Transformer operation

A transformer transfers electrical energy from a primary coil to a secondary coil by electromagnetic induction. The coils are wound around a magnetic core, which links them magnetically.

An alternating current in the primary coil produces a changing magnetic flux in the core. This changing flux links the secondary coil and induces an alternating voltage across it.

The flux must change for a voltage to be induced in the secondary coil. A steady direct current produces a steady flux, so it does not sustain transformer action.

Transformer voltage and turns ratios

For an ideal transformer, the ratio of the secondary voltage to the primary voltage equals the ratio of their numbers of coil turns:

Here $$N_p$$ and $$N_s$$ are the numbers of turns in the primary and secondary coils, and $$V_p$$ and $$V_s$$ are the primary and secondary voltages.

A step-up transformer has more turns on the secondary coil than on the primary, so its secondary voltage is greater. A step-down transformer has fewer secondary turns, so its secondary voltage is lower.

Eddy currents

When the magnetic flux through a path within a conductor changes, it induces a current around that path. Currents that circulate in this way within the conductor are called eddy currents.

The eddy currents produce their own magnetic effects, which oppose the change in flux that induced them. Because the conductor has resistance, the currents also transfer energy to the conductor as thermal energy, causing it to heat up.

Eddy currents in a transformer core

A transformer core carries changing magnetic flux. This changing flux induces eddy currents in the core material, so some energy is dissipated as heat in the core.

To reduce this heating, the core is made from thin conducting sheets called laminations, separated by insulating layers. The insulation restricts the paths through which eddy currents can circulate, reducing the currents and the associated heating.

Transformer efficiency

A transformer receives electrical power through its primary coil and delivers electrical power through its secondary coil. Its efficiency is the fraction of the input power delivered as useful electrical output.

For voltages and currents whose products represent real electrical power, the efficiency is:

Here, $$I$$ is current and $$V$$ is voltage. The subscripts $$p$$ and $$s$$ refer to the primary and secondary coils respectively. Multiply the efficiency ratio by 100 to express it as a percentage.

The voltage–current products in this equation must represent real power. If they do not, the ratio of the products cannot be used directly as the efficiency.

Transformer losses

A real transformer is less than 100% efficient because not all the input power reaches the secondary as useful electrical output. The main losses and ways to reduce them are:

• Heating in the windings: The resistance of the coil wires causes them to heat as current flows. Lower-resistance windings reduce this loss.
• Eddy-current heating in the core: Changing magnetic flux induces currents in the core, which heat it. A laminated core, made of insulated layers, restricts these currents and reduces the heating.
• Repeated magnetisation of the core: Repeatedly changing the core’s magnetisation transfers energy to the core as heat. A suitable core material that is easily magnetised and demagnetised reduces this loss.
• Magnetic flux leakage: Some magnetic flux produced by one coil does not link the other coil, so it does not transfer energy between them effectively.

High-voltage power transmission

Transmission lines have electrical resistance, so some of the power they carry is dissipated as heat. Transmitting power at a higher voltage allows the same real power to be carried by a lower current, reducing this heating loss.

When the voltage and current are such that their product gives the real power, their relationship is:

Here $$P$$ is real power, $$V$$ is voltage and $$I$$ is current. For a given power, increasing the voltage reduces the current: for example, doubling the voltage halves the current.

The power dissipated as heat in the transmission path is:

Here $$R$$ is the total resistance of the conducting transmission path carrying the current. Because the current is squared, reducing it has a strong effect: if the current is halved and the resistance stays the same, the heating loss falls to one quarter.

Calculating transmission loss

Suppose $$100,\text{kW}$$ is transmitted at $$10,\text{kV}$$, with voltage and current whose product gives the real power. The total resistance of the conducting path is $$2,\Omega$$.

The current is $$I=P/V=100,000/10,000=10,\text{A}$$. The heating loss is therefore $$P_{\text{loss}}=10^2\times 2=200,\text{W}$$.

If the same power is transmitted at $$100,\text{kV}$$ instead, the current is $$1,\text{A}$$. With the same path resistance, the loss is $$1^2\times 2=2,\text{W}$$. A tenfold increase in voltage has reduced the current to one tenth and the loss to one hundredth.

Transformers can raise the voltage before transmission and reduce it again for use.