AC, transformers and the grid

Mains electricity is alternating current, which can be stepped up or down by transformers. Power stations generate at about 25 kV; step-up transformers raise this to hundreds of kilovolts for the National Grid so the current – and the energy wasted heating the cables – is small. Step-down transformers then bring it safely to 230 V (or 120 V) for homes.

Key facts

  • Transformer equation: Vs ÷ Vp = Ns ÷ Np. More secondary turns = step-up.
  • Ideal (100% efficient) transformer: Vp Ip = Vs Is.
  • Transformers only work with a.c.: a changing current makes a changing field in the core.
  • Cable losses P = I²R: raising V by a factor k cuts the current by k and the losses by k².
  • r.m.s. values: V_rms = V₀ ÷ √2; average power P = V_rms I_rms = V_rms² ÷ R.
  • On an oscilloscope: peak = divisions × volts/div; period = divisions × time/div.

Worked examples

Example 1

A transformer has 2000 primary turns and 100 secondary turns. The input is 230 V. Find the output voltage.

  1. Vs = Vp × Ns ÷ Np Vs = 230 × 100 ÷ 2000 = 11.5 V

Answer: 11.5 V

Example 2

10 MW is sent through cables of resistance 5 Ω at 400 kV. Find the current and the power wasted in the cables. What would the loss be at 25 kV?

  1. Current I = 10 × 10⁶ ÷ 400 × 10³ = 25 A
  2. Loss P = 25² × 5 = 3125 W ≈ 3.1 kW
  3. At 25 kV I = 400 A, P = 400² × 5 = 800 000 W

Answer: 25 A, 3.1 kW; 800 kW at 25 kV

Example 3

The UK and South African mains is 230 V r.m.s. Find the peak voltage and the average power in a 46 Ω heater.

  1. V₀ = V_rms × √2 V₀ = 230 × 1.414 = 325 V
  2. P = V_rms² ÷ R P = 230² ÷ 46 = 1150 W

Answer: 325 V; 1150 W