Sinusoidal alternating current and voltage, peak and root mean square values, the oscilloscope, the transformer equation, transformer efficiency and the transmission of electrical power.

What this dot point is asking

AQA specification points 3.7.5.5 and 3.7.5.6 want you to describe sinusoidal alternating current and voltage, calculate peak and root mean square values, interpret oscilloscope traces, use the transformer equation, and explain how transformers allow efficient power transmission.

Alternating current and peak values

An ac source produces a current and voltage that vary sinusoidally with time, reversing direction each cycle: $V = V_0 \sin(2\pi f t)$, where $V_0$ is the peak value and $f$ the frequency. UK mains has a frequency of $50 \text{ Hz}$, so the current reverses one hundred times a second. The peak value is the maximum displacement from zero, reached twice per cycle (once positive, once negative).

Root mean square values

The average current over a complete cycle is zero, because it spends equal time positive and negative, so the mean current is useless for describing the power delivered. Power depends on $I^2$, which is always positive, so we use the root mean square.

The factor $\dfrac{1}{\sqrt{2}}$ arises because the mean of $\sin^2$ over a cycle is exactly $\tfrac{1}{2}$. The average power delivered is $P = I_{rms}V_{rms} = \tfrac{1}{2}I_0V_0$. Quoted mains voltages are always rms values: UK mains is $230 \text{ V}$ rms, corresponding to a peak of $230\sqrt{2} \approx 325 \text{ V}$.

The oscilloscope

To read a trace: count the vertical divisions from the centre line to a peak and multiply by the volts-per-division to get $V_0$; count the horizontal divisions for one cycle and multiply by the time-per-division to get $T$.

The transformer equation

A transformer has two coils wound on a soft-iron core. An alternating current in the primary produces a changing magnetic flux in the core, which links the secondary and induces an alternating EMF in it (electromagnetic induction).

For an ideal (100 percent efficient) transformer, power is conserved: $V_pI_p = V_sI_s$. This means a step-up transformer that raises the voltage must lower the current in the same ratio, and vice versa.

Efficiency and power transmission

The voltage is then stepped back down by substation transformers before reaching homes, ending at $230 \text{ V}$ rms for safety.

Try this

Q1. A sinusoidal supply has a peak voltage of $325 \text{ V}$. Calculate the rms voltage. [2 marks]

• Cue. $V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{325}{\sqrt{2}} = 230 \text{ V}$.

Q2. Explain why electrical power is transmitted at high voltage. [2 marks]

• Cue. Higher voltage means lower current for the same power, and power loss is $I^2R$, so losses are reduced.

Q3. State one cause of energy loss in a real transformer core and how it is reduced. [2 marks]

• Cue. Eddy currents, reduced by laminating the core into thin insulated layers.