Step-Up and Step-Down Transformers: Power and Efficiency – Electrical Power Transmission (Senior 3 Physics) | Ganzaa

Electrical Power Transmission

Lessons

Senior 3 · Physics

Electrical Power Transmission

4 lessons

Lesson 2 of 4

Step-Up and Step-Down Transformers: Power and Efficiency

25 min

Something for nothing?

A step-up transformer can turn 230 V into 2300 V. Does that mean it gives out ten times as much energy as it takes in? If it did, we could connect its output back to its input and run the country's lights for free. Of course it does not. A transformer can change the voltage, but it cannot create energy: whatever it gains in voltage, it pays for in current, and it always loses a little energy as heat. This lesson follows that energy through the transformer.

Step-up and step-down

Type	Turns	Voltage	Current	Examples
Step-up	$N_{s} > N_{p}$	$V_{s} > V_{p}$	$I_{s} < I_{p}$	At power stations, raising the generator's voltage to 110 kV or 220 kV for the transmission lines; in microwave ovens and old television tubes
Step-down	$N_{s} < N_{p}$	$V_{s} < V_{p}$	$I_{s} > I_{p}$	At substations and on poles, lowering the line voltage to 30 kV, 15 kV and then 400 V and 230 V; in phone and laptop chargers; in welding machines, which need a very large current at low voltage

Definition

Step-up transformer: a transformer with more turns on the secondary than on the primary, so that $V_{s} > V_{p}$ . Step-down transformer: one with fewer turns on the secondary, so that $V_{s} < V_{p}$ .

Ideal transformer: one that wastes no energy, so the power output equals the power input.

Power in, power out

Electrical power is $P = V I$ (S1). In an ideal transformer no energy is wasted, so

V_p\,I_p = V_s\,I_s \qquad\Rightarrow\qquad \frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}

The currents are in the inverse ratio of the turns. A step-up transformer that multiplies the voltage by 10 divides the current by 10; a step-down transformer that divides the voltage by 10 can deliver 10 times the current.

Worked example 1: at the power station

A generator gives 11 kV (a typical value) and a current of 1000 A. An ideal transformer steps the voltage up to 110 kV for the transmission line. Find (a) the turns ratio; (b) the current in the line; (c) the power transmitted.

\text{(a) } \frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{110\,000}{11\,000} = 10

\text{(b) } I_s = I_p \times \frac{V_p}{V_s} = 1000 \times \frac{11\,000}{110\,000} = 100\ \text{A}

\text{(c) } P = V_s\,I_s = 110\,000 \times 100 = 1.1 \times 10^{7}\ \text{W} = 11\ \text{MW}

The power is the same on both sides: $11\,000 \times 1000 = 11\ \text{MW}$ too.

Worked example 2: how much current does a charger draw?

An ideal step-down transformer in a charger gives 5.0 V at 2.0 A from the 230 V mains. Find the current in its primary coil.

I_p = \frac{V_s\,I_s}{V_p} = \frac{5.0 \times 2.0}{230} \approx 0.043\ \text{A} = 43\ \text{mA}

Efficiency and where the energy goes

Real transformers are not ideal, though large ones come close: they can convert more than 99% of the input power. The efficiency is

\eta = \frac{\text{power output}}{\text{power input}} \times 100\% = \frac{V_s\,I_s}{V_p\,I_p} \times 100\%

The missing power becomes heat in the coils and the core. Engineers attack each cause separately:

Energy loss	Cause	How it is reduced
Heating of the coils ("copper loss")	The currents flow through the resistance of the wire: power lost $= I^{2} R$	Thick copper wire of low resistance, especially for the coil that carries the larger current
Eddy currents	The changing field also induces currents that swirl round inside the iron core and heat it	A laminated core of thin iron sheets insulated from each other
Hysteresis	Energy is needed to magnetise the core one way and then the other, 50 times a second	A core of soft iron (or special steel), which is easily magnetised and demagnetised (S2)
Flux leakage	Some of the primary's magnetic field misses the secondary	A closed core, with the coils wound close together or one on top of the other

The heat that remains must be removed: large transformers sit in tanks of insulating oil with cooling fins, which is why they feel warm. (Their steady hum comes from the iron core, which vibrates slightly as it is magnetised back and forth.)

Worked example 3: finding the efficiency

A transformer takes 2.0 A from the 230 V mains and gives 4.0 A at 110 V. Find its efficiency and the power wasted.

P_{\text{in}} = 230 \times 2.0 = 460\ \text{W} \qquad P_{\text{out}} = 110 \times 4.0 = 440\ \text{W}

\eta = \frac{440}{460} \times 100\% \approx 96\% \qquad \text{wasted} = 460 - 440 = 20\ \text{W}

Worked example 4: a real transformer

A transformer of efficiency 92% lights a 12 V, 36 W lamp at full brightness from the 230 V mains. Find the current in the primary.

P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{36}{0.92} \approx 39.1\ \text{W} \qquad I_p = \frac{39.1}{230} \approx 0.17\ \text{A}

Warning

The power equation needs the efficiency. $V_{p} I_{p} = V_{s} I_{s}$ holds only for an ideal transformer. If an efficiency is given, use $V_s I_s = \eta \times V_p I_p$ . And remember: a step-up transformer raises the voltage but lowers the current; it never raises the power.

Tip

Voltage goes with turns, current goes against them. $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ but $\frac{I_s}{I_p} = \frac{N_p}{N_s}$ . Check every answer: $V_{p} I_{p}$ should equal $V_{s} I_{s}$ (ideal) or be a little larger (real).

How transformers work (Cognito). Watch the step-by-step explanation of the structure, and how the number of turns decides whether a transformer steps the voltage up or down.

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Practice

State two differences between a step-up and a step-down transformer.
An ideal transformer steps 230 V down to 23 V. The secondary current is 3.0 A. Find the primary current.
A transformer has 500 turns on its primary and 2500 on its secondary, and the primary is connected to 230 V. Find the output voltage. Is it step-up or step-down?
A transformer's input power is 500 W and its output power is 470 W. Find its efficiency. What happens to the other 30 W?
Explain why a transformer core is laminated.
Why is the core made of soft iron, not of hard steel?
A transformer of efficiency 90% supplies 9.0 A at 12 V from the 230 V mains. Find the current in its primary.
A pupil says: "A step-up transformer gives out more energy than it takes in, because it increases the voltage." Evaluate this claim.

Show answersHide answers

A step-up transformer has more secondary turns than primary turns and raises the voltage (lowering the current); a step-down transformer has fewer secondary turns and lowers the voltage (raising the current).
$I_p = \frac{V_s I_s}{V_p} = \frac{23 \times 3.0}{230} = 0.30\ \text{A}$ .
$V_s = 230 \times \frac{2500}{500} = 1150\ \text{V}$ ; step-up.
$\eta = \frac{470}{500} \times 100\% = 94\%$ . The 30 W becomes heat in the coils (resistance) and in the core (eddy currents and hysteresis).
The changing magnetic field induces eddy currents in the iron core, which heat it. Thin sheets insulated from one another break up the current paths, so the eddy currents and the energy they waste are much smaller.
Soft iron is easily magnetised and demagnetised, so little energy is lost each time the field reverses (low hysteresis loss). Hard steel keeps its magnetism and would waste much more energy.
$P_{\text{out}} = 12 \times 9.0 = 108\ \text{W}$ ; $P_{\text{in}} = \frac{108}{0.90} = 120\ \text{W}$ ; $I_p = \frac{120}{230} \approx 0.52\ \text{A}$ .
Wrong. Energy is conserved: the output power $V_{s} I_{s}$ can be at most equal to the input power $V_{p} I_{p}$ . When the voltage is raised, the current falls by the same factor (or more), and some energy is always lost as heat, so the output energy is slightly less than the input.

Key points

Step-up: $N_{s} > N_{p}$ , voltage up, current down. Step-down: the reverse.
Ideal transformer: $V_{p} I_{p} = V_{s} I_{s}$ , so $\frac{I_s}{I_p} = \frac{N_p}{N_s}$ .
Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$ ; large transformers exceed 99%.
Losses: coil resistance (thick copper), eddy currents (laminated core), hysteresis (soft iron), flux leakage (closed core, coils close together).

Sources and credits

23.10: Transformers and 23.7: Eddy Currents and Magnetic Damping, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Rwandan voltages: Transmission, Rwanda Energy Group (110 kV and 220 kV lines), and Reticulation Standards, Version VI, REG, August 2020 (30 kV and 15 kV; 400 V and 230 V).
Video: GCSE Physics, How Transformers Work, Cognito, embedded from YouTube.

Written by Ganzaa: the lesson text, the eddy-current diagram, the tables, worked examples and practice exercises.

1Something for nothing?

2Step-up and step-down

3Power in, power out

4Efficiency and where the energy goes

Something for nothing?

Step-up and step-down

Power in, power out

Efficiency and where the energy goes