Coils and turns

Definition

Primary coil

The input coil of a transformer, which is connected to the alternating supply.

Definition

Secondary coil

The output coil of a transformer, across which the changed alternating potential difference is produced.

A transformer has two coils of insulated wire wrapped around a shared iron core.
The primary coil is connected to the alternating supply, so it is the coil that power enters.
The secondary coil is connected to the output circuit, so it is the coil that power leaves.
Each coil is made of many loops of wire, and each complete loop is called one turn.
The number of turns on the primary coil is written $NpN_{\text{p}}$ and the number on the secondary coil is written $NsN_{\text{s}}$ .
The potential difference across the primary coil is written $VpV_{\text{p}}$ and the one across the secondary coil is written $VsV_{\text{s}}$ .

The turns ratio equation

Definition

Turns ratio

The number of turns on the primary coil of a transformer compared with the number of turns on its secondary coil.

The ratio of the two potential differences is equal to the ratio of the numbers of turns on the two coils.
This is written as $VpVs=NpNs\dfrac{V_{\text{p}}}{V_{\text{s}}} = \dfrac{N_{\text{p}}}{N_{\text{s}}}$ , where both potential differences are in volts and the turns are plain numbers with no unit.
If the secondary coil has more turns than the primary coil ( $Ns>NpN_{\text{s}} > N_{\text{p}}$ ), the secondary potential difference is larger, so the transformer is step-up.
If the secondary coil has fewer turns than the primary coil ( $Ns<NpN_{\text{s}} < N_{\text{p}}$ ), the secondary potential difference is smaller, so the transformer is step-down.
A secondary coil with $16$ times as many turns as the primary therefore gives a secondary potential difference $16$ times larger.
Both potential differences must be in the same unit before you substitute, so either convert both to volts or keep both in kilovolts.
Rearranging for the secondary potential difference gives $Vs=Vp×NsNpV_{\text{s}} = V_{\text{p}} \times \dfrac{N_{\text{s}}}{N_{\text{p}}}$ .

Example

Finding the output of a grid transformer

A power station transformer has $1000$ turns on its primary coil and $00016\,000$ turns on its secondary coil.
The generators supply a primary potential difference of $V25\,000\ \text{V}$ .

V_{\text{s}} = V_{\text{p}} \times \dfrac{N_{\text{s}}}{N_{\text{p}}}

V_{\text{s}} = 25\,000 \times \dfrac{16\,000}{1000} = 400\,000\ \text{V}

The secondary potential difference is $kV400\ \text{kV}$ , so this is a step-up transformer.

Power in a transformer

A transformer that wastes no energy is described as $100%100\%$ efficient.
In such a transformer, the power supplied to the primary coil equals the power delivered by the secondary coil, because energy is conserved.
Power is potential difference multiplied by current, so this gives $VpIp=VsIsV_{\text{p}} I_{\text{p}} = V_{\text{s}} I_{\text{s}}$ .
In this equation $IpI_{\text{p}}$ is the current in the primary coil and $IsI_{\text{s}}$ is the current in the secondary coil, both in amperes.
The product of potential difference and current stays the same, so if a transformer raises the potential difference by a factor, it lowers the current by the same factor.
A step-up transformer with $16$ times as many secondary turns therefore gives $16$ times the potential difference and one sixteenth of the current.
A step-down transformer does the reverse, lowering the potential difference and raising the current.
Real grid transformers are slightly less than $100%100\%$ efficient, so their output power is a little smaller than their input power.

Example

Finding the current after stepping up

The same transformer receives a primary current of $A4000\ \text{A}$ at $V25\,000\ \text{V}$ and gives out $V400\,000\ \text{V}$ .
Assume the transformer is $100%100\%$ efficient and rearrange the power equation for the secondary current.

I_{\text{s}} = \dfrac{V_{\text{p}} I_{\text{p}}}{V_{\text{s}}}

I_{\text{s}} = \dfrac{25\,000 \times 4000}{400\,000} = 250\ \text{A}

The power transferred is $W25\,000 \times 4000 = 100\,000\,000\ \text{W}$ on both sides of the transformer.
The potential difference has risen by a factor of $16$ and the current has fallen by a factor of $16$ .

Advantages of high-voltage transmission

The power dissipated as heat in a transmission cable is $P = I^2 R$ , so it depends on the square of the current.
A step-up transformer that raises the potential difference by a factor $n$ lowers the current by a factor $n$ , which reduces the heating in the cables by a factor $n^2$ .
In the grid transformer above, the current falls by a factor of $16$ , so the heating in the cables falls by a factor of $16^2 = 256$ .
Less energy is dissipated, so more of the energy generated reaches consumers and the transmission is more efficient.
Less fuel has to be burned to deliver the same energy to homes, which lowers costs and carbon dioxide emissions.
A smaller current can be carried safely by thinner, lighter cables, which allows lighter pylons spaced further apart.
The drawbacks are that transformers, tall pylons and long insulators are expensive, and a very high potential difference is dangerous, but the energy saved over the life of the grid outweighs these costs.

Example

Comparing cable losses before and after stepping up

The $W100\,000\,000\ \text{W}$ from the power station travels through cables with a total resistance of $Ω4\ \Omega$ .
Without stepping up, the current would be $A4000\ \text{A}$ .

I^2 R = 4000^2 \times 4 = 64\,000\,000\ \text{W}

Almost two-thirds of the power would be dissipated in the cables.
After stepping up to $kV400\ \text{kV}$ , the current is $A250\ \text{A}$ .

I^2 R = 250^2 \times 4 = 250\,000\ \text{W}

Only $0.25%0.25\%$ of the power is now dissipated, which is $256$ times less than before.

Common Mistake

Do not put a primary value with a secondary value on the same side of the turns ratio equation, because each side must compare primary with secondary in the same order.
Do not assume that a step-up transformer increases the power, because the current falls by the same factor as the potential difference rises.
Do not mix kilovolts and volts in one calculation, because the answer will be wrong by a factor of $1000$ .
Do not forget to square the current when you work out how much the heating in the cables changes.

Exam technique

Tackling transformer calculations

Write down the equation you are using before you substitute, because the equation and the substitution often earn marks even if the final arithmetic slips.
List the known values with labels such as $VpV_{\text{p}}$ , $NsN_{\text{s}}$ and $IsI_{\text{s}}$ so that you cannot swap primary and secondary values.
Use the turns ratio equation when the question gives numbers of turns, and the power equation when it gives currents.
Check your answer makes sense, since a step-up transformer must give a larger potential difference and a smaller current than it receives.
When asked why high-voltage transmission is better, link the smaller current to $P = I^2 R$ and finish with less energy dissipated in the cables.

Self review

State the equation that links the potential differences and numbers of turns of a transformer.
Is a transformer with more turns on its secondary coil than on its primary coil step-up or step-down?
State the equation that links the potential differences and currents of a $100%100\%$ efficient transformer.
By what factor does the current change when a transformer raises the potential difference by a factor of $20$ ?
Explain why reducing the current by a factor of $10$ reduces the heating in the cables by a factor of $100$ .

Coils and turns

Definition

Primary coil

The input coil of a transformer, which is connected to the alternating supply.

Definition

Secondary coil

The output coil of a transformer, across which the changed alternating potential difference is produced.

A transformer has two coils of insulated wire wrapped around a shared iron core.
The primary coil is connected to the alternating supply, so it is the coil that power enters.
The secondary coil is connected to the output circuit, so it is the coil that power leaves.
Each coil is made of many loops of wire, and each complete loop is called one turn.
The number of turns on the primary coil is written $NpN_{\text{p}}$ and the number on the secondary coil is written $NsN_{\text{s}}$ .
The potential difference across the primary coil is written $VpV_{\text{p}}$ and the one across the secondary coil is written $VsV_{\text{s}}$ .

The turns ratio equation

Definition

Turns ratio

The number of turns on the primary coil of a transformer compared with the number of turns on its secondary coil.

The ratio of the two potential differences is equal to the ratio of the numbers of turns on the two coils.
This is written as $VpVs=NpNs\dfrac{V_{\text{p}}}{V_{\text{s}}} = \dfrac{N_{\text{p}}}{N_{\text{s}}}$ , where both potential differences are in volts and the turns are plain numbers with no unit.
If the secondary coil has more turns than the primary coil ( $Ns>NpN_{\text{s}} > N_{\text{p}}$ ), the secondary potential difference is larger, so the transformer is step-up.
If the secondary coil has fewer turns than the primary coil ( $Ns<NpN_{\text{s}} < N_{\text{p}}$ ), the secondary potential difference is smaller, so the transformer is step-down.
A secondary coil with $16$ times as many turns as the primary therefore gives a secondary potential difference $16$ times larger.
Both potential differences must be in the same unit before you substitute, so either convert both to volts or keep both in kilovolts.
Rearranging for the secondary potential difference gives $Vs=Vp×NsNpV_{\text{s}} = V_{\text{p}} \times \dfrac{N_{\text{s}}}{N_{\text{p}}}$ .

Example

Finding the output of a grid transformer

A power station transformer has $1000$ turns on its primary coil and $00016\,000$ turns on its secondary coil.
The generators supply a primary potential difference of $V25\,000\ \text{V}$ .

V_{\text{s}} = V_{\text{p}} \times \dfrac{N_{\text{s}}}{N_{\text{p}}}

V_{\text{s}} = 25\,000 \times \dfrac{16\,000}{1000} = 400\,000\ \text{V}

The secondary potential difference is $kV400\ \text{kV}$ , so this is a step-up transformer.

Power in a transformer

A transformer that wastes no energy is described as $100%100\%$ efficient.
In such a transformer, the power supplied to the primary coil equals the power delivered by the secondary coil, because energy is conserved.
Power is potential difference multiplied by current, so this gives $VpIp=VsIsV_{\text{p}} I_{\text{p}} = V_{\text{s}} I_{\text{s}}$ .
In this equation $IpI_{\text{p}}$ is the current in the primary coil and $IsI_{\text{s}}$ is the current in the secondary coil, both in amperes.
The product of potential difference and current stays the same, so if a transformer raises the potential difference by a factor, it lowers the current by the same factor.
A step-up transformer with $16$ times as many secondary turns therefore gives $16$ times the potential difference and one sixteenth of the current.
A step-down transformer does the reverse, lowering the potential difference and raising the current.
Real grid transformers are slightly less than $100%100\%$ efficient, so their output power is a little smaller than their input power.

Example

Finding the current after stepping up

The same transformer receives a primary current of $A4000\ \text{A}$ at $V25\,000\ \text{V}$ and gives out $V400\,000\ \text{V}$ .
Assume the transformer is $100%100\%$ efficient and rearrange the power equation for the secondary current.

I_{\text{s}} = \dfrac{V_{\text{p}} I_{\text{p}}}{V_{\text{s}}}

I_{\text{s}} = \dfrac{25\,000 \times 4000}{400\,000} = 250\ \text{A}

The power transferred is $W25\,000 \times 4000 = 100\,000\,000\ \text{W}$ on both sides of the transformer.
The potential difference has risen by a factor of $16$ and the current has fallen by a factor of $16$ .

Advantages of high-voltage transmission

The power dissipated as heat in a transmission cable is $P = I^2 R$ , so it depends on the square of the current.
A step-up transformer that raises the potential difference by a factor $n$ lowers the current by a factor $n$ , which reduces the heating in the cables by a factor $n^2$ .
In the grid transformer above, the current falls by a factor of $16$ , so the heating in the cables falls by a factor of $16^2 = 256$ .
Less energy is dissipated, so more of the energy generated reaches consumers and the transmission is more efficient.
Less fuel has to be burned to deliver the same energy to homes, which lowers costs and carbon dioxide emissions.
A smaller current can be carried safely by thinner, lighter cables, which allows lighter pylons spaced further apart.
The drawbacks are that transformers, tall pylons and long insulators are expensive, and a very high potential difference is dangerous, but the energy saved over the life of the grid outweighs these costs.

Example

Comparing cable losses before and after stepping up

The $W100\,000\,000\ \text{W}$ from the power station travels through cables with a total resistance of $Ω4\ \Omega$ .
Without stepping up, the current would be $A4000\ \text{A}$ .

I^2 R = 4000^2 \times 4 = 64\,000\,000\ \text{W}

Almost two-thirds of the power would be dissipated in the cables.
After stepping up to $kV400\ \text{kV}$ , the current is $A250\ \text{A}$ .

I^2 R = 250^2 \times 4 = 250\,000\ \text{W}

Only $0.25%0.25\%$ of the power is now dissipated, which is $256$ times less than before.

Common Mistake

Do not put a primary value with a secondary value on the same side of the turns ratio equation, because each side must compare primary with secondary in the same order.
Do not assume that a step-up transformer increases the power, because the current falls by the same factor as the potential difference rises.
Do not mix kilovolts and volts in one calculation, because the answer will be wrong by a factor of $1000$ .
Do not forget to square the current when you work out how much the heating in the cables changes.

Exam technique

Tackling transformer calculations

Write down the equation you are using before you substitute, because the equation and the substitution often earn marks even if the final arithmetic slips.
List the known values with labels such as $VpV_{\text{p}}$ , $NsN_{\text{s}}$ and $IsI_{\text{s}}$ so that you cannot swap primary and secondary values.
Use the turns ratio equation when the question gives numbers of turns, and the power equation when it gives currents.
Check your answer makes sense, since a step-up transformer must give a larger potential difference and a smaller current than it receives.
When asked why high-voltage transmission is better, link the smaller current to $P = I^2 R$ and finish with less energy dissipated in the cables.

Self review

State the equation that links the potential differences and numbers of turns of a transformer.
Is a transformer with more turns on its secondary coil than on its primary coil step-up or step-down?
State the equation that links the potential differences and currents of a $100%100\%$ efficient transformer.
By what factor does the current change when a transformer raises the potential difference by a factor of $20$ ?
Explain why reducing the current by a factor of $10$ reduces the heating in the cables by a factor of $100$ .

8.2.6 Transformers and power transmission at high voltage

Coils and turns

The turns ratio equation

Finding the output of a grid transformer

Power in a transformer

Finding the current after stepping up

Advantages of high-voltage transmission

Comparing cable losses before and after stepping up

Tackling transformer calculations

Review 8.2.6 Transformers and power transmission at high voltage by teaching Genie

8.2.6 Transformers and power transmission at high voltage

Coils and turns

The turns ratio equation

Finding the output of a grid transformer

Power in a transformer

Finding the current after stepping up

Advantages of high-voltage transmission

Comparing cable losses before and after stepping up

Tackling transformer calculations

Review 8.2.6 Transformers and power transmission at high voltage by teaching Genie

8.2.6 Transformers and power transmission at high voltage

Coils and turns

The turns ratio equation

Finding the output of a grid transformer

Power in a transformer

Finding the current after stepping up

Advantages of high-voltage transmission

Comparing cable losses before and after stepping up

Tackling transformer calculations

Review 8.2.6 Transformers and power transmission at high voltage by teaching Genie

8.2.6 Transformers and power transmission at high voltage Revision Guide

8.2.6 Transformers and power transmission at high voltage

Coils and turns

The turns ratio equation

Finding the output of a grid transformer

Power in a transformer

Finding the current after stepping up

Advantages of high-voltage transmission

Comparing cable losses before and after stepping up

Tackling transformer calculations

Review 8.2.6 Transformers and power transmission at high voltage by teaching Genie

8.2.6 Transformers and power transmission at high voltage Revision Guide