4.5.6 The transformer
- Syllabus
- 0625–2026–2027
- Topic
- 4.5.6
- Level
- —
Learning objectives
4.5.6.1Construction of a simple transformer• Describe the construction of a simple transformer with a soft-iron core, as used for voltage transformations4.5.6.2Terms primary, secondary, step-up and• Use the terms primary, secondary, step-up and step-down4.5.6.3Transformer relation: Vp/Vs = Np/Ns• Recall/use transformer relation: Vp/Vs = Np/Ns, where p and s mean primary and secondary4.5.6.4Use of transformers in high- voltage• Describe the use of transformers in high- voltage transmission of electricity4.5.6.5Advantages of high-voltage• State the advantages of high-voltage transmission4.5.6.6Principle of operation of a simple• Explain the principle of operation of a simple iron-cored transformer4.5.6.7100% transformer efficiency: IpVp =• Recall/use 100% transformer efficiency: IpVp = IsVs, where p and s mean primary and secondary4.5.6.8Recall/use: P = I2R to explain why• Recall/use: P = I2R to explain why power losses in cables are smaller when the voltage is greater
Build a simple transformer
A simple transformer has two separate coils of insulated copper wire wound around the same closed soft-iron core. The coils are not electrically connected: energy passes from one circuit to the other through a changing magnetic field in the core.
| Part | Construction and purpose |
|---|---|
| primary coil | copper winding connected to the input a.c. supply |
| secondary coil | separate copper winding connected to the output circuit |
| soft-iron core | shared magnetic path that is easily magnetised and demagnetised |
| insulation | keeps neighbouring turns and the two circuits electrically separate |
Copper is used for the coils because its low resistance limits heating. Soft iron is used instead of permanently magnetised material because its magnetisation follows the changing primary current readily.
A transformer requires two coils linked magnetically by a core. Current does not travel through the iron core or directly from the primary wire into the secondary wire.
Name the coils and the voltage change
| Term | Meaning |
|---|---|
| primary | the input coil, connected to the a.c. supply |
| secondary | the output coil, connected to the load |
| step-up | secondary voltage is greater than primary voltage |
| step-down | secondary voltage is less than primary voltage |
| Transformer | Turns comparison | Voltage comparison |
|---|---|---|
| step-up | N_s > N_p | V_s > V_p |
| step-down | N_s < N_p | V_s < V_p |
| unchanged voltage | N_s = N_p | V_s = V_p |
First identify which coil is connected to the source: that is the primary. Then compare the output voltage with the input voltage to name the transformer. The physical left or right side of a drawing does not decide the terms.
Step-up and step-down describe voltage, not energy creation. A transformer operates continuously with a changing current such as a.c.; a steady d.c. supply does not provide continuous transformation.
Use the transformer turns relation
For an ideal transformer, the voltage ratio equals the turns ratio: V_p / V_s = N_p / N_s. The subscript p means primary and s means secondary.
| Find | Rearranged relation |
|---|---|
| V_s | V_s = V_p × N_s / N_p |
| V_p | V_p = V_s × N_p / N_s |
| N_s | N_s = N_p × V_s / V_p |
| N_p | N_p = N_s × V_p / V_s |
Example: a primary has 800 turns at 240 V and the secondary has 40 turns. V_s = 240 × 40 / 800 = 12 V, so it is a step-down transformer.
Check the direction before accepting a number: fewer secondary turns must give a smaller secondary voltage, while more secondary turns must give a larger secondary voltage.
Keep each voltage paired with the turns on the same coil. Do not invert only one ratio, and do not use the turns relation to compare currents.
Transform voltage across the electricity grid
The electricity grid uses transformers at different stages so alternating electrical energy can be transmitted efficiently and then supplied at suitable voltages.
| Stage | Transformer action | Purpose |
|---|---|---|
| generator at a power station | produces a.c. at a moderate voltage | supplies the grid |
| power-station transformer | steps voltage up | prepares power for long-distance transmission |
| high-voltage cables | carry the power at high voltage | reduce transmission losses |
| substation transformers | step voltage down in stages | make distribution and final use suitable |
| consumers | receive the required lower supply voltage | operate homes and other loads |
Follow the energy route: generator → step-up transformer → transmission cables → step-down transformer or transformers → consumers. The transformer beside the power station is step-up; the transformer before consumers is step-down.
Transformers change alternating voltage; they do not generate the electrical energy carried by the grid. A transformer alone cannot continuously step a steady d.c. voltage up or down.
State the advantages of high-voltage transmission
For a given transmitted power, using a higher voltage allows a smaller current. The cables then waste less electrical energy as thermal energy, so a larger fraction reaches consumers.
| Advantage | Practical consequence |
|---|---|
| less heating and energy loss in cables | transmission is more efficient |
| smaller required current | thinner, lighter or cheaper cables can be used for an acceptable loss |
| less wasted generation | less fuel and fewer generating resources are required for the same delivered energy |
| lighter cable system | supports and pylons can be less costly or spaced further apart where engineering constraints allow |
An examination answer should link the electrical advantage to an economic result: less energy wasted lowers operating cost, while thinner cables and fewer supports can lower construction cost.
High voltage is not itself the useful advantage. The advantage comes from the lower current and therefore lower cable heating for the same power transfer; high-voltage systems also require suitable insulation.
Explain how an iron-cored transformer works
| Stage | What happens |
|---|---|
| 1 | an alternating voltage drives an alternating current in the primary coil |
| 2 | the alternating current produces a changing magnetic field in the soft-iron core |
| 3 | the core concentrates and links the changing field through the secondary coil |
| 4 | the changing field induces an alternating e.m.f. across the secondary coil |
| 5 | if the secondary circuit is complete, the induced e.m.f. drives an alternating current |
Soft iron magnetises and demagnetises readily, so the field follows the primary a.c. and provides a strong changing magnetic link between the two coils.
A steady d.c. current produces a steady magnetic field after switching. With no continuing change of field, there is no continuing induced e.m.f. in the secondary. A brief effect may occur only while the d.c. is switched on or off.
The core carries changing magnetic flux, not an electric current from primary to secondary. The secondary voltage is induced; the two electrical circuits remain separate.
Use ideal transformer power
A 100% efficient transformer has no power loss, so input power equals output power: I_p V_p = I_s V_s.
| Find | Rearranged relation |
|---|---|
| I_s | I_s = I_p V_p / V_s |
| I_p | I_p = I_s V_s / V_p |
| V_s | V_s = I_p V_p / I_s |
| V_p | V_p = I_s V_s / I_p |
Example: an ideal transformer takes 0.20 A at 240 V and gives 12 V. I_s = 0.20 × 240 / 12 = 4.0 A. The lower output voltage is paired with a higher output current.
For the same ideal power, current changes inversely with voltage: a step-up transformer steps current down, while a step-down transformer steps current up.
Use I_p V_p = I_s V_s only when the transformer is stated or assumed to be 100% efficient. A real transformer has output power below input power because some energy is dissipated.
Use I²R to explain cable losses
For a fixed power delivered by a transmission system, P = VI shows that raising the transmission voltage lowers the cable current. The cable heating loss is then found from P_loss = I²R.
| Change, with transmitted power and cable resistance fixed | Consequence |
|---|---|
| transmission voltage increases | cable current decreases |
| cable current decreases | I² decreases by the square of that factor |
| I² decreases | cable power loss and heating decrease |
| cable power loss decreases | transmission efficiency increases |
The square makes current especially important. If current is halved while cable resistance is unchanged, P_loss becomes (1/2)² = 1/4 of its original value. If current is one tenth, the loss is one hundredth.
A complete explanation states the fixed transmitted power, uses high V → low I from P = VI, and then uses low I → much smaller I²R heating loss.
Do not say that high voltage directly lowers cable resistance. For the same cable, R is treated as fixed; the voltage increase reduces current, and the smaller current reduces I²R loss.