## Episode 123: Alternating current

Lesson for 16-19

- Activity time 85 minutes
- Level Advanced

The aims are to distinguish alternating from direct currents and to remind your students of why AC is so important (they should already have met this at pre-16 level).

Lesson Summary

- Demonstration: The output of a generator (10 minutes)
- Student experiment: Measuring AC with a CRO (20 minutes)
- Discussion: r.m.s values as equivalent DC values (20 minutes)
- Demonstration: Finding the r.m.s value (15 minutes)
- Student questions: Practice on AC (20 minutes)

#### Demonstration: The output of a generator

A simple hand-turned generator can be used to create an alternating voltage that can be seen on an analogue demonstration voltmeter (needle swaying back and forth). This can be used to emphasise the ideas of frequency and peak value. Follow up by connecting the generator to a lamp and showing how the brightness depends on frequency of rotation.

Don’t get bogged down with electromagnetic induction but do compare what charges are doing inside a lamp filament when it is connected (a) to a DC supply and (b) to an AC supply (in the first case they drift in one direction in the second they move back and forth).

Episode 123-1: Generating alternating current (Word, 24 KB)

#### Student experiment: Measuring AC with a CRO

Measuring the peak value (amplitude), peak-to-peak value, and frequency of an AC supply using an oscilloscope.

This exercise builds on their introduction to oscilloscopes and sets the scene for the next demonstration. Make sure they are able to identify and measure peak and peak-to-peak values and to work out frequency from the time period of the voltage variation.

Episode 123-2: Measuring the peak and peak-to-peak values, and frequency (Word, 35 KB)

#### Beware:

a lot of oscilloscopes have calibration positions

on their variable y-gain and time base settings. Students will need to be reminded to set these prior to making measurements otherwise they will get systematic errors throughout.

#### Discussion: r.m.s values as equivalent DC values

Ask the class what the average value of an AC voltage or current is over a whole number of cycles. It is obviously zero. So how can AC transfer energy? Remind them that power is calculated by
*P* = *I* × *V*
and point out that both *I* and *V* change sign together, so power is always positive but varies over the cycle, having a maximum value of *I*_{p}*V*_{p} and a minimum of zero. This should also make it clear that the average power is *less* than *I*_{p}*V*_{p} . Good mathematicians may know that the average value of a sine or cosine squared over a whole number of cycles is just 12. Weaker students might be persuaded by the symmetry (either side of *V* = 12) of a graph of sine-squared or cosine-squared against time. Either way you need to lead them to the idea that average power delivered by AC is:

*P* = 12 *I*_{p} *V*_{p}
or,
*P* = 12 (*I*_{p})^{2} *R*
or,
*P* = 12 (*V*_{p})^{2}*R*

(using *V* = *I* × *R*)

The same power would be delivered by a DC with values *I* and *V* if:

*P* = *I*^{ 2} × *R*

*P* = 12(*I*_{p})^{2} *R*

, or

*P* = *V*^{ 2}*R*

*P* = 12(*I*_{p})^{2}*R*

These lead to the equations:

*I*^{ 2} = *I*_{p}2

giving

*I* = *I*_{p}√ 2

*and*

*V*^{ 2} = *V*_{p}2

in turn giving

*V* = *V*_{p}√ 2

This links the DC equivalent values to AC peak values. The point is that a sinusoidal AC supply of peak value *V*_{p} delivers the same average power as steady DC of value *V*_{p}√ 2 .

We call these DC equivalent values

the r.m.s values for AC and we can use them in the same way as steady DC values: e.g. average AC power
*P* = *I*_{r m s} × *V*_{r m s}

You ought to say that r.m.s stands for root-mean-square

but it is only really worth going into the meaning of this in detail with groups who can handle the mathematics.

#### Demonstration: Finding the r.m.s value

To reinforce the idea of DC equivalence show them a demonstration in which a filament lamp is lit from first a DC and then an AC supply and the supplies are adjusted to make the lamp equally bright (equal powers). From the previous discussion you should be able to coax them to predict that the peak value of the AC (shown on an oscilloscope) is root-two times the steady DC value.

Episode 123-3: Showing the equivalence of AC and DC (Word, 29 KB)

#### Student questions: Practice on AC

End the session by discussing the mains supply. AC currents and voltages are usually quoted as r.m.s values so 230 V 50 Hz mains AC varies from + 325 V to − 325 V and has a period of 150 (0.02) second. 230 V is its r.m.s value, 325 V is its peak value and 650 V is its peak to peak value.

#### Questions on AC signals

Episode 123-4: Questions on alternating current (Word, 24 KB)