Learn5054Paper 1 · Multiple ChoiceElectricity and magnetismTransformers

Electricity and magnetism · Subtopic 8 of 8

Transformers

A transformer has two coils wound on a soft iron core. An alternating current in the primary coil makes a changing magnetic field that induces an alternating e.m.f. in the secondary. The voltages are in the same ratio as the turns: $\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}$ . Power stations step the voltage up for transmission so the current, and the energy wasted heating the cables, is small.

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Section 1 of 14

Learning objectives

Describe the structure of a transformer and explain how it works.
Use $\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}$ and identify step-up and step-down transformers.
Use $I_pV_p=I_sV_s$ for an ideal transformer.
Explain why high voltages are used for transmission, using $P=I^2R$ .

Pause and think

Power stations generate electricity at about $25\,000$ V, transmit it at up to $400\,000$ V, and your home uses about $230$ V. Why not just send it at $230$ V all the way?

Section 2 of 14

Core idea

Start hereWhy pylons carry 400 000 volts

Power lines run at enormous voltages, then step down to 230 V in homes. Why bother? At high voltage the current is small, and heating losses in the cables depend on current squared.

The big idea

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$ ; ideal: $V_pI_p=V_sI_s$ . High-voltage transmission cuts $I^2R$ losses. Works only with a.c.

1. How it works

A step-up transformer

Changing current in the primary makes a changing magnetic field in the core, which induces a voltage in the secondary. It only works with a.c.

Turns ratio

$\text{a.c. in primary}\Rightarrow\text{changing field in core}$
Soft iron links both coils.
$\text{e.m.f. per turn is the same in both coils}$
Same changing field through each turn.
$\frac{V_s}{V_p}=\frac{N_s}{N_p}$
Total e.m.f. ∝ number of turns.

d.c. gives no changing field, so no output.

Worked exampleA step-down transformer

230 V to 11.5 V; primary 2000 turns; $I_s=2$ A.

Goal: $N_s$ ; $I_p$ .
Strategy: Ratio; power conservation.

$\frac{11.5}{230}=\frac1{20}$
Voltage ratio.
$N_s=\frac{2000}{20}=100$
Same ratio for turns.
$I_p=\frac{11.5\times2}{230}=0.1\text{ A}$
$V_pI_p=V_sI_s$ .

Answer

100

turns;

0.1

✓ CheckVoltage down 20×, current up 20× ✓.

2. Transmission

Power lost in cables vs transmission voltage (same power sent)

Power lost

=I^2R

. Doubling the voltage halves the current and cuts the loss to a quarter. Hence the National Grid uses up to 400 000 V.

Worked example1 MW through 5 Ω cables

At 10 kV and at 100 kV.

Goal: Losses.
Strategy: $I=P/V$ , loss $=I^2R$ .

$I=100\text{ A}:\ 100^2\times5=50\text{ kW}$
10 kV.
$I=10\text{ A}:\ 10^2\times5=500\text{ W}$
100 kV.

Answer

50

kW vs

500

✓ Check10× the voltage, 100× less loss ✓.

Common misconceptionMore power from step-up A transformer can't create energy: higher voltage means lower current.

Pause and think

Step-up or step-down near homes?

Summary

Turns ratio = voltage ratio.
High voltage, low current, low loss.

Step-up: More secondary turns; raises voltage.

HistoryThe War of the Currents. In the

1880

s Thomas Edison backed d.c. power, while Nikola Tesla and George Westinghouse backed a.c. Because a.c. can be transformed to high voltages for efficient long-distance transmission, a.c. won, and almost every power grid today uses it.

Fun factTransformers hum. The deep hum near a substation or a phone charger is the transformer's iron core vibrating as it's magnetised back and forth

100

times a second on a

50

Hz supply. In countries using

60

Hz, the hum is slightly higher in pitch.

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