TransformersAQA A-Level Physics: Revision notes

A transformer changes the size of an alternating p.d. using electromagnetic induction. An alternating current in the primary coil produces a changing magnetic flux in a shared soft-iron core. This flux links the secondary coil, where it induces an alternating emf. A transformer only works with alternating (or changing) current, because a steady current gives no changing flux.

For an ideal transformer: $\frac{N_s}{N_p} = \frac{V_s}{V_p}$

A step-up transformer has $N_s > N_p$ and raises the p.d.; a step-down transformer has $N_s < N_p$ and lowers it.

An ideal transformer is 100% efficient, so power in equals power out: $I_pV_p = I_sV_s$. A step-up transformer therefore gives a smaller current in the secondary.

Real transformers lose some energy, so:

$\text{efficiency} = \frac{I_sV_s}{I_pV_p}$

This is often quoted as a percentage. Large grid transformers reach over 98%.

The core of a transformer is a conductor and sits in a changing magnetic flux. By Faraday's law, emfs are induced in the core, and these drive circulating eddy currents. The currents heat the core ($I^2R$), so energy is transferred thermally to the surroundings.

The effect is reduced by laminating the core: it is built from thin sheets of iron separated by insulating layers. This raises the resistance of the loops that the currents can follow, so the eddy currents are much smaller.

Energy is wasted in real transformers because of:

• Resistance of the windings: the coils heat up. Thicker, low-resistance copper wire reduces this.
• Eddy currents in the core: reduced by lamination.
• Flux leakage: not all the flux from the primary links with the secondary. Winding the coils closely on a continuous core reduces this.
• Hysteresis: energy is used in repeatedly magnetising and demagnetising the core. A soft-iron core, which magnetises and demagnetises easily, keeps this small.

All of these lead to energy being transferred thermally, so the output power is less than the input power.

The power delivered to a line is $P = IV$, so for a fixed power a high voltage means a small current. The power wasted as heat in the transmission lines is $P_{loss} = I^2R$.

Stepping the p.d. up by a factor of 10 reduces the current by 10 and the power loss by a factor of 100. This is why the National Grid transmits at 275 kV or 400 kV and uses step-down transformers near towns.

A 1.0 MW supply is sent along lines of resistance 5.0 Ω at 20 kV.

$I = P/V = 1.0\times10^6 / 2.0\times10^4 = 50$ A

$P_{loss} = I^2R = 50^2 \times 5.0 = 1.25\times10^4$ W, which is 1.3% of the supply.

At 2.0 kV the current would be 500 A and the loss $1.25\times10^6$ W, more than the power supplied, which shows why high voltage is essential.