distances and low voltage for the final short distances gives the most efficient and safe system overall.

Common Mistake

Do not write that a high voltage reduces the resistance of the cables, because the resistance stays the same and it is the current that falls.
Do not say that a high voltage on its own stops energy being wasted, because the saving comes from the smaller current that it allows.
Do not describe energy as "used up" or "destroyed" in the cables, because it is dissipated to the thermal energy store of the surroundings.
Do not claim that the grid is $100%100\%$ efficient, because the cables always have some resistance and always heat up a little.

Exam technique

Building the efficiency chain

Explain grid efficiency as a linked chain: high potential difference, then low current, then less heating of the cables, then less energy dissipated, then higher efficiency.
Each link usually earns a mark, so missing out the step about current costs marks even when the rest is correct.
Mention $P = V I$ to justify why the current falls and $P = I^2 R$ to justify why the heating falls.
Use the word dissipated and name the thermal energy store of the surroundings when you describe where the wasted energy goes.

Self review

Why do the cables of the national grid heat up?
State the equation that gives the power dissipated in a cable.
For a fixed power, what happens to the current when the potential difference is increased?
By what factor does the heating of a cable change if the current is halved?
Explain why the potential difference is reduced before the power reaches homes.

Energy dissipated in the cables

Definition

Dissipation

The spreading of energy into the thermal energy stores of the surroundings, where it is stored in a less useful way.

Every cable in the national grid has resistance, even though the cables are thick and made of good conductors such as aluminium.
When a current passes through a resistance, the cable heats up, so energy is transferred to the thermal energy store of the cable.
This energy then spreads into the air around the cable, so it is dissipated to the surroundings and never reaches a consumer.
The rate at which energy is dissipated in a cable is given by $P = I^2 R$ , where $P$ is power in watts ( $W\text{W}$ ), $I$ is current in amperes ( $A\text{A}$ ) and $R$ is resistance in ohms ( $Ω\Omega$ ).
The current appears squared in this equation, so doubling the current makes the heating four times greater.
Halving the current cuts the heating to a quarter, and cutting the current to a tenth reduces the heating to one hundredth.
The current in the cables is therefore the single biggest factor that decides how much energy the grid wastes.

Why high voltage improves efficiency

Definition

Efficiency

The useful output energy transfer divided by the total input energy transfer, which gives the proportion of energy that is transferred usefully.

The power transferred along a cable is linked to the potential difference and the current by $P = V I$ .
For a fixed amount of power, a larger potential difference means a smaller current, because $V$ and $I$ multiply together to give the same $P$ .
Raising the potential difference by a factor of $10$ therefore lowers the current by a factor of $10$ .
A current ten times smaller makes the heating in the cables $10^2 = 100$ times smaller.
Less energy is dissipated to the surroundings, so a greater proportion of the input energy reaches homes and businesses.
A greater useful share of the input energy is what a higher efficiency means, since $transfer\text{efficiency} = \dfrac{\text{useful output energy transfer}}{\text{input energy transfer}}$ .
This is why the national grid transmits power at hundreds of thousands of volts, and only a few per cent of the energy is dissipated in the transmission cables.

Example

Comparing the losses at two potential differences

A power station sends $W2\,000\,000\ \text{W}$ along cables with a total resistance of $Ω5\ \Omega$ .
At a potential difference of $V10\,000\ \text{V}$ , rearrange $P = V I$ to find the current.

\dfrac{P}{V} = \dfrac{2\,000\,000}{10\,000} = 200\ \text{A}

The power dissipated in the cables at this current is found with $P = I^2 R$ .

200^2 \times 5 = 200\,000\ \text{W}

This wastes $10%10\%$ of the power sent.
At a potential difference of $V100\,000\ \text{V}$ , the current is ten times smaller.

\dfrac{2\,000\,000}{100\,000} = 20\ \text{A}

20^2 \times 5 = 2000\ \text{W}

This wastes only $0.1%0.1\%$ of the power sent, so raising the potential difference ten times has cut the wasted power a hundred times.

Why the voltage is lowered again

A potential difference of $kV400\ \text{kV}$ is lethal, and it can even make a spark jump through the air to a nearby object.
Step-down transformers therefore reduce the potential difference before the power reaches each locality, so it is safe to bring into streets and homes.
The cables after a step-down transformer carry a larger current, but they are short, so the extra energy dissipated in them stays small.
Transformers waste a small amount of energy themselves, but far less than would be wasted by sending a large current over hundreds of kilometres.
Using high voltage for long distances and low voltage for the final short distances gives the most efficient and safe system overall.

Common Mistake

Do not write that a high voltage reduces the resistance of the cables, because the resistance stays the same and it is the current that falls.
Do not say that a high voltage on its own stops energy being wasted, because the saving comes from the smaller current that it allows.
Do not describe energy as "used up" or "destroyed" in the cables, because it is dissipated to the thermal energy store of the surroundings.
Do not claim that the grid is $100%100\%$ efficient, because the cables always have some resistance and always heat up a little.

Exam technique

Building the efficiency chain

Explain grid efficiency as a linked chain: high potential difference, then low current, then less heating of the cables, then less energy dissipated, then higher efficiency.
Each link usually earns a mark, so missing out the step about current costs marks even when the rest is correct.
Mention $P = V I$ to justify why the current falls and $P = I^2 R$ to justify why the heating falls.
Use the word dissipated and name the thermal energy store of the surroundings when you describe where the wasted energy goes.

Self review

Why do the cables of the national grid heat up?
State the equation that gives the power dissipated in a cable.
For a fixed power, what happens to the current when the potential difference is increased?
By what factor does the heating of a cable change if the current is halved?
Explain why the potential difference is reduced before the power reaches homes.

8.2.5 Efficiency of the national grid

Energy dissipated in the cables

Why high voltage improves efficiency

Comparing the losses at two potential differences

Why the voltage is lowered again

Building the efficiency chain

Review 8.2.5 Efficiency of the national grid by teaching Genie

8.2.5 Efficiency of the national grid

Energy dissipated in the cables

Why high voltage improves efficiency

Comparing the losses at two potential differences

Why the voltage is lowered again

Building the efficiency chain

Review 8.2.5 Efficiency of the national grid by teaching Genie

8.2.5 Efficiency of the national grid

Energy dissipated in the cables

Why high voltage improves efficiency

Comparing the losses at two potential differences

Why the voltage is lowered again

Building the efficiency chain

Review 8.2.5 Efficiency of the national grid by teaching Genie

8.2.5 Efficiency of the national grid Revision Guide

8.2.5 Efficiency of the national grid

Energy dissipated in the cables

Why high voltage improves efficiency

Comparing the losses at two potential differences

Why the voltage is lowered again

Building the efficiency chain

Review 8.2.5 Efficiency of the national grid by teaching Genie

8.2.5 Efficiency of the national grid Revision Guide