The Physics Question Bank

Ramped Questions for GCSE Physics

Transformer Power (VpIp = VsIs)

✏️ Paper first! Work out every question on paper before you tap Show solution. Write down every step – the equation, the numbers with units, the rearranging and the answer with its unit. In the exam, if your final answer is wrong you can still get marks for correct working, but only if the examiner can see it.

Course: Separate Physics only | Tier: Higher only | Equation sheet: given in the exam

The equation

p.d. across primary × current in primary = p.d. across secondary × current in secondary (for a 100% efficient transformer) V_p × I_p = V_s × I_s

Quantity	Symbol	Unit
primary p.d.	V_p	volts (V)
primary current	I_p	amperes (A)
secondary p.d.	V_s	volts (V)
secondary current	I_s	amperes (A)

Rearranged: I_s = V_pI_p ÷ V_s | I_p = V_sI_s ÷ V_p | V_s = V_pI_p ÷ I_s | Power in = power out

Use FIFA for every answer

F – Formula: write the equation as it appears on the sheet.
I – Insert: put in the numbers (convert to standard units first).
F – Fix: rearrange to make the unknown the subject.
A – Answer: calculate and always give the unit.

Higher tier only: work through all 10. Every question needs rearranging.

Questions

Q1 A transformer takes 0.5 A from the 230 V mains and gives out 23 V. Calculate the secondary current.

Show solution

F: V_p I_p = V_s I_s
I: 230 × 0.5 = 23 × I_s
F: I_s = 115 ÷ 23
A: 5 A

Q2 A transformer supplies 2 A at 12 V to a model railway. The primary is at 240 V. Calculate the primary current.

Show solution

F: V_p I_p = V_s I_s
I: 240 × I_p = 12 × 2
F: I_p = 24 ÷ 240
A: 0.1 A

Q3 A transformer takes 0.2 A from 230 V and supplies 4 A. Calculate the secondary p.d.

Show solution

F: V_p I_p = V_s I_s
I: 230 × 0.2 = V_s × 4
F: V_s = 46 ÷ 4
A: 11.5 V

Q4 A power station generates 400 A at 25 kV. A transformer steps this up to 400 kV. Calculate the current in the transmission cables.

Show solution

Convert first: 25 kV = 25 000 V; 400 kV = 400 000 V
F: V_p I_p = V_s I_s
I: 25 000 × 400 = 400 000 × I_s
F: I_s = 10 000 000 ÷ 400 000
A: 25 A

Q5 A charger gives 2 A at 4.6 V. It is plugged into the 230 V mains. Calculate the primary current in mA.

Show solution

F: V_p I_p = V_s I_s
I: 230 × I_p = 4.6 × 2
F: I_p = 9.2 ÷ 230 = 0.04 A
A: 40 mA

Q6 A transformer connected to 230 V mains runs a 12 V, 69 W garden lamp. Calculate the primary current.

Show solution

Think: V_s × I_s is the output power = 69 W
F: V_p I_p = V_s I_s
I: 230 × I_p = 69
F: I_p = 69 ÷ 230
A: 0.3 A

Q7 1.0 MW of power is transmitted. Calculate the current if it is sent at (a) 25 kV and (b) 400 kV.

Show solution

F: V I = power (same on both sides of an ideal transformer)
I: (a) 25 000 × I = 1 000 000; (b) 400 000 × I = 1 000 000
F: (a) I = 1 000 000 ÷ 25 000; (b) I = 1 000 000 ÷ 400 000
A: (a) 40 A (b) 2.5 A – a higher p.d. means a much smaller current.

Q8 A transformer has V_p = 2.5 × 10⁴ V and I_p = 4.0 × 10³ A. The secondary p.d. is 4.0 × 10⁵ V. Calculate the secondary current.

Show solution

F: V_p I_p = V_s I_s
I: 2.5 × 10⁴ × 4.0 × 10³ = 4.0 × 10⁵ × I_s
F: I_s = 1.0 × 10⁸ ÷ 4.0 × 10⁵
A: 2.5 × 10² A

Q9 1.0 MW is sent along cables with a total resistance of 5 Ω. Calculate the power wasted in the cables if it is sent at (a) 25 kV and (b) 400 kV.

Show solution

Step 1 – currents: I = P ÷ V; (a) 1 000 000 ÷ 25 000 = 40 A; (b) 1 000 000 ÷ 400 000 = 2.5 A
Step 2 – power wasted
F: P = I² R
I: (a) P = 40² × 5; (b) P = 2.5² × 5
A: (a) 8000 W (b) 31.25 W – this is why the National Grid uses step-up transformers.

Q10 A transformer takes 0.35 A from 230 V and gives out 9.0 V. Calculate the secondary current, to 2 significant figures.

Show solution

F: V_p I_p = V_s I_s
I: 230 × 0.35 = 9.0 × I_s
F: I_s = 80.5 ÷ 9.0 = 8.94… A
A: 8.9 A (2 s.f.)