Transformer Efficiency and Power Revision Notes | GCSE AQA Higher Physics

1. Why Transformer Efficiency Matters

Power stations generate electricity at around 25,000V, but your home receives 230V. Between the two, the voltage gets stepped up to 400,000V for transmission across the country, then stepped down again for local distribution. This voltage juggling act happens through transformers - and every step wastes some energy.

In an ideal world, transformers would be 100% efficient - all the electrical power going in would come out the other side. Reality is messier. Real transformers lose energy as heat in the coils and magnetic core, typically achieving 95-99% efficiency. That missing 1-5% might sound small, but when you're transmitting gigawatts of power across a national grid, those losses add up to millions of pounds.

Understanding transformer efficiency isn't just academic - it's the key to explaining why we use high voltages for power transmission and how engineers minimise energy waste in electrical systems. The maths behind it reveals something counterintuitive: sometimes you need to increase voltage to save energy.

2. The Perfect Transformer Model

Before tackling real-world efficiency problems, we need to understand the ideal case. In a perfect transformer, energy conservation rules everything. The electrical power flowing into the primary coil must equal the electrical power flowing out of the secondary coil - no energy disappears.

This gives us the fundamental power equation for transformers: Vp × Ip = Vs × Is. The subscript 'p' stands for primary (input side) and 's' for secondary (output side). Notice what this tells us: if voltage goes up, current must go down proportionally to keep the power constant.

**Trap**: Many students think transformers can create energy. They can't - they only change the voltage-current combination while keeping power roughly the same. A step-up transformer that doubles voltage will halve the current, assuming perfect efficiency.

3. Calculating Input Current for Required Power Output

Real exam questions often give you the power output needed and ask you to find the input current. This is where the power equation becomes a practical tool rather than just theory. You need to work backwards from what the load requires to what the supply must provide.

The key insight is that power output determines secondary current (P = VI), and the transformer ratio then determines primary current. For ideal transformers, you can use either the power equation directly or combine it with the voltage transformation ratio.

Remember that the input current depends on both the power demand and the input voltage. A higher input voltage means lower input current for the same power transfer - this is why power stations use high voltages for transmission.

Worked example: Finding Input Current from Power Output

A transformer supplies 5kW of power to a load at 250V. The primary voltage is 11kV. Calculate the primary current, assuming 100% efficiency. [4 marks]

1. $I_s = \frac{P}{V_s} = \frac{5000}{250} = 20\text{ A}$ — Power = Voltage × Current, so Current = Power ÷ Voltage [M1]
2. $V_p \times I_p = V_s \times I_s$ — For ideal transformers, power input = power output [M1]
3. $11000 \times I_p = 250 \times 20$ — Convert 11kV to 11,000V for consistency [M1]
4. $I_p = \frac{250 \times 20}{11000} = \frac{5000}{11000} = 0.45\text{ A}$ — Notice: higher input voltage means much lower input current [A1]

Answer: Primary current = 0.45 A

4. Real Transformer Efficiency

Real transformers aren't perfect. Energy gets lost as heat in the copper windings (I²R losses), as heat in the iron core due to changing magnetic fields (eddy currents), and as sound energy from mechanical vibrations. These losses mean that power output is always less than power input.

Efficiency is calculated as a percentage: (Useful Power Output ÷ Total Power Input) × 100%. A 95% efficient transformer wastes 5% of the input energy as heat. This might seem small, but in a power station transformer handling 500MW, that's 25MW of waste heat - enough to power a small town.

The efficiency affects the current calculations we've been doing. If a transformer is only 90% efficient, the input power must be higher than the output power to compensate for losses. This means higher input current than the ideal case would predict.

Worked example: Efficiency Calculation with Real Losses

A transformer has an input power of 12kW and delivers 11.4kW to the load. Calculate its efficiency and the power lost as heat. [3 marks]

1. $\text{Efficiency} = \frac{\text{Power Output}}{\text{Power Input}} \times 100\%$ — Standard efficiency formula [M1]
2. $\text{Efficiency} = \frac{11400}{12000} \times 100\% = 95\%$ — Convert kW to W for consistency [A1]
3. $\text{Power Lost} = 12000 - 11400 = 600\text{ W}$ — Power lost = Input - Output [A1]

Answer: Efficiency = 95%, Power lost = 600W

5. Why High Voltage Transmission Saves Energy

Here's the counterintuitive bit: to save energy, power companies increase voltage to dangerous levels. The National Grid transmits electricity at 400kV - high enough to kill you from several metres away. Why take this risk when your home runs perfectly well on 230V?

The answer lies in power losses during transmission. When current flows through transmission cables, energy is lost as heat according to P = I²R. The power loss depends on the square of the current - double the current, and you get four times the heat loss. This is why minimising current is crucial for efficient transmission.

Transformers make this possible. By stepping up the voltage before transmission, they step down the current proportionally. The same power gets transmitted (P = VI), but at high voltage and low current instead of low voltage and high current. The result: dramatically reduced I²R losses in the cables.

Consider transmitting 1MW of power over a cable with 10Ω resistance. At 1000V, you need 1000A, giving losses of I²R = 1000² × 10 = 10MW - you lose ten times more power than you deliver! At 100kV, you only need 10A, giving losses of 10² × 10 = 1kW - just 0.1% of the transmitted power.

6. Exam Technique and Common Mistakes

Transformer efficiency questions follow predictable patterns in GCSE exams. You'll typically need to calculate input current from output power, work out efficiency percentages, or explain why high voltage transmission is more efficient. The key is recognising which equation to use and keeping your units consistent.

Watch out for the most common mistakes: forgetting to convert kW to W, mixing up primary and secondary values, and assuming 100% efficiency when the question gives you real efficiency data. Always check whether the question asks for ideal or real transformer behaviour.

For 'explain' questions about high voltage transmission, structure your answer around the I²R relationship. Start with power conservation (P = VI), explain how high voltage means low current, then link to reduced I²R losses. This logical chain will earn you full marks.