Real transformers waste energy, and efficiency tells you how much

Transformers — AQA A-Level Physics

No real transformer is 100% efficient. Some energyThe capacity to do work. Measured in joules (J). is always lost as heat in the coils and the core. The efficiencyThe ratio of useful energyThe capacity to do work. Measured in joules (J). output to total energyThe capacity to do work. Measured in joules (J). input, expressed as a fraction or percentage. tells you what fraction of the input powerThe rate of energy transfer. Measured in watts (W). is usefully transferred to the output.

\text{EfficiencyThe ratio of useful energy output to total energy input, expressed as a fraction or percentage.} = \frac{V_s I_s}{V_p I_p} \times 100\%

$V_s I_s$: output powerThe rate of energy transfer. Measured in watts (W). (W)
$V_p I_p$: input powerThe rate of energy transfer. Measured in watts (W). (W)

EfficiencyThe ratio of useful energy output to total energy input, expressed as a fraction or percentage. is always less than 100% for a real transformer. Large power transformers can reach 99%+, but exam questions typically use 90-98%.
You can also write this as: $Efficiency = Ps / Pp x 100$%, where $Ps = output power$ and $Pp = input power$.
When efficiency is given, the output power is: $P_s = \eta \times P_p$, where $\eta$ is efficiency as a decimal.

Sources of energy loss in a real transformer

Resistive heating in the coils (I²R losses): CurrentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flowing through the copper wire generates heat. The primary and secondary coils both have resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω)., and energy is dissipated as thermal energy. Larger currents mean larger losses.
Eddy currents in the core: The changing magnetic fluxThe product of magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³. and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). doesn't just induce an EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). in the secondary coil. It also induces currents within the iron core itself. These circulating currents (eddy currents) dissipate energy as heat in the core.
Hysteresis losses: The iron core is repeatedly magnetised and demagnetised by the alternating flux. Each cycle, some energy is wasted in realigning the magnetic domains. This energy appears as heat.
Flux leakage: Not all the magnetic fluxThe product of magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³. and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). produced by the primary coil passes through the secondary coil. Some flux 'leaks' through the air instead of staying in the core. This reduces the EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). induced in the secondary.

Worked Example

A transformer has a primary voltage of 230 V and a primary current of 2.0 A. The secondary voltage is 12 V. The transformer is 92% efficient. Calculate the secondary current.

Show Solution

List known values

Primary voltageThe energy transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference.: $V_p = 230 \text{ V}$
Primary current: $I_p = 2.0 \text{ A}$
Secondary voltage: $V_s = 12 \text{ V}$
Efficiency: $\eta = 92\% = 0.92$

Calculate input power

$$P_p = V_p I_p = 230 \times 2.0 = 460 \text{ W}$$

Calculate output power using efficiency

$$P_s = \eta \times P_p = 0.92 \times 460 = 423.2 \text{ W}$$

Find secondary current from output power

$$P_s = V_s I_s$$ $$I_s = \frac{P_s}{V_s} = \frac{423.2}{12} = 35.3 \text{ A (3 s.f.)}$$

Note that the current is much larger on the secondary side because the voltage is much lower. But the product $V_s I_s$ is still less than $V_p I_p$ because the transformer is not 100% efficient.

Answer

$I_s = 35$ A (2 s.f.)

Examiner Tips and Tricks

A common 6-mark question asks you to describe and explain the energy losses in a transformer and how each is reduced.
Structure your answer: name the loss, explain the physics behind it, then state the fix.
Four $losses = potentially 6$ or more marks.