Define peak and root-mean-square values for alternating current, relate them to power, and explain the operation of an ideal transformer

What this dot point is asking

SEAB wants you to define peak and root-mean-square (r.m.s.) values for an alternating current, to use r.m.s. values in power calculations, and to explain the ideal transformer through the turns ratio and power conservation. This connects induction to the practical business of distributing electrical power.

The answer

Peak and root-mean-square values

An alternating current (a.c.) varies sinusoidally: $I = I_0\sin(\omega t)$, where $I_0$ is the peak value. Because the current is sometimes large and sometimes zero, the meaningful average for power is the root-mean-square value, the steady d.c. value that would dissipate the same mean power.

For a sinusoidal a.c.:

The r.m.s. value is about $0.707$ of the peak. Mains supplies are always quoted as r.m.s. values.

Power in an a.c. circuit

The mean power dissipated in a resistor uses r.m.s. values:

The mean power is exactly half the peak power for a resistive load, which is the reason the $\sqrt{2}$ factor appears: $\langle P \rangle = \tfrac{1}{2}V_0 I_0$.

The ideal transformer

A transformer has a primary coil of $N_p$ turns and a secondary of $N_s$ turns wound on a common iron core. An alternating current in the primary produces a changing flux in the core, which links the secondary and induces an e.m.f. there (Faraday's law). For an ideal transformer:

A step-up transformer ($N_s > N_p$) increases voltage; a step-down transformer decreases it. Transformers work only with a.c., because they need a changing flux.

Power conservation in an ideal transformer

An ideal transformer is 100 percent efficient, so the power out equals the power in:

Hence stepping the voltage up steps the current down in the same ratio. A real transformer is slightly less efficient due to resistive heating in the windings, eddy currents and hysteresis in the core, and flux leakage.

Examples in context

Example 1. The national grid. Power is generated at a moderate voltage, stepped up to hundreds of kilovolts for transmission to slash $I^2 R$ cable losses, then stepped down in stages for safe domestic use. The whole strategy depends on transformers, which is the practical reason mains electricity is a.c. rather than d.c.

Example 2. A phone charger. A phone charger contains a step-down transformer (plus rectifier) that reduces the $230\ \text{V}$ r.m.s. mains to a few volts. The turns ratio sets the output voltage, and power conservation means the low-voltage output can supply a larger current than flows in the primary.

Try this

Q1. State the relationship between the peak and r.m.s. values of a sinusoidal voltage. [1 mark]

• Cue. $V_{\text{rms}} = V_0/\sqrt{2}$, about $0.707$ of the peak.

Q2. A transformer steps $240\ \text{V}$ down to $12\ \text{V}$. If the primary has $1000$ turns, find the number of secondary turns. [2 marks]

• Cue. $N_s = N_p \dfrac{V_s}{V_p} = 1000 \times \dfrac{12}{240} = 50$ turns.

Q3. Explain why electrical power is transmitted at high voltage. [3 marks]

• Cue. For a given power, higher voltage means lower current; cable loss is $I^2 R$, so reducing the current sharply cuts the wasted energy, allowing efficient long-distance transmission.