Alternating currentsEdexcel A-Level Physics: Revision notes

A direct current (d.c.) flows in one direction only. An alternating current (a.c.) repeatedly reverses direction, and the p.d. driving it varies in a regular way. The mains supply is a sinusoidal alternating supply.

An oscilloscope shows how the p.d. varies with time. The time-base setting converts horizontal divisions into time and the Y-gain converts vertical divisions into volts.

• The period $T$ is the time for one complete cycle, in seconds.
• The frequency $f$ is the number of complete cycles per second, in hertz (Hz).

$f = \frac{1}{T}$

UK mains has $f = 50$ Hz, so $T = 1/50 = 0.020$ s.

Oscilloscope example. One cycle spans 4.0 divisions at 5.0 ms per division, so $T = 20$ ms and $f = 50$ Hz.

The peak value ($V_0$ or $I_0$) is the maximum p.d. or current reached in a cycle, measured from zero. The peak-to-peak value is twice this.

On an oscilloscope, $V_0$ is the number of divisions from the centre line to the top of the trace multiplied by the Y-gain. For a trace reaching 3.0 divisions at 2.0 V per division, $V_0 = 6.0$ V.

An alternating p.d. is at its peak only briefly, so its peak value overstates the heating effect. The root-mean-square (r.m.s.) value is the value of the steady direct p.d. (or current) that would transfer the same mean power to a resistor.

For a sinusoidal supply:

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \qquad I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

Mean power in a resistor $R$:

$P = \frac{V_{\text{rms}}^2}{R} = I_{\text{rms}}^2 R = V_{\text{rms}}I_{\text{rms}}$

The peak power is $V_0^2/R$, which is twice the mean power.

Mains of 230 V r.m.s. has a peak of $230\sqrt{2} = 325$ V.

A heater of resistance 46 Ω is connected to 230 V r.m.s. mains.

$I_{\text{rms}} = 230/46 = 5.0$ A

$I_0 = 5.0 \times \sqrt{2} = 7.1$ A

$P = V_{\text{rms}}I_{\text{rms}} = 230 \times 5.0 = 1150$ W

A lamp of resistance 12 Ω on an a.c. supply of peak 8.5 V: $V_{\text{rms}} = 8.5/\sqrt{2} = 6.0$ V, so the lamp is as bright as on 6.0 V d.c., and the mean power is $6.0^2/12 = 3.0$ W.