This episode revises the idea of EMF, and introduces the idea of internal resistance and energy dissipation.

Lesson Summary

• Demonstration and discussion: Revising ideas about energy in the context of electric circuits (20 minutes)
• Student experiment: Leading to the idea of lost volts (30 minutes)
• Demonstration: (Optional) Drop in terminal PD of a source on load (5 minutes)

Discussion and demonstrations: Revising ideas about energy in the context of electric circuits

Here you have the opportunity to check and reinforce students’ basic ideas about current, voltage and energy in electrical circuits. This could be truncated or skipped with an able group.

Episode 120-1: Energy transfer from a chemical cell (Word, 24 KB)

Use any suitable battery-driven device as a prop – e.g. toy car, MP3 player, torch – and ask the class a number of questions. (For the sake of illustration, we have assumed a supply voltage of 3.0 V from two AA cells, but this should be adapted to the particular example used.)

Some questions to ask:

• How is energy stored in the batteries? (Chemically.)
• This is powered by two 1.5 V AA cells in series – what is the supply voltage? (3.0 V)
• What is the EMF of the supply? (3.0 V – remind them that EMF is the name given to the potential difference across the supply when it is not loaded.)
• After a while the battery needs to be replaced; why? (Steer them away from the idea that it runs out of charge – you are looking for an answer based on chemical reactions. The reactions that allow the battery to do work on charge have completed – like burning fuel.)
• What determines how quickly it runs down? (Look for answers linked to power, the rate at which energy is transferred and link this to current drawn and supply voltage.)
• How could the design be modified so it runs for longer before batteries need replacing? (Use two pairs of two AA batteries in parallel – effectively more chemical fuel but running at the same voltage so each cell provides half the previous power and current and lasts twice as long. This is unwise in practice though, if one pair of cells runs down faster than the other current can be forced backwards through the cells and they get very hot.)
• What determines how much current is drawn from the supply? (Load resistance – the lower the load resistance the greater the current and power and the sooner the battery needs to be replaced.)

Now open up the discussion to include what is happening within the cells. Point out that current flows through all parts of the circuit, including the cells. Since the cells themselves are made of material with resistance there will be some heating in the cell. This will increase with current, so there is more dissipation at higher currents. Although the chemicals that produce the current are the same ones that are responsible for the cell’s internal resistance (this term can be introduced quite naturally through discussion) it is convenient to separate these two roles so that they end up with an idea of heating as a way that energy is dissipated. We represent a real cell as two components in a circuit diagram:

• an ideal cell, with no internal resistance, and
• a small resistor, representing the internal resistance.

It can make things clearer if you draw a box or circle around these two, to indicate that they are aspects of the same component.

Energy is conserved, so the energy that was stored chemically that is transferred by the current must be equal to the energy dissipated by the components in the circuit, both th external load and the internal resistance of the cell.

Clearly the heating in the cell is not doing useful work and so is in a sense lost. This sets up the idea of lost volts that will be referred to later, though the energy is not lost, merely dissipated.

Student experiment: Leading to the idea of lost volts

This could be carried out as a demonstration but, if resources are available, the point comes across more strongly as a quick class practical. It can be advisable to do these experiments using rechargeable cells otherwise it can be quite expensive! However, rechargeable cells have a very low internal resistance so the experiment doesn’t work well. One could always cheat and place a rechargeable cell in a box with a series resistor.

Episode 120-2: Loading a battery (Word, 24 KB)

The idea is simple. Students add lamps in parallel to a suitable battery. Introduce the idea of terminal voltage (terminal pd) and get them to measure it. They could also measure the current drawn from the cell, although this is less important.

Ask them to record the terminal voltage of the battery with no lamps connected and then as the number of lamps in parallel is increased.

Can they explain what is happening? Draw out the following ideas:

• As the number of parallel lamps increases, the terminal voltage drops and the current increases.
• This is because there is more heating of the internal resistance of the cell.
• The terminal voltage of the cell is equal to only its EMF when no current is drawn (no lamps attached).
• The difference between the battery EMF and its terminal voltage is a lost voltage across the internal resistance of the battery, given by Ir.
• The more current that is drawn from the battery the larger the lost voltage.

Class discussion after the experiment should prepare the way for the theory session which follows. It is worth pointing out that the EMF of a cell can be measured by connecting a high resistance voltmeter directly across its terminals (high here means high compared to the internal resistance of the cell).

Demonstration: (optional) Drop in terminal pd of a source on load (5 minutes)

This shows the effect of a high internal resistance and an EHT supply.

Episode 120-3: Drop in terminal pd of a source on load (Word, 26 KB)

The starting point for the theory can be either Kirchhoff’s second law or conservation of energy in the circuit (the same thing really) but a general discussion based on the circuit diagram below should use a variety of approaches.

Lesson Summary

• Discussion: Deriving an equation (15 minutes)
• Discussion: Practical effects of internal resistance (10 minutes)
• Student questions: Internal resistance of a power supply (20 minutes)
• Student experiment: Measuring internal resistance and EMF (45 minutes)
• Student questions: Practice questions (30 minutes)
• Discussion: More about the practical importance of internal resistance (10 minutes)

Discussion: Deriving an equation

There are three ways to arrive at the equation relating EMF, terminal PD, current and internal resistance. It is worth discussing all three, to show their equivalence. The order you take will depend on the approach used previously with the class:

Kirchhoff’s 2^nd Law: As charge goes around the circuit the sum of EMFs must equal the sum of voltage drops leading to:

E = IR + Ir

The terminal voltage is equal to IR so this can be rearranged to give:

V = E − Ir

and interpreted as terminal voltage = EMF − lost volts

Energy is conserved. Imagine a unit of charge, Q, moving around the circuit:

QE = QIR + QIr

This leads to the same equations as above.

Use Ohm’s law with E driving current through the combined resistance (R + r):

I = ER + r

Multiplying throughout by (R + r) leads to the same equations and conclusions as in (1).

Discussion: Practical effects of internal resistance

At this point it might be worth pausing to illustrate the effects. Take a car as an example. The headlamps are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a low resistance it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01  Ω ). The headlamps themselves draw a much lower current. Ask them what happens when the engine is started (switch to starter motor closed for a short time). Look for an answer in general terms:

• sudden demand for more current
• large lost volts (around 0.05  Ω   ×  60 A = 3 V)
• terminal voltage drops to 12 V – 3 V = 9 V
• headlamps dim

When the engine fires, the starter motor switch is opened and the current drops. The terminal voltage rises and the headlamps return to normal. It’s better to turn the headlamps off when starting the car.

As an aside, a lot of students seem to think the engine is powered by the battery! Point out that its main purpose while the engine is running is to provide the sparks for ignition and that while the car is driving the alternator continually recharges the battery, the energy for both headlamps and driving comes ultimately from the fuel that is burnt (since the car has to work a little bit harder to turn the alternator).

Student questions: Internal resistance of a power supply

Some simple questions about the internal resistance of a power supply.

Episode 121-1: Internal resistance of power supplies (Word, 30 KB)

Student experiment: Measuring internal resistance and EMF

There are two experiments here, in which students determine the EMF E and internal resistance r of cells – one involving a potato cell (leading to a high internal resistance) and one involving a normal C cell (much lower internal resistance). You could get them to do both or ask some students to do one and some the other. Beware that, if you use an alkaline, high power C cell, it will run down quickly when there is a low load resistance, so you are advised to use cheap, low power cells which polarise quickly, they will depolarise over night. An alternative is to construct an artificial cell with a larger internal resistance by adding a higher series resistance (e.g. 100  Ω ) to a standard cell.

Episode 121-2: Internal resistance of a source of EMF (Word, 48 KB)

Episode 121-3: Internal resistance of a C cell (Word, 28 KB)

To determine E and r from the experimental results, there are various approaches. The simplest is to measure terminal voltage (V) and current (I) and to plot V against I. This gives an intercept at V = E on the y-axis and has a gradient of −  r.

Student questions: Practice questions

Questions on EMF and internal resistance.

Episode 121-4: Questions on EMF and internal resistance (Word, 29 KB)

Discussion: More about the practical importance of internal resistance

Sometimes it is desirable to have a high internal resistance. Ask the class what happens if a 5 V cell is shorted – i.e. its terminals are connected together by a wire of zero resistance? Some might think I = VR with R = 0 should mean that an infinite current would flow (limited by other physical factors!)

Remind them of the internal resistance r. This limits the cell to a maximum (short-circuit) current of:

I = Er

We can use this to prevent EHT supplies giving the user an unpleasant shock. Take an EHT power supply off the shelf and show the connections for the series internal resistance. It is usually 5 M Ω .

These supplies are designed to provide a high voltage to a high resistance load (e.g. cathode ray tube) but if the terminals or wires connected to them were accidentally touched this could provide a nasty shock (lower resistance in the load and higher current). One way to deal with this is to connect a large resistance in series with the output (positive) terminal. If the terminals are shorted (e.g. by contact through a person) the current drawn is limited to I = Er. A typical EHT supply (up to 5000 V) is protected by a 5 M Ω  resistor so the maximum current if shorted is just 1 mA. That shouldn’t kill you! Be aware however that HT supplies (0-300 V) have a much lower internal resistance, and could kill you, so special shrouded leads should be used.

EHT supplies often have a further safety resistor (e.g. 10 M Ω ) to reduce the maximum current still further. This resistor can be by-passed when necessary. No school EHT supply is allowed to provide more than 5 mA.

Students learn to use an oscilloscope to measure voltages.

Lesson Summary

• Demonstration and/or student experiment (30 minutes)
• Student questions: Understanding the CRO (20 minutes)

Demonstration and/or student experiment: Learning to use a CRO

Bear in mind that most students (and teachers?) find oscilloscopes daunting at first. If they are not already familiar with them then try to use the simplest ones available (single beam, non-storage). Whatever you use, be prepared to spend a lot of time helping them with settings and familiarising them with the adjustments.

It may help if you start by demonstrating the experiment first, and then allow your students time to repeat it for themselves. The most important ideas to get across are:

• Used normally, an oscilloscope plots a graph of voltage (y-axis) against time (x-axis).
• The scales on the x- and y- axes can be adjusted using the timebase (time per division) and voltage gain controls (voltage per division). (Check what these are called on the oscilloscopes you use.)

Episode 122-1: The cathode ray oscilloscope (Word, 32 KB)

There are two main aims here:

• Familiarity with the oscilloscope so that they can get a stable trace and adjust settings.
• Taking measurements of voltage and time from the screen.

Student questions: Understanding the CRO

Questions on use of the oscilloscope.

Episode 122-2: Questions on use of the oscilloscope (Word, 26 KB)

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_pV_p and a minimum of zero. This should also make it clear that the average power is less than I_pV_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)²R

(using V = I × R)

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

P = I² × R

P = 12(I_p)² R

, or

P = V²R

P = 12(I_p)²R

These lead to the equations:

I² = I_p2

giving

I = I_p√ 2 

and

V² = V_p2

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)