10 Higher-tier AQA GCSE Physics questions on the transformer turns ratio (Vp/Vs = np/ns) that get harder as you go, each with a full worked solution.
Course: Separate Physics only | Tier: Higher only | Equation sheet: given in the exam
📌 KEY DEFINITIONS – learn them word for word
How a transformer works: an alternating current in the primary coil creates a changing magnetic field in the iron core. This changing field induces a potential difference across the secondary coil.
Step-up: more turns on the secondary → p.d. increases. Step-down: fewer turns on the secondary → p.d. decreases.
The equation
p.d. across primary ÷ p.d. across secondary = turns on primary ÷ turns on secondary Vp / Vs = np / ns
| Quantity | Symbol | Unit |
|---|---|---|
| primary p.d. | Vp | volts (V) |
| secondary p.d. | Vs | volts (V) |
| turns on primary coil | np | no unit |
| turns on secondary coil | ns | no unit |
Rearranged: Vs = Vp × ns ÷ np | ns = np × Vs ÷ Vp | np = ns × Vp ÷ Vs | Step-up: more secondary turns; step-down: fewer.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (both p.d.s in the same unit).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and give the unit (turns have none).
Higher tier only: work through all 10. Most questions need rearranging.
Questions
Q1 A transformer has 200 turns on the primary and 20 on the secondary. The primary is connected to 230 V. Calculate the secondary p.d.
Show solution
F: Vp / Vs = np / ns
I: 230 ÷ Vs = 200 ÷ 20 = 10
F: Vs = 230 ÷ 10
A: 23 V
Q2 A transformer steps 230 V down to 12 V. The primary has 1150 turns. How many turns are on the secondary?
Show solution
F: Vp / Vs = np / ns
I: 230 ÷ 12 = 1150 ÷ ns
F: ns = 1150 × 12 ÷ 230
A: 60 turns
Q3 A transformer gives 6 V from its 30-turn secondary when the primary is connected to 240 V. How many turns are on the primary?
Show solution
F: Vp / Vs = np / ns
I: 240 ÷ 6 = np ÷ 30
F: np = 30 × 240 ÷ 6
A: 1200 turns
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Q4 A power station transformer steps 25 kV up to 400 kV. The primary has 500 turns. How many turns are on the secondary?
Show solution
F: Vp / Vs = np / ns
I: 25 ÷ 400 = 500 ÷ ns (both p.d.s in kV)
F: ns = 500 × 400 ÷ 25
A: 8000 turns
Q5 A doorbell transformer has 2300 primary turns and 80 secondary turns. It is connected to the 230 V mains. Calculate the p.d. for the bell.
Show solution
F: Vp / Vs = np / ns
I: 230 ÷ Vs = 2300 ÷ 80
F: Vs = 230 × 80 ÷ 2300
A: 8 V
Q6 A substation transformer steps 400 kV down to 132 kV. The secondary has 3300 turns. How many turns are on the primary?
Show solution
F: Vp / Vs = np / ns
I: 400 ÷ 132 = np ÷ 3300
F: np = 3300 × 400 ÷ 132
A: 10 000 turns
Q7 A phone charger transformer changes 230 V to 5.0 V. Its secondary has 40 turns. How many turns are on the primary?
Show solution
F: Vp / Vs = np / ns
I: 230 ÷ 5.0 = np ÷ 40
F: np = 40 × 230 ÷ 5.0
A: 1840 turns
Q8 A step-up transformer has 2.0 × 103 primary turns and 3.2 × 104 secondary turns. The primary p.d. is 2.5 × 104 V. Calculate the secondary p.d.
Show solution
F: Vp / Vs = np / ns
I: 2.5 × 104 ÷ Vs = 2.0 × 103 ÷ 3.2 × 104
F: Vs = 2.5 × 104 × 3.2 × 104 ÷ 2.0 × 103
A: 4.0 × 105 V
Q9 A transformer with 460 primary turns and 24 secondary turns is connected to 230 V. The secondary powers a 6 Ω lamp. Calculate the current in the lamp.
Show solution
Step 1 – secondary p.d.
F: Vp / Vs = np / ns
I: 230 ÷ Vs = 460 ÷ 24
F: Vs = 230 × 24 ÷ 460 = 12 V
Step 2 – current: I = V ÷ R = 12 ÷ 6
A: 2 A
Q10 A transformer has 1500 primary turns and 85 secondary turns. It is connected to 230 V. Calculate the secondary p.d., to 2 significant figures.
Show solution
F: Vp / Vs = np / ns
I: 230 ÷ Vs = 1500 ÷ 85
F: Vs = 230 × 85 ÷ 1500 = 13.03 V
A: 13 V (2 s.f.)
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