GRIFFIN EDUCATION working with
EDU-LAB and The RADMASTE Centre
UNIT P1a TOPIC 9 PRODUCING AND MEASURING
ELECTRICITY
LET THERE BE LIGHT!
TEACHER NOTES
OVERVIEW OF THE ACTIVITY
The first part of this activity is to consolidate learners’ understanding of the concept of resistance. They will
recall the properties of conductors and insulators, and the factors that affect resistance.
The second part of the activity will concentrate on electrical circuits. Learners will investigate the effects of
resistors (light bulbs) in series and in parallel. They will use two equations which calculate resistance.
MR DLAMINI’S QUESTIONS
1
The learners should be discussing the concept of resistance in conductors. How different conductors
offer different resistances. Also, the effects of current moving in a conductor that offers little or no
resistance, compared to a conductor that offers lots of resistance.
They may bring into the discussions how we use specifically designed devices, like resistors, just for
their effects. For example, we use the elements (plates) on a stove or in a kettle for their heating
effect. A light bulb, is a resistor. The tungsten wire (filament) in a light bulb has a high resistance of 5.6
o
C. At this high temperature the filament glows red, and
then white, and produces light.
Resistance is a property of a substance which slows down the movement of electrons in it. When the
loosely bound outer (valence) electrons move in a conductor they collide with the atoms of the
conductor. Work is done to overcome these collisions. This means that the electrons transfer energy to
the atoms. The more work that is done the greater the amount of energy transferred. This energy is in
the form of heat. The wires of the house offer little or no resistance to the moving electrons so they will
not get hot. However, the tungsten wire in the light bulb is designed to offer a high resistance. The
moving electrons transfer energy to the wire which in turn transfers energy to the surroundings in the
form of heat and light.
2
Two factors that affect resistance are the length and thickness of the conductor. The longer the
conductor the greater the resistance; the thicker the conductor the lower the resistance. By making
the resistance wire of a light bulb very long and very thin the desired effect of light is much greater.
3
The glass portion of the bulb is filled with unreactive gases, usually argon and nitrogen. In the presence
of air the tungsten filament oxidises, ie reacts with oxygen, and breaks. The gases prolong the ‘lifetime’
of the bulb. Unfortunately, however, the tungsten filament also evaporates very slowly. It gets thinner
and thinner and eventually breaks. When the filament breaks we say the bulb has ‘blown’.
4
Substances with very high resistances, eg plastics, are called insulators. These substances are nonmetals. They do not have loosely bound valence electrons that can move freely and form an electrical
current.
At the end of their discussions your learners should have covered the concept of resistance and also mentioned
some factors that affect resistance. There are four factors that affect resistance. They will most likely discuss
two of factors which are the length and thickness of the conductor. The other two factors are,
(a)
the material of the conductor - copper and silver have very little resistance whereas tungsten has a
much higher resistance; and
(b)
temperature - an increase in temperature increases the resistance of the conductor.
If the latter two factors are not discussed, mention them to your learners.
When your learners have finished this exercise ask each group to stick their flipchart paper on one of the
classroom walls. The whole class will now distill out what is important. These important issues are to be written
on a fresh piece of flip chart paper and stuck on the wall until it is no longer necessary. The issues should relate
Let there be light
Page 1 of 8
to: resistance being an inherent property of conductors; how a substance resists moving charges; the factors
that affect resistance.
Prediction 1 - (Resistors in series):
Here are some examples of learner predictions. With the addition of each extra bulb;
 there will be an increase in resistance
 there will be a decrease in the current in the circuit
 the brightness of the first bulb will decrease.
2
Light Bulbs
Ammeter Reading (A)
Brightness of bulb(s)
L1
L1 + L 2
L3 + L2 + L 3
eg 0.24
eg 0.20
eg 0.08
very bright
both the same brightness
all the same dullness
Resistors in series behave like a longer resistor. They increase the resistance of the circuit. This in turn results in
a decrease in the current in the circuit. The brightness of the bulbs also decreases. If there is a decrease in the
current then there will be fewer charges transferring energy to the bulbs. The bulbs will give out less light
energy and so glow less brightly.
What to discuss
1
The current strength and / or the brightness of a light bulb. The ammeter readings reflect a change
every time resistance changes. Each time we add a light bulb the ammeter reading decreases.
The brightness of the bulbs also reflect changes in resistance. The greater the current the brighter the
light bulb. Each time we add a light bulb in series the brightness of the bulbs decreases. This means that
when each bulb is added there is an increase in the circuit’s resistance. The presence of the increased
resistance is shown by the decrease in the brightness of the bulbs. With an increase in the total
resistance of a circuit there will be a decrease in the current in the circuit.
2
An ohm.
4
RT = R1 + R2 + R3
5
R  1/I or R 1/ I. With an increase in resistance there is a decrease in current.
6
With an increase in the resistance of the circuit there is a decrease in the current in the circuit. This in
turn means that fewer coulombs transfer energy to the bulbs so less energy is given out as light.
Prediction 2 - (Resistors in parallel)
Learners should mention that they would expect the changes in resistance to be different to those of the series
circuit. Learners should not give the same predictions as before because the circuits are very different.
Most will say that with the addition of each extra bulb;
 there will be an decrease in resistance.
 there will be a increase in the current in the circuit.
 the brightness of the first bulb will either stay the same or be a little less bright.
What to do
5
As each bulb is connected in parallel, another current path is formed. Because of the indirect
relationship between current and resistance this means that the resistance of the circuit decreases and
the current in the circuit increases. Putting resistors in parallel is like making a thicker resistor. The
thicker the resistor the lower its resistance.
What to discuss
1
The aim of this little exercise is for learners to recognise the importance of;
 setting out an answer neatly
 units as part of an answer
 including the formula used
A common error learners make is to leave the answer as 1/R eq. The final resistance value is Req or Rparallel
= 2  and not 1/R = ½ .
2
Learners will find this useful. It is a way of checking a calculation of parallel resistors. If the value they
obtain is greater than the value of the smallest resistance used they have made an error. It also shows
that the resistance decreases quite substantially when we connect resistors in parallel.
Let there be light
Page 2 of 8
/Req
3
Let your pupils discuss Lebala’s observation as several will have experienced it as well. They may not be
able to find a solution. However, someone may discuss potential difference across the parallel bulbs.
There is a small drop in the potential difference across the bulbs. The increased current leads to some of
the emf (maximum potential output of an energy source) being used to move the extra charge through
the cell. Because the potential difference across a circuit is kept constant, with an increase in “the lost
volts” there will be a decrease in the potential difference across the parallel bulbs. This phenomenon
will be covered in depth in the next activity.
4
Rtotal = Rseries + Rparallel or Rtotal = Rseries + Requivalent
HOMEWORK
At this point it is a good idea to discuss the good practise of writing answers. Learners should write each step
under the other; they should include the formula to be used. If there is a lot of information given encourage
your learners to write a summary before they start their calculations. They write down the symbols and the
amounts of the given quantities, eg, R = 2  or I = 5 A.
1
Rseries
=
=
=
R1 + R2 + R3
2+2+2
6
2(a)
= 1/R1 + 1/R2
=
1/3  + 1/3 
=
2/3 
Req
=
3/2 
Req
=
1.5 
2(b)
1/Req
=
=
=
Req
= 1/R1 + 1/R2 + 1/R3
1/2  + 1/3  + 1/4 
6 + 4 + 3/12 
13/12 
= 12/13 
2(c)
Rtotal
=
=
=
=
=
=
=
Rseries + Requivalent
2 + [1/(2 + 2) + 1/4 ]
2 + [1/4  + 1/4 ]
2 + [2/4 ]
2 + [4/2 ]
2 +2
4
Several learners will calculate the equivalent resistance separate from the series resistance and then
they will add them together.
3b
(i)When S1 is closed, the light bulbs are in parallel so the total resistance will decrease.
Rbefore = 2 
Rafter = 1/Req
= 1/R1 + 1/R2
= 1/2 + 1/2 
= 2/2 
Req
=1
(ii)
There is no current flow as both switches S1 and S2 are open, therefore there is no resistance. When S1 is
closed there will be an increase in resistance . However, the current will be in just one light bulb.
Rbefore = 0 
Rafter = 2 
(iii)
The resistance of the circuit is equal to the equivalent resistance of the two light bulbs in parallel.
When S1 is closed a short circuit is formed and the current by passes both light bulbs. This means the
light bulbs offer no resistance to the current. Therefore, the resistance decreases.
Rbefore = 1/Req = 1/R1 + 1/R2
= 1/2 + 1/2 
= 2/2 
Req
=1
Rafter = 0  because of the short circuit.
4(i)
The resistors have equal resistances so the reading on each ammeter will be 2 A. The current splits
equally.
Let there be light
Page 3 of 8
(ii)
The total resistance of the branch with the two 2 ohm resistors equals that of the branch with the single
4 ohm resistor. The current will divide equally between the two branches so the readings on A 2 and A4
will be the same, 2 A each.
The current in the two ammeters in series in the branch is the same. The ammeter readings on A2 and A3
will thus be 2 A each. Current does not get “used up” when electrical charges move through resistors in
series.
(iii)
The current will divide so that most of the charges pass through the resistor with low resistance and
fewer charges pass through the resistor with a high resistance. The reading on A 2 will be 3 A and the
reading on A3 will be 1 A. The current is divided proportionally between the two resistors (3A:1A).
Let there be light
Page 4 of 8
STUDENT WORKSHEET
LET THERE BE LIGHT!
What you need: a domestic light bulb, a basic micro-electricity kit
What to do: You will examine the light bulb to answer some of Mr Dhlamini’s questions.
MR DHLAMINI’S QUESTIONS
1
Why does the wire connecting up the lights in the house not get very hot but the wire in a light bulb gets
so hot that you can get burnt? (Is this how a light bulb works?)
Below, there are a few terms to help you in your discussions.
heat
work done
resistors
2
3
4
valence electrons
collide
collisions
resistance
atoms
composition
glass bulb
filament
filament support
Why is the wire (filament) inside the bulb very thin, very long and coiled? Why is it not short, thick and
straight?
What is the use of the glass bulb part of a light bulb?
Why can’t electricity go through the plastic that surrounds the electrical wire we use to connect the
lights together?
You discussed how the factors, length and thickness, affect the resistance of a conductor. What change will
there be in the total resistance of the circuit if another resistor, eg. a second identical light bulb, is added:
a
in series with the first light bulb?
b
in parallel with the first light bulb?
Prediction 1 - (Resistors in series) : Predict the changes, if any, in resistance that will
occur when a second light bulb (L2), and then a third light bulb (L3), are connected in
series with the first light bulb (L1) shown in the circuit on the right.
1
Take your micro-electricity kit and make a circuit of the circuit diagram given
alongside. Unscrew the light bulb until you are ready to work with the circuit.
This prolongs the life of your battery and the bulb.
2
When your circuit is set up correctly, close the circuit and take an ammeter
reading. Also observe the brightness of the bulb. Record your results in the
table given below.
Light Bulbs
3
4
5
Ammeter Reading
(A)
A
L1
Brightness of bulb/s
(very bright / bright / dull)
Add light bulb L2 in series with light bulb L1. Repeat step 2.
Add light bulb L3 in series with light bulbs L1 and L2. Repeat step 2.
How accurate is your prediction? Consider the factors that affect resistance to help explain what
happens when you add resistors in series.
What to discuss
1
2
3
At the moment you do not have an instrument to measure the total resistance of your circuit. What
other factors show an increase or decrease in the total resistance of the circuit?
As resistance is a measurable quantity it must have a unit. What is the unit for resistance?
One of the aims of a scientist is to prove that relationships exist between things. As you are learning to
be scientists identify two important relationships in this activity. (Note: use the term “resistors” instead
of light bulbs as these relationships exist between any objects designed to resist an electric current.)
Let there be light
Page 5 of 8
4
The one relationship is between the number of resistors and the total resistance of the series circuit.
Represent this relationship using the symbols, R T to represent total resistance, and R1, R2 and R3 to
represent the resistance of each individual light bulb. (Note: Rseries can be used rather than RT.)
5
The second relationship is between the total resistance of the circuit and the total current in the circuit.
Discuss this relationship and represent it using the symbols R (resistance);
I (current);
and select any of the following symbols: = or  (directly proportional) or 1/ (inversely or indirectly
proportional).
6
If we consider the brightness of the bulbs in terms of energy transfer. We see that every time we add
another light bulb in series the light bulbs shine less brightly. Discuss the effect of an increase of
resistance of a series circuit in terms of current and energy transfer.
Prediction 2 - (Resistors in parallel): Predict the changes in resistance that will occur when you connect a
second (L2) and then a third light bulb (L3) in parallel with light bulb L1.
What to do
1
2
3
4
5
Start with the original circuit. Draw up a new table.
Close the circuit, take an ammeter reading and observe the brightness of light bulb L 1. Record your
results on your table.
Connect light bulb L2 in parallel with light bulb L1. Repeat step 2.
Connect light bulb L3 in parallel with light bulbs L1 and L2. Repeat step 2.
How accurate was your prediction? Consider the factors that affect resistance to help explain what
happens when you add resistors in parallel in a circuit.
What to discuss
1
We can also use a formula to find the total resistance of a circuit with parallel resistors.
We represent the total resistance of a parallel circuit with the symbols Rparallel or Req (‘eq’ represents
‘equivalent’. The formula is 1/Rparallel = 1/R1 + 1/R2 + 1/R3
Below are the answers of three Grade 12 learners to the question: “What is the total resistance of two 4
The resistances of the connecting wires and cell are negligible.”
Criticise the answers given below and select the answer which is not only correct but is also the most
informative.
a
1/Rparallel = 1/4 + 1/4 = 2/4 = ½
b
1/Req = 1/R1 + 1/R2 = 1/4 + 1/4 = 2/4, 1/Req= ½ , Req= 2/1 = 2 
c
1/Req = 1/R1 + 1/R2 = 1/4 + 1/4 = 2/4 = ½ 
2
The following quote is from a Physical Science Grade 12 text book. The same principle is found in any
Grade 12 Physical Science text book.
“The equivalent resistance of two or more resistances in parallel is less than the least of the original
resistances.”
Discuss the meaning of this quotation, and give an example using resistance values to show that the
quote is true.
3
Consider this discussion between two Grade 12 learners, Lebala and Phoka.
Lebala: “ I just don’t understand. The ammeter reading shows that the current increases each time we
add a bulb in parallel so why do the bulbs shine dimmer?”
Phoka: “ They must have more resistance when they are in parallel.”
Lebala: “I don’t think so, Phoka. We have already shown that there is an indirect relationship between
current and resistance, it has to be something else. “
How can you explain the phenomenon that Lebala has observed?
4
The circuit alongside is known as a series - parallel circuit.
Use some of the symbols you have met in this activity and give an equation which
you could use to calculate the total resistance of this series - parallel circuit.
A
L2
L
Let there be light
L1
3
Page 6 of 8
HOMEWORK
One of the aims of this activity was to get you to understand the relationships between the total resistance of a
circuit and the resistances of the light bulbs when they are connected differently. Here are a few problems for
you to solve using the equations you derived in the activity.
Note: the resistances of the connecting wires and the cell are negligible.
1
2
Give the resistance of the circuit on the right.
2
2
2
L1
L2
L3
Calculate the resistance of each of the circuits given below.
2
2
3
2
2
3
3
(a)
3
4
4
(b)
(c)
When switches are closed or opened they cause changes to the circuit.
a
Study each of the following circuits and predict whether the total circuit resistance increases /
stays the same / decreases when switch S1 is closed.
b
Calculate the total circuit resistance before and after the changes and see whether your results
support your predictions.
2
2
S1
2
S1
S1
2
2
2
S2
(i)
Let there be light
(ii)
(iii)
Page 7 of 8
4
Understanding the indirect relationship between current and resistance is very important.
(a)
Without the use of any equations, but using ratios, determine the readings on the ammeters in
the circuits. The reading on A1 in each circuit is 4 A. Enter the values onto your sheet.
(i)
A2 = ____________
A1
A3 = ____________
2 ohms
A2
2 ohms
A3
(ii)
A2 = ____________
A1
2 ohms
A1
2 ohms
A3
A3 = ____________
A4 = ____________
A
2
1ohm
4 ohms
A2
4
3Aohms
(
A3
iii)
A1
A2 = ____________
A3 = ____________
1ohm
A2
3 ohms
A3
Let there be light
Page 8 of 8