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7.3.4 Transformers (HT only)

7.3.4a Transformer construction and the turns ratio

Transformer construction

Definition

Transformer

A transformer is a device that uses electromagnetic induction to change an alternating potential difference.

  1. The primary coil is connected to the input alternating potential difference.
  2. The secondary coil is connected to the output circuit.
  3. Both coils are wound on an iron core.
  4. The two coils are separate, with no direct electrical connection; energy passes between them through a changing magnetic field in the core.
  5. Iron is used because it is easily magnetised, so the changing field from the primary coil passes effectively through the secondary coil.

How a transformer works

  1. A transformer needs an alternating current in the primary coil, because it continually changes direction and size.
  2. The alternating current in the primary coil produces a changing magnetic field.
  3. The iron core becomes magnetised and carries the changing field through the secondary coil.
  4. The changing field induces an alternating potential difference across the secondary coil.
  5. If the secondary circuit is complete, this drives an alternating current in it.
  6. A steady direct current gives no continuous transformer action, because a constant current's field is not changing.
Key Idea

Current is not passed directly from one coil to the other; a changing magnetic field links the coils and induces a potential difference across the secondary coil.

The transformer turns ratio

  1. The potential difference across each coil depends on its number of turns, given by VpVs=npns\dfrac{V_p}{V_s} = \dfrac{n_p}{n_s}Vs​Vp​​=ns​np​​.
  2. VpV_pVp​ is the potential difference across the primary coil, in volts, V\text{V}V.
  3. VsV_sVs​ is the potential difference across the secondary coil, in volts, V\text{V}V.
  4. npn_pnp​ is the number of turns on the primary coil.
  5. nsn_sns​ is the number of turns on the secondary coil.
  6. Write both ratios in the same order: if primary is on top for the potential differences, primary must be on top for the turns.
  7. A coil with more turns has the greater potential difference, so a secondary coil with five times as many turns has five times the potential difference.
Example

Question: A transformer has 400400400 turns on the primary coil and 200020002000 turns on the secondary coil, with 12 V12\ \text{V}12 V across the primary. Calculate the secondary potential difference.

  1. Write the equation: VpVs=npns\dfrac{V_p}{V_s} = \dfrac{n_p}{n_s}Vs​Vp​​=ns​np​​.
  2. Substitute the values: 12Vs=4002000\dfrac{12}{V_s} = \dfrac{400}{2000}Vs​12​=2000400​.
  3. Rearrange: Vs=12×2000400V_s = \dfrac{12 \times 2000}{400}Vs​=40012×2000​.
  4. Calculate: Vs=60 VV_s = 60\ \text{V}Vs​=60 V, which is greater than the primary, so this is a step-up transformer.

Step-up and step-down transformers

  1. A step-up transformer increases the potential difference, so Vs>VpV_s > V_pVs​>Vp​ and it has more secondary turns, ns>npn_s > n_pns​>np​.
  2. A step-down transformer decreases the potential difference, so Vs<VpV_s < V_pVs​<Vp​ and it has fewer secondary turns, ns<npn_s < n_pns​<np​.
Common Mistake
  • The word primary does not mean the larger potential difference; the primary is just the input coil and the secondary is the output coil.
  • A changing field induces a potential difference across the secondary coil; a current flows only if the secondary circuit is complete.
Exam technique
  • For how a transformer works, give the chain: alternating current in the primary, changing field in the iron core, induced potential difference in the secondary, current if the secondary circuit is complete.
  • In calculations keep primary quantities together and secondary quantities together, then compare VsV_sVs​ with VpV_pVp​ to decide step-up or step-down.
Self review
  • What are the three main parts of a basic transformer?
  • Why is iron used for the core?
  • Why must the primary current be alternating?
  • State the transformer turns-ratio equation.
  • How do the numbers of turns compare in a step-up transformer?
  • What extra condition is needed for a current to flow in the secondary circuit?

7.3.4b Transformers and power transmission

Power input and power output

Definition

Ideal transformer

An ideal transformer is 100%100\%100% efficient, so its electrical power output equals its electrical power input.

  1. For an ideal transformer, Pinput=PoutputP_{input} = P_{output}Pinput​=Poutput​.
  2. Electrical power is P=VIP = VIP=VI, so for a transformer VpIp=VsIsV_p I_p = V_s I_sVp​Ip​=Vs​Is​.
  3. VpV_pVp​ and IpI_pIp​ are the potential difference and current in the primary coil, in V\text{V}V and A\text{A}A.
  4. VsV_sVs​ and IsI_sIs​ are the potential difference and current in the secondary coil, in V\text{V}V and A\text{A}A.
  5. The primary coil takes the input power and the secondary coil gives the output power; if the transformer is 100%100\%100% efficient, no power is wasted and the two are equal.
  6. To find the current drawn from the input supply for a given output, use Ip=PoutputVpI_p = \dfrac{P_{output}}{V_p}Ip​=Vp​Poutput​​, assuming the transformer is ideal.

Linking potential difference, turns and current

  1. The potential differences depend on the turns, VsVp=nsnp\dfrac{V_s}{V_p} = \dfrac{n_s}{n_p}Vp​Vs​​=np​ns​​.
  2. For an ideal transformer the power stays constant, so increasing the potential difference decreases the current, from VpIp=VsIsV_p I_p = V_s I_sVp​Ip​=Vs​Is​.
  3. Combining the relationships gives IsIp=VpVs=npns\dfrac{I_s}{I_p} = \dfrac{V_p}{V_s} = \dfrac{n_p}{n_s}Ip​Is​​=Vs​Vp​​=ns​np​​, so the current ratio is the inverse of the potential difference and turns ratios.
  4. In a step-up transformer, Vs>VpV_s > V_pVs​>Vp​, so Is<IpI_s < I_pIs​<Ip​.
  5. In a step-down transformer, Vs<VpV_s < V_pVs​<Vp​, so Is>IpI_s > I_pIs​>Ip​.
Example

Question: A step-up transformer has 500500500 primary turns and 500050005000 secondary turns. The input is 25 kV25\ \text{kV}25 kV and the required power output is 2.0 MW2.0\ \text{MW}2.0 MW. Assuming it is 100%100\%100% efficient, find the secondary potential difference, the input current and the output current.

  1. Convert units: 25 kV=25 000 V25\ \text{kV} = 25\,000\ \text{V}25 kV=25000 V and 2.0 MW=2.0×106 W2.0\ \text{MW} = 2.0 \times 10^6\ \text{W}2.0 MW=2.0×106 W.
  2. Use the turns equation: VsVp=nsnp\dfrac{V_s}{V_p} = \dfrac{n_s}{n_p}Vp​Vs​​=np​ns​​, so Vs25 000=5000500\dfrac{V_s}{25\,000} = \dfrac{5000}{500}25000Vs​​=5005000​.
  3. Find the secondary potential difference: Vs=25 000×10=250 000 V=250 kVV_s = 25\,000 \times 10 = 250\,000\ \text{V} = 250\ \text{kV}Vs​=25000×10=250000 V=250 kV.
  4. Find the input current: Ip=PoutputVp=2.0×10625 000=80 AI_p = \dfrac{P_{output}}{V_p} = \dfrac{2.0 \times 10^6}{25\,000} = 80\ \text{A}Ip​=Vp​Poutput​​=250002.0×106​=80 A.
  5. Find the output current: Is=PoutputVs=2.0×106250 000=8.0 AI_s = \dfrac{P_{output}}{V_s} = \dfrac{2.0 \times 10^6}{250\,000} = 8.0\ \text{A}Is​=Vs​Poutput​​=2500002.0×106​=8.0 A.
  6. The potential difference has risen by a factor of 101010 while the current has fallen by a factor of 101010.

High-potential-difference power transmission

  1. The National Grid uses step-up transformers to raise the potential difference before power is sent through the transmission cables.
  2. For a fixed power, I=PVI = \dfrac{P}{V}I=VP​, so a higher potential difference gives a lower current.
  3. A lower current causes less heating in the cables, which have resistance, so less energy is transferred to the thermal energy stores of the cables and surroundings.
  4. Near consumers, step-down transformers reduce the potential difference again to safer, usable levels.
  5. The reasoning chain is: step-up transformer raises the potential difference, the current falls for the same power, less heating in the cables, so less energy is wasted and transmission is more efficient.
Common Mistake
  • A step-up transformer does not create energy; raising the potential difference makes the current fall so the input and output powers stay equal.
  • The current ratio is the inverse of the turns ratio, so do not apply the turns ratio to the current in the same direction.
Exam technique
  • Label every value as primary or secondary, use the turns equation for a missing potential difference, then use VpIp=VsIsV_p I_p = V_s I_sVp​Ip​=Vs​Is​ to link the currents and power.
  • For the National Grid, give the full chain: higher potential difference, lower current, less cable heating, less energy wasted.
Self review
  • What equation links the input and output powers of a 100% efficient transformer?
  • How are the potential difference ratio and turns ratio related?
  • Why does a step-up transformer give a smaller secondary current?
  • How would you calculate the current drawn from the input supply for a known power output?
  • Why does the National Grid transmit power at a high potential difference?
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Labelled transformer showing separate primary and secondary coils, an iron core, changing magnetic field, and the National Grid transmission chain

A transformer uses electromagnetic induction to change an alternating potential difference. The primary coil is connected to the input supply, and the separate secondary coil is connected to the output circuit.

Both coils are wound around an iron core. Iron is used because it is easily magnetised, allowing the changing magnetic field from the primary coil to pass effectively through the secondary coil.

There is no direct electrical connection between the coils. Energy is transferred by the changing magnetic field in the core.

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Why must a transformer use an alternating current (AC) supply?

7.3.4 Transformers (HT only) Revision Guide

  1. GCSE
  2. /Physics
  3. /7.3.4 Transformers (HT only)

Revision notes for AQA GCSE Physics 7.3.4 Transformers (HT only). Open the guide for explanations and worked examples. Written against the AQA GCSE Physics (8463) specification, so the content matches what's examinable rather than general Physics background.

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