How is it measured?
The resistance of a length of wire is calculated by measuring the current present in the circuit (series) and dividing it by the voltage across the wire (parallel) . These measurements are then applied to the formula:
Voltage (Volts)
Current (Amps)
How Temperature affects resistance.
If a metallic wire is heated up, the atoms will start to vibrate more rapidly because of this increase in energy. This causes more collisions between the atoms, as the atoms move in the path of the electrons. This slows down the current, increasing the resistance.
Increase temp……
If temperature increase means resistance increase, temperature decrease must mean resistance decrease. Metals such as lead and aluminium become super conductors when cooled to near absolute zero. Current will continue to flow through it for ever as no energy is turned to heat ( i.e. no resistance) so long as the metals stays near absolute zero.
How the type of material affects resistance.
The type of material will affect the amount of free electrons which are able to flow through the wire. The number of free electrons depends on the outer shell of the atoms, so if there are more, there must be more electrons available. If the material has a high number of atoms there will be a high number of electrons causing a lower resistance due to the increase in electrons. Also if the material’s atoms are closely packed then the electrons will collide more frequently therefore resistance will increase.
How wire length affects resistance.
As wire length is increased the electrical resistance is also increased.
Why?
Electricity is the flow of electrons. As electrons pass through a wire , they collide with the atoms of the wire. This slows the flow of electricity, creating resistance. Also if the length of the wire doubles the resistance should double. This is because there will be roughly twice as many atoms to collide with. Increasing length is like increasing resistance in series. ( )
If we double the length……
How Cross Sectional Area affects Resistance
According to one of Ohm’s laws , resistance is inversely proportional to the cross sectional area. This means that as the cross sectional ( or you could say diameter or radius) increases the resistance decreases. This is because if we increase the cross sectional area the electrons will have more space to flow through and will be less restricted therefore will encounter less resistance. Look at the following diagram:
If we decrease the diameter we restrict the flow of the electrons therefore more resistance…..
If we increase the diameter the electrons are less restricted therefore less resistance……
However this does not mean that as diameter doubles, resistance doubles. This is because if diameter is doubled , the surface area is roughly increased by a factor of 3. (area = Πr2).
The effects of length , diameter and type of wire can be checked using ICT, but I am going to investigate how length affects resistance manually.
Introduction
As the problem says, we are supplied with reels of wire out of which we have to create a resistance of 1.9Ω and then 28.5Ω.
I intend to basically pass the current and find the corresponding voltages through wires of constantan of length 100cm,80cm,60cm and 40cm. For each I will plot a voltage against current graph and find the resistances of each length. I will then use these resistances to plot a graph of resistance against length. I will then read off the length needed to create 1.9Ω. I will then find the equation of the graph and use it to find the length needed to create 28.5Ω. I will do this for three different diameters to back up my results and also as an extension, to investigate how diameter affects resistance.
- 15 connecting wires
- 2 crocodile clips
- Mains power pack
- Electronic ammeter
- Electronic voltmeter
- Rheostat
- 3 coils of constantan ( 24swg,28swg and 32swg)
- wire cutters
- metre stick
- heat proof mat
- Don’t use lengths of wire under 40 cm, as they might become too hot and melt.
- Don’t use too large of a voltage
- Change the current not the voltage
- Leave any wire encountering electrical current too cool down, as they might become hot
- Use a heat proof mat under the wire, so that any heat from the fire won’t set anything alight
- Follow all other obvious safety precautions
Independent variables
The wire length is to be changed to investigate its effect on resistance. So wire length is the independent variable.
Dependant variables
I am going to take the voltage and currents through the wire to draw voltage against current graphs. The currents and corresponding voltages should vary as wire length varies so voltage and current are the dependant variables. Or you could say the resistance (V/I) is the dependant variable.
Controlled variables
The diameter will be controlled i.e. I will take voltage and current readings of 100cm, 80cm, 60cm and 40 cm of 1 diameter then do the same for a different diameter. Diameter needs to be controlled because as I explained previously, resistance decreases as diameter increases. This would have a negative effect on my results.
The type of will also be controlled. I will only use constantan wire. Different materials will have different densities which could mean more or less collisions between the electrons and the atoms wires which would increase or decrease the resistance, affecting my results.
Before starting the experiment read the WHOLE method
- Set up the following circuit diagram:
* Wires inserted here, clipped on by crocodile clips.
This is what it should look similar to:
-
From the reel of 32 swg (0.28mm dia.) constantan wire , measure out 100cm (using a metre stick) as accurately as possible and cut it using the wire cutters.
Try to make the wire as straight as possible by pulling form both ends to ensure that it is as close to 100cm as possible. This will improve the accuracy of results.
-
Attach the 100cm wires in between the crocodile clips. Make sure that the crocodile clips are attached to the end of the constantan wire.
- Set the power pack to 4v and then switch on the mains electricity. Make sure the ammeter and voltmeter are on and reading 0A/0V. If a negative value appears, the voltmeter/ammeter may be connected in the in the wrong way. (fix this if need be)
-
Set the rheostat to its lowest value i.e. slide it to the very left. Now the ammeter should read the lowest possible current that will go through the 100cm wire at 4v. Note this value down ( this will help to find the intervals of current to use). Then set the rheostat to its highest value by sliding it to the very right. The ammeter will now show the highest current that will go through the 100cm wire at 4v. Using the lowest and highest current select 4-5 intervals ( at which the voltage will be found at).
e.g. lowest value = 0.2A , highest value = 0.8A
The intervals would be 0.2A, 0.4A, 0.6A, 0.8A
- Using the rheostat , change the current to each of these values and record its corresponding voltage on the voltmeter. Then fill in the following table:
- After recording the results, turn off the power pack then turn the mains electricity off. Leave the wire too cool down for about 30 seconds , then remove the wire.
-
Now , cut this wire down to 80cm again using the metre stick and repeat the steps 2-6 (inclusive), using 80cm.
- Then size the 80cm wire to 60cm and repeat steps 2-6
- The size the 60cm wire to 40cm and repeat steps 2-6
- Now repeat steps 1-9(inclusive) using 28 swg ( 0.37mm dia.) constantan wire.
- Repeat steps 1-9(inclusive) using 24swg (0.56mm dia.) constantan wire.
- Make sure all the electricity is switched of then clear equipment.
You should now have a total of 12 tables of current against voltage. i.e. a table for 100cm, 80cm, 60cm,40cm for the 3 diameters.
What am I going to do with my results after I complete the experiment?
I have written my method. Now I m going to discuss what I will do with my results after I have completed the experiment.
Using 100cm, 80cm, 60cm, and 40cm of constantan wire, I will find voltages at certain current intervals. I will do this for 32swg , 28swg and 24swg.
I will then fill in12 tables of voltage against current i.e. a table for 100cm, 80cm, 60cm, 40cm for the 3 diameters.
With each diameter I will have 4 tables (voltage against current for 100cm, 80cm, 60cm, 40cm). With these four tables in each diameter I will plot graphs of voltage against current. I will then find the resistance of each of the four graphs by calculating the gradient of the line ( which should be straight as voltage is directly proportional to current.) So for each diameter I should have a resistance for 100cm, 80cm, 60cm and 40cm. I should then , for each diameter, be able to plot a graph of resistance against length.
(y)
Resistance (Ω)
Length ( cm) (x)
Using the graph I will be able to read off what length of wire will be needed to create a resistance of 1.9Ω.
Then I will find the equation of the line using the rule y=mx+c. I will do this by finding the gradient and the intercept. I will then substitute the value for y as 28.5, and work out x (the length).
I predict that as length is increased , resistance will also be increased. This is because as current flows through a wire, the electrons collide with the wire’s atoms, hindering the flow of electrons, causing resistance. If the wire is increased in length, the electrons will encounter more collisions with the atoms therefore hindering the current even more therefore increasing resistance.
Furthermore, if the length doubles/triples etc. then the resistance should also double/triple etc . I think this, because by doubling the length there will be roughly twice more atoms that the electrons may collide with therefore twice the resistance.
If resistance does double as length is doubled, I predict that my graphs of resistance against length will be directly proportional. i.e. the best fit line will be straight and cross through the origin. And when length is doubled resistance should double (on the graph) So the graph should look like this:
When I did the experiment I obtained the following results:
100 cm
80cm
60 cm
40cm
I will now convert thee four tables into graphs of voltage against current and find the resistance of each one (to enable me to draw a master graph of resistance against length) . I will do this for diameters 28 swg and 32 swg as well.
Using the resistances of these four graphs I will draw a graph of resistance against length.
From the graph I will read off what length is needed to create 1.9Ω. So I will be able to say to the electrician ‘ to achieve 1.9Ω using 24swg constantan, you need …. Cm’
I will also, using the equation of the graph find what length is needed to create 28.5Ω.
Graph
Using the data from the graph I can say that , to achieve 1.9Ω using 24swg constantan wire you need roughly 95cm.
Now I will find what length is needed to create a resistance of 28.5Ω:
Equation of line = y = m x + c
y co-ordinate
(resistance)
So of 24swg constantan wire you need roughly 14.25 m to create a resistance of 28.5Ω.
100cm
80cm
60cm
40cm
Graphs of voltage against current follow:
Now using these resistances I will draw a master graph of resistance against length.
From the graph I will read off what length is needed to create 1.9Ω, and using the equation of the graph , I will find what length is needed to create 28.5Ω (28.5 is too big a value to read off the graph).
Graph follows:
Now I can say that , to create a resistance of 1.9Ω using 28swg wire , you need a length of roughly 41 cm.
Now by using the formula of the graph, I will calculate what length is needed to create 28.5 Ω.
To create28.5Ω by using 28swg constantan wire requires a length of 6.59m.
100cm
80cm
40cm
Due to some circumstances I was not able to take a set of readings for 60cm. This , however , will not affect the graph of resistance against length too much , as 3 points should be enough to draw a line of best fit.
Graphs follow:
I am now going to plot a graph of resistance against length. I should then be able to read off, from the graph, the length needed to create 1.9Ω.
I will then find and use the equation of the graph to find what length is needed to create 28.5Ω (using 32 swg constantan wire).
Graph follows:
This time a problem arises because we cannot read 1.9Ω off the graph . So instead we will use the equation of the graph to find what length of 32swg constantan wire is needed to create 28.5Ω and 1.9Ω.
Summary of results
These are in essence the answers to the Electricians Dilemma:
24swg ( 0.56mm dia.)
-
of 24 swg Constantan 95 cm is needed to create a resistance of 1.9Ω
- of 24 swg Constantan 14.25m is needed to create a resistance of 28.5Ω
28swg (0.37mm dia.)
- of 28 swg Constantan, 41 cm is needed to create a resistance of 1.9Ω.
- of 28 swg Constantan,6.59m is needed to create a resistance of 28.5Ω
32swg (0.28mm dia.)
- of 32 swg Constantan, 23.85cm is needed to create a resistance of 1.9Ω.
- of 32 swg Constantan, 3.56m is needed to create a resistance of 28.5Ω
1)
I noticed clear patterns from my results ( in the form of graphs) for 24swg, 28swg and 32swg.
I noticed that all my Voltage against resistance graphs were directly proportional i.e. the line was a straight one going through the origin and that as current doubled voltage doubled. Their general shape was like this:
2)
I noticed that my resistance against length graphs showed that as length increased resistance increased. They also showed that as length was doubled resistance was roughly doubled. I also found that the intercept of that graph was 0 or very close to 0. This all shows that my graphs show that resistance is directly proportional to length.
Explanation of results
All the voltage against resistance
