GCSE Physics - How to Solve: Transformer Calculations (Turns Ratio, Power) – Complete Guide

How to Solve: Transformer Calculations (Turns Ratio, Power) – Complete Guide

Introduction "Mastering transformer calculations unlocks 6–8 marks on your GCSE/A-Level Physics exam—enough to boost your grade by a full level—and lets you design real-world power grids, phone chargers, and electric vehicle systems!"

WHAT YOU NEED TO KNOW FIRST

Before tackling transformers, you must understand:
1. Electromagnetic induction – A changing magnetic field induces a voltage.
2. Power in circuits – Power (P) = Voltage (V) × Current (I).
3. Ratio basics – How to simplify and compare fractions (e.g., 2:1 = 2/1).

If you’re shaky on any of these, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Turns Ratio Formula [ \frac{V_p}{V_s} = \frac{N_p}{N_s} ]
2. (V_p) = Primary voltage (V)
3. (V_s) = Secondary voltage (V)
4. (N_p) = Number of turns on primary coil
5. (N_s) = Number of turns on secondary coil

6. MEMORISE THIS – It’s the foundation of all transformer questions.

7. Power in an Ideal Transformer [ P_p = P_s \quad \text{(Input power = Output power)} ] [ V_p \times I_p = V_s \times I_s ]

8. (I_p) = Primary current (A)
9. (I_s) = Secondary current (A)

10. MEMORISE THIS – Assumes 100% efficiency (no energy loss).

11. Efficiency Formula (Real Transformers) [ \text{Efficiency} = \frac{P_s}{P_p} \times 100\% ]

12. Given on exam sheet (but know how to use it).

STEP-BY-STEP METHOD

Follow these exact steps for every transformer question:

1. Identify what’s given – Write down all known values (e.g., (V_p), (N_s), (I_s)).
2. Determine what’s asked – Are you finding voltage, current, turns, or power?
3. Choose the right formula – Use the turns ratio for voltage, power formula for current.
4. Rearrange if needed – Solve for the unknown (e.g., (V_s = V_p \times \frac{N_s}{N_p})).
5. Calculate – Plug in numbers and solve.
6. Check units – Ensure volts (V), amps (A), and turns are consistent.
7. Verify logic – Does the answer make sense? (e.g., step-up = higher voltage).

WORKED EXAMPLES

Example 1 – Basic (Turns Ratio)

Question: A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. If the primary voltage is 230 V, what is the secondary voltage?

Step-by-Step Solution:
1. Given: - (N_p = 500) turns - (N_s = 100) turns - (V_p = 230) V
2. Asked: (V_s = ?)
3. Formula: (\frac{V_p}{V_s} = \frac{N_p}{N_s})
4. Rearrange: (V_s = V_p \times \frac{N_s}{N_p})
5. Calculate: [ V_s = 230 \times \frac{100}{500} = 230 \times 0.2 = 46 \text{ V} ]
6. Check: Fewer turns on secondary → lower voltage (step-down). ✔️

What we did and why: We used the turns ratio to find the secondary voltage because the number of turns directly affects voltage. Since the secondary has fewer turns, the voltage drops—this is a step-down transformer.

Example 2 – Medium (Current Calculation)

Question: A step-up transformer has a primary voltage of 12 V and a secondary voltage of 240 V. If the primary current is 5 A, what is the secondary current? Assume 100% efficiency.

Step-by-Step Solution:
1. Given: - (V_p = 12) V - (V_s = 240) V - (I_p = 5) A
2. Asked: (I_s = ?)
3. Formula: (V_p \times I_p = V_s \times I_s) (power conservation)
4. Rearrange: (I_s = \frac{V_p \times I_p}{V_s})
5. Calculate: [ I_s = \frac{12 \times 5}{240} = \frac{60}{240} = 0.25 \text{ A} ]
6. Check: Step-up transformer → higher voltage → lower current. ✔️

What we did and why: We used the power formula because in an ideal transformer, input power equals output power. Since voltage increases, current must decrease to keep power constant.

Example 3 – Exam-Style (Efficiency & Power)

Question: A transformer has 200 turns on the primary coil and 1000 turns on the secondary coil. The primary voltage is 20 V, and the primary current is 2 A. The transformer is 80% efficient. Calculate: a) The secondary voltage. b) The secondary current. c) The output power.

Step-by-Step Solution:

Part a) Secondary Voltage
1. Given: - (N_p = 200) - (N_s = 1000) - (V_p = 20) V
2. Asked: (V_s = ?)
3. Formula: (\frac{V_p}{V_s} = \frac{N_p}{N_s})
4. Rearrange: (V_s = V_p \times \frac{N_s}{N_p})
5. Calculate: [ V_s = 20 \times \frac{1000}{200} = 20 \times 5 = 100 \text{ V} ]
6. Check: More turns on secondary → higher voltage (step-up). ✔️

Part b) Secondary Current (Ideal First)
1. Given: - (V_p = 20) V - (I_p = 2) A - (V_s = 100) V (from part a)
2. Asked: (I_s = ?) (ideal)
3. Formula: (V_p \times I_p = V_s \times I_s)
4. Rearrange: (I_s = \frac{V_p \times I_p}{V_s})
5. Calculate: [ I_s = \frac{20 \times 2}{100} = \frac{40}{100} = 0.4 \text{ A} ]
6. Adjust for efficiency (80%): - Real output power = 80% of ideal output power. - Ideal output power = (V_s \times I_s = 100 \times 0.4 = 40) W - Real output power = (40 \times 0.8 = 32) W - Real (I_s = \frac{\text{Real output power}}{V_s} = \frac{32}{100} = 0.32) A

Part c) Output Power
1. From part b: Real output power = 32 W.

What we did and why: - For voltage, we used the turns ratio (step-up = higher voltage). - For current, we first calculated the ideal current, then adjusted for efficiency because real transformers lose some power as heat. - Output power is lower than input power due to inefficiency.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Right, listen up—this is your last-minute transformer cheat sheet. First, turns ratio = voltage ratio: (\frac{V_p}{V_s} = \frac{N_p}{N_s}). More turns on secondary? Step-up = higher voltage. Fewer turns? Step-down = lower voltage. Second, power in = power out (if ideal): (V_p \times I_p = V_s \times I_s). Higher voltage means lower current, and vice versa. Third, if efficiency is given, adjust output power: real power = ideal power × efficiency. Finally, label your coils—primary is input, secondary is output. Now go smash those exam questions!"