Transformers calculations (HT, physics only) — AQA GCSE Physics

Module 7 · Lesson 7.08 · Spec 4.7.3.4

Transformer Calculations

Higher Tier only

Physics only

Two equations on the Physics Equations Sheet. One links pd ratio to turns ratio. One links power input to power output assuming 100% efficiency. Together they solve every GCSE transformer calculation.

Two Equations · Both Given on the Sheet

You don't need to recall these equations — they appear on the Physics Equations Sheet. But you must know which to use, when, and how to orient them.

1

Turns Ratio

V_p / V_s = N_p / N_s

V in volts (V) · N is a count (no units)

2

Power Conservation (assuming 100% efficient)

V_s × I_s = V_p × I_p

Power input = power output, both in watts (W) · I in amperes (A)

How to pick which equation

Question gives you turns (and/or both pds) → use Equation 1. Question gives you currents (and pds) → use Equation 2. Some questions need both, chained together — the highest-marked transformer pattern.

Equation 1 · Turns Ratio Detail

V_p / V_s = N_p / N_s

The pd ratio across the two coils equals the turns ratio. Both ratios are the same number.

Rearranging — three useful forms

Find V_s: V_s = V_p × (N_s / N_p)

Find V_p: V_p = V_s × (N_p / N_s)

Find N_s: N_s = N_p × (V_s / V_p)

Sanity checks

Step-up: N_s > N_p → V_s > V_p. Your answer for V_s should be larger than V_p.

Step-down: N_s < N_p → V_s < V_p. Your answer for V_s should be smaller than V_p.

If you got the opposite, you almost certainly inverted the ratio.

Worked Example · Find the Secondary pd

Question3 marks

A transformer has 150 turns on the primary coil and 600 turns on the secondary coil. The primary coil is connected to a 9.0 V alternating supply. Calculate the output potential difference across the secondary coil.

E

V_p / V_s = N_p / N_s, rearranged: V_s = V_p × (N_s / N_p)

S

V_s = 9.0 × (600 / 150) = 9.0 × 4

U

volts (V)

A

V_s = 36 V

✓ Check

N_s > N_p (600 > 150), so this is a step-up transformer. Output V_s (36 V) is bigger than input V_p (9.0 V). ✓ Consistent.

Equation 2 · Power Conservation Detail

V_s × I_s = V_p × I_p

Power output (secondary) = Power input (primary), both in watts (W)

Spec wording (4.7.3.4)

"If transformers were 100% efficient, the electrical power output would equal the electrical power input." Unless the question states the efficiency, assume 100% efficient — this is the default for GCSE.

Rearranging

Find I_s: I_s = (V_p × I_p) / V_s

Find I_p: I_p = (V_s × I_s) / V_p

Find a pd: V_s = (V_p × I_p) / I_s, etc.

Equivalent shortcut

You can also calculate power directly: P = V × I (from Module 2). The output power equals the input power, so once you know one, you know both. The transformer equation just says "Pᵢₙ = P_out" in coil-specific form.

Worked Example · Find the Secondary Current

Question3 marks

A step-down transformer is connected to a 230 V mains supply and draws a primary current of 0.40 A. The output potential difference is 12 V. Assuming the transformer is 100% efficient, calculate the current in the secondary coil.

E

V_s × I_s = V_p × I_p, rearranged: I_s = (V_p × I_p) / V_s

S

I_s = (230 × 0.40) / 12 = 92 / 12

U

amperes (A)

A

I_s = 7.7 A (2 sig fig)

✓ Check

Step-down transformer: V went DOWN (230 → 12, ratio 19.2×), so I goes UP (0.40 → 7.7, ratio 19.2×). Same ratio in the inverse direction. ✓ Consistent with power conservation.

The Inverse V–I Relationship

Combining the two equations gives the deep result that powers the National Grid.

Derivation

Start with both equations:

V_p / V_s = N_p / N_s AND V_s × I_s = V_p × I_p

Rearrange the power equation:

V_p / V_s = I_s / I_p

So the pd ratio equals the inverse of the current ratio:

V_p / V_s = N_p / N_s = I_s / I_p

Step-up · V goes up

V_s > V_p by factor k

I_s < I_p by the same factor k

Voltage up → current down. The output coil has bigger pd but smaller current.

Step-down · V goes down

V_s < V_p by factor k

I_s > I_p by the same factor k

Voltage down → current up. The output coil has smaller pd but bigger current.

Physical reasoning

Power is conserved (assuming 100% efficient). Power = V × I. If you multiply V by a factor, you must divide I by the same factor to keep V × I constant. Conservation forces the inverse relationship.

Quantities on the Transformer · Diagram

Before you substitute numbers, label every quantity on the diagram. Which is V_p? Which is I_s? Which side has the load? Mistakes happen when students mix up sides.

Quantities for the primary (input) side: V_p, I_p, N_p. Quantities for the secondary (output) side: V_s, I_s, N_s.

Worked Example · Chained 5-Mark Calculation

Question5 marks

A transformer is used in a soldering iron. The primary coil has 800 turns and is connected to a 230 V alternating supply. The secondary coil has 80 turns. The soldering iron has a power output of 23 W. Assuming the transformer is 100% efficient, calculate the current drawn from the primary supply.

Step 1

Find V_s using turns ratio:

E

V_p / V_s = N_p / N_s → V_s = V_p × (N_s / N_p)

S

V_s = 230 × (80 / 800) = 230 × 0.1 = 23 V

Step 2

Find I_s using output power P = V × I:

E

P_output = V_s × I_s → I_s = P_output / V_s

S

I_s = 23 / 23 = 1.0 A

Step 3

Find I_p using power conservation (100% efficient):

E

V_s × I_s = V_p × I_p → I_p = (V_s × I_s) / V_p

S

I_p = (23 × 1.0) / 230 = 23 / 230

A

I_p = 0.10 A

✓ Shortcut

Because the transformer is 100% efficient, P_input = P_output = 23 W. So you can skip directly to: V_p × I_p = 23, hence I_p = 23 / 230 = 0.10 A. Same answer, fewer steps.

Application · Why High-Voltage Transmission?

This is the central reason the National Grid uses transformers, and a recurring exam stem.

The logic — three steps

Step-up transformer at power station: raises pd (e.g. 25 kV → 400 kV). Since V_s × I_s = V_p × I_p, raising V by a factor of 16 drops I by the same factor of 16.

Long-distance transmission at high pd, low I: the power dissipated as heat in the transmission cables is P_loss = I²R. Lower I → much lower I²R losses (halve I → quarter the wasted power).

Step-down transformer near homes: reduces pd back to safe 230 V mains (and current rises proportionally for end-use).

Mark-scheme phrasing for 3-mark explain

Reward phrases: "step-up transformer increases the pd" · "current is therefore reduced (for the same power)" · "less power lost as heat in the cables" or "lower I²R losses". The chain of reasoning must include reduced current as the link — not just "less power lost".

Watch the language

Don't say "no energy is lost" — energy is conserved, but some is still lost as heat. The point is to reduce the losses, not eliminate them. Also: the transformer itself is assumed 100% efficient — it's the cables that dissipate energy via I²R.

Examiner Alert · The Inversion Trap

The most common single-mark loss on transformer calculations: inverting the turns ratio. Half of one famous past paper got zero on this calculation alone.

55%

scored ZERO on Nov 2020 Q7.2 — often due to incorrect substitution into the chosen equation

55%

scored ALL 5 MARKS on Jun 2023 Q8.2 chained calculation — equations sheet helps

55% of students scored zero, often due to incorrect substitution into the chosen equation. — AQA Examiner Report, Nov 2020 Paper 2 Q7.2 (transformer turns calculation)

55% of students gained all 5 marks. Most of the other students scored at least 3 marks for a partially correct answer. There were alternative ways of calculating the correct answer using the transformer and power equations, and it was pleasing to see so many students having the confidence to use them appropriately. — AQA Examiner Report, Jun 2023 Paper 2 Q8.2 (chained transformer calculation)

The mechanical fix · "Primary on top, both sides"

Write the equation as:

V_p / V_s = N_p / N_s

Both V_p and N_p on the TOP. Both V_s and N_s on the BOTTOM. Don't try to be clever and "flip" one side — keep them both oriented the same way and rearrange algebraically.

The Orientation Rule · Step by Step

A foolproof routine to set up any turns-ratio calculation. Follow it every time and the inversion error becomes impossible.

Routine

Write the equation as given: V_p / V_s = N_p / N_s. Don't rewrite it.

Label what's known on the diagram: V_p, V_s, N_p, N_s. Identify which one is the unknown.

Substitute the three known values into the equation. Keep the unknown as a symbol on the side where it naturally sits.

Rearrange algebraically using cross-multiplication. Don't try to "predict" the rearrangement before substituting.

Sanity check: is this step-up or step-down? Does the answer have the right direction (bigger/smaller than the input pd)?

Why this works

Substituting BEFORE rearranging eliminates the inversion ambiguity. If you write 9.0 / V_s = 150 / 600, cross-multiplication gives V_s = 9.0 × 600 / 150 = 36 — only one possible answer. If instead you try to predict "do I multiply or divide?" before substituting, you risk flipping.

Weak vs Strong · The Inversion Mistake

Same question: primary 150 turns, secondary 600 turns, V_p = 9.0 V. Calculate V_s.

Weak · 1/3

V_p / V_s = N_s / N_p

9.0 / V_s = 600 / 150

V_s = 9.0 × 150 / 600

V_s = 2.25 V

Loses 2 marks: the inverted equation V_p/V_s = N_s/N_p is wrong. The answer 2.25 V is the reciprocal of the right answer × V_p. A sanity check would catch it: 600 turns secondary > 150 primary, so this must be step-up — output should be BIGGER than 9.0 V, not smaller. Earns 1 mark for the right equation choice, 0 for substitution and answer.

Strong · 3/3

V_p / V_s = N_p / N_s

9.0 / V_s = 150 / 600

V_s = 9.0 × 600 / 150

V_s = 36 V

Earns all 3 marks: equation written correctly (primary on top, both sides). Substitution preserves the orientation. Cross-multiplication gives a clean answer. Sanity check: step-up transformer (more turns on secondary), so V_s > V_p. ✓ 36 > 9.0. ✓

Common Errors

Students often writeCorrect versionWhy it matters

V_p/V_s = N_s/N_p (inverted ratio)V_p/V_s = N_p/N_s — primary on top, both sidesGives the reciprocal of the right answer. Nov 2020 Q7.2: 55% scored zero from this

Forgetting to assume 100% efficient when applying V_sI_s = V_pI_pThe spec equation requires 100% efficiency. State the assumption or use efficiency = useful out / total inThe equation is only valid if all the input power transfers to the output

"A step-up transformer increases the power"Power is conserved (assuming 100% efficient). Step-up increases the pd; current decreases by the same factorConservation of energy — no free power gain

"No energy is lost in the National Grid"Energy is conserved ; some is still dissipated as heat in the cables. High-V transmission reduces losses, doesn't eliminate themAbsolute statements get marked down

Using V_p in the power output calculationPower output = V_s × I_s (secondary side). Power input = V_p × I_p. Don't mix sidesWrong side → wrong answer

Forgetting to convert kV to V or kW to W1 kV = 1000 V · 1 kW = 1000 W. Substitute in SI base unitsFactor-of-1000 errors are common in National Grid questions

✓ Checkpoint & Module 7 Complete

You can now

State both transformer equations from the Physics Equations Sheet

Apply V_p/V_s = N_p/N_s to find any unknown pd or turns

Apply V_s × I_s = V_p × I_p (assuming 100% efficient) to find any unknown current

Chain both equations together for 5-mark calculations

Avoid the inversion trap using the "primary on top, both sides" rule

Apply the inverse V–I relationship to explain why high-voltage transmission reduces I²R losses in the National Grid

Recognise step-up (V_s > V_p, I_s < I_p) vs step-down (V_s < V_p, I_s > I_p) from the numbers

Module 7 complete · what you've covered

Magnetism (poles, fields, magnetic materials) · electromagnetism (wires producing fields, solenoids) · the motor effect (F = BIl, motors, loudspeakers) · the generator effect (alternators, dynamos, microphones) · transformers (structure, operation, calculations). Module 7 done — onto Module 8: Space Physics, starting with the solar system, then orbital motion, then red-shift and the Big Bang.