Transformers: practice problems with step-by-step solutions

Free Transformers practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: turns ratio, currents and power, efficiency and losses.

Free practice problemEasy
N1​ = 5000N2​ = ?11500 V230 V

An ideal transformer must step 11500 V down to 230 V. Its primary winding has 5000 turns. How many turns N2 does the secondary winding need?

V1 = 11500 VV2 = 230 VN1 = 5000
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What you practise

  1. Easyturns ratio
  2. Mediumcurrents and power
  3. Hardefficiency and losses

Key formulas

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

Learn Transformers from scratch

Before this: AC impedance & phasors

1What a transformer is

A transformer changes an AC voltage to a higher or lower AC voltage. Power stations step the voltage up to send power far; local transformers step it down to 230 V for homes.

It has two coils of wire wound on one iron core. The coil connected to the supply is the primary (1). The coil connected to the load is the secondary (2).

RL​N1​N2​V1​V2​

The AC current in the primary makes a changing magnetic field in the core. That changing field induces a voltage in every turn of the secondary.

A steady DC current makes a steady field, and a steady field induces nothing. So a transformer only works with AC.

Self-check What does an ideal transformer give at its secondary when its primary is connected to a 12 V battery (DC)?
  1. 0 V, because a steady field induces no voltage
  2. 12 V, the same as the primary
  3. A voltage set by the turns ratio
  4. An AC voltage of 12 V

Answer: 0 V, because a steady field induces no voltage. Only a changing magnetic field induces a voltage. A DC current makes a constant field, so the secondary gives nothing once the current has settled.

2The turns ratio

Every turn of wire on the core gets the same voltage. So the voltage of each coil is proportional to its number of turns N.

a = N1 / N2 = V1 / V2

An ideal transformer loses no power, so power in equals power out: V1·I1 = V2·I2. The currents therefore go the other way round:

I1 / I2 = N2 / N1 = 1/a
  • Step-down (a > 1): lower voltage, higher current on the secondary.
  • Step-up (a < 1): higher voltage, lower current on the secondary.
Self-check An ideal transformer has 400 primary turns and 200 secondary turns. The primary is on 240 V. What is the secondary voltage?
  1. 120 V
  2. 240 V
  3. 480 V
  4. 60 V

Answer: 120 V. V2 = V1·N2/N1 = 240·200/400 = 120 V. Turning the ratio upside down gives 480 V.

3The method step by step

  1. Find the turns ratio a = N1/N2, or a = V1/V2 from the rated voltages.
  2. Find the voltage on the other side: V2 = V1/a.
  3. If a load is connected, find its current with Ohm’s law: I2 = V2/RL.
  4. Find the current on the other side: I1 = I2/a. The high-voltage side carries the small current.
  5. Check with power: V1·I1 must equal V2·I2.

A quick sense check: the side with more turns has the higher voltage and the lower current.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

An ideal transformer with a turns ratio N1:N2 = 10:1 has 2300 V on its primary. A 23 Ω resistor is connected to the secondary. Find V2, I2 and the primary current I1.

1

Look at the circuit

23 ΩN1​ = 10N2​ = 12300 VV2​ ?I1​I2​

The turns ratio is a = 10, so this is a step-down transformer.

2

Secondary voltage

V2 = V1 / a = 2300 / 10 = 230 V
3

Secondary current

I2 = V2 / RL = 230 / 23 = 10 A
4

Primary current

The current ratio is the inverse of the voltage ratio.

I1 = I2 / a = 10 / 10 = 1 A
5

Check the answer

Power in: 2300·1 = 2300 W. Power out: 230·10 = 2300 W. They are equal, as they must be for an ideal transformer.

Answer

V2 = 230 V, I2 = 10 A and I1 = 1 A.

5Ratings, referred loads and losses

A transformer is rated in kVA, for example “10 kVA, 7200/240 V”. The rated current of each winding is the rating divided by that winding’s voltage:

I = S / V

Seen from the primary, a load RL on the secondary looks larger by a²:

RL′ = a²·RL

In the example, 23 Ω looks like 10²·23 = 2300 Ω from the primary. Then I1 = 2300/2300 = 1 A, the same answer.

A real transformer has two kinds of loss:

  • Core loss Pcore: in the iron. It is the same at every load.
  • Copper loss Pcu: in the windings. It grows with the square of the load fraction x (x = 0.5 at half load): Pcu = x²·Pcu,fl.
η = x·S·pf / (x·S·pf + Pcore + x²·Pcu,fl)

The efficiency is highest at the load where copper loss equals core loss:

x = √(Pcore / Pcu,fl)
Self-check A 25 kVA transformer has a core loss of 75 W and a full-load copper loss of 300 W. At what load is its efficiency highest?
  1. 12.5 kVA
  2. 25 kVA
  3. 50 kVA
  4. 6.25 kVA

Answer: 12.5 kVA. x = √(75/300) = √0.25 = 0.5, so 0.5·25 = 12.5 kVA. Forgetting the square root gives x = 0.25 and 6.25 kVA.

6Common mistakes

  • Turning the ratio upside down. The side with more turns has the higher voltage.
  • Using the voltage ratio for the currents. Currents go the other way: the high-voltage side has the small current.
  • Using the wrong winding voltage for the rated current. Each winding uses its own voltage in I = S/V.
  • Scaling the copper loss with x instead of x². At half load it is a quarter of the full-load value.
Self-check A 10 kVA, 7200/240 V transformer runs at rated load. What current flows in the low-voltage winding?
  1. 0.0417 A
  2. 1.39 A
  3. 30 A
  4. 41.7 A

Answer: 41.7 A. I = S/V = 10000/240 = 41.7 A. Using 7200 V gives the high-voltage current, 1.39 A. Leaving S in kVA gives 0.0417 A, and 30 is only the turns ratio.

7How exam questions look

  • Easy: the turns ratio. You find the secondary voltage, or the number of secondary turns for a given step-down.
  • Medium: currents and power. You find the rated current of a winding, the primary current with a resistor on the secondary, the resistance seen from the primary, or the turns ratio that matches a loudspeaker to an amplifier.
  • Hard: efficiency and losses. You find the total losses at part load, the load for maximum efficiency, or the full-load copper loss from a given efficiency.

For matching, the load must look like the source resistance: a²·RL = Rsource, so a = √(Rsource/RL).

For losses at part load, add the fixed core loss and x²·Pcu,fl. The power factor matters for the output power, not for the losses.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
N1​ = 60N2​ = 3000240 VV2​ = ?

An ideal transformer has 60 turns on its primary winding and 3000 turns on its secondary winding. The primary is connected to a 240 V AC supply. Find the secondary voltage V2.

N1 = 60N2 = 3000V1 = 240 V

Find V2 (V)

Show hints
  1. The voltage per turn is the same in both windings.
  2. V₂/V₁ = N₂/N₁.
  3. N₂/N₁ = 50.
Show the answer and the step-by-step solution

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Voltage ratio equals turns ratio

V2 = V1·N2/N1 = 240·3000/60 = 12000 V

Answer

V2 = 12000 V
Example 2Medium
4 ΩN1​N2​256 Ω

An amplifier with an output resistance of 256 Ω drives a 4 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 256 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 256 ΩRL = 4 Ω

Find a

Show hints
  1. Seen from the primary, the load looks like a²·RL.
  2. Set a²·RL equal to the amplifier’s output resistance.
  3. a² = 256/4.
Show the answer and the step-by-step solution

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Match the referred load

a²·4 = 256
a = √(256/4) = √64 = 8

Answer

a = 8
Example 3Hard
pf 0.9525 kVA, 25% load

A single-phase 25 kVA transformer has a core loss of 75 W and a copper loss of 300 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.95 lagging. Find the total losses Ploss in watts at this load.

S = 25 kVAPcore = 75 WPcu,fl = 300 Wx = 0.25pf = 0.95

Find Ploss (W)

Show hints
  1. Which loss changes with the load, and which does not?
  2. Copper loss ∝ I², and the current is proportional to the load.
  3. x² = 0.0625.
Show the answer and the step-by-step solution

Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.

Copper loss at this load

Pcu = x²·Pcu,fl = 0.0625·300 = 18.75 W

Total losses

The core loss is the same at any load.

Ploss = 75 + 18.75 = 93.75 W

Answer

Ploss = 93.75 W

The efficiency at this load is η = 98.45%.

Questions and answers

What is Transformers?

A transformer changes an AC voltage to a higher or lower AC voltage. Power stations step the voltage up to send power far; local transformers step it down to 230 V for homes.

It has two coils of wire wound on one iron core. The coil connected to the supply is the primary (1). The coil connected to the load is the secondary (2).

How do I solve Transformers problems step by step?

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

What are common mistakes in Transformers?
  • The turns ratio is upside down. The secondary has more turns, so this transformer steps the voltage up: V2 = V1·N2/N1.
  • Impedance transforms with the square of the turns ratio, so take the square root: a = √(Rs/RL).
  • The ratio is upside down. The primary faces the larger resistance, so it needs more turns.
What do the Transformers exercises cover?
  • Easy: turns ratio
  • Medium: currents and power
  • Hard: efficiency and losses

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

AC impedance & phasors

Calculators

Transformer turns ratio calculator

Courses with this topic

Electrical Engineering FundamentalsPower Systems and Electrical MachinesFE Electrical and Computer exam practice

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Printable worksheet

Transformers worksheet with answers
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