Investigating resistance when altering thickness of wire / length of wire
Investigating resistance when altering thickness of wire / length of wire
The circuit:
Diameter:
* Wires of different thicknesses are used.
* In order to produce a straight line graph: plot Resistance (y axis) against (1/A) x-axis where A is the area of the wire. To do this:
*
* Measure the diameter of each wire, work out the area by doing (pi x d2)/4 then do 1/A where A = area of cross section.
* Record the current and voltage for each of the wires used keeping the same circuit for each wire. Work out resistance by doing V/I where V is the voltage and I is the current.
* For different wires of the same material, the resistance R of each wire is directly proportional to 1/A where A is the cross section area of the wire or R is inversely proportional to A. This means that as area is doubled, resistance is halved and when area is tripled the resistance is a third of what it was before. A wider wire has lower resistance because more electrons can get through per second (current is higher) than if the wire was thin.
Length:
* Different lengths of the same type of wire are tested.
* Voltage and current is measured for each length of wire. In your table put voltage, current, resistance, and length (in metres). Work out resistance by doing V/I where V is voltage and I is current.
* The graph shows that resistance is directly proportional to length so as length is doubled. resistance doubles.
* Wires behave like resistors only they have very very low resistance compared to resistors. The total resistance of resistors in series is the sum of the resistance of each one. Each cm of wire has a particular resistance, if you double the length of wire, it is like having two of the shorter wires in series.
* If the small resistor represents a short wire and the large resistor is a long wire of double the length of a short one. One short wire has a resistance of 1ohm, 2 short wires have a resistance of 2ohms when connected in series. The long wire is just like two short wires put together.
Planning
Some variables that will be relevant to this investigation are:
* Length
* Thickness
* Temperature
* Voltage
* Resistance
* Material
Of these the variables will be input and output voltages in experiment one, and length and resistance in experiment two. The other variables (temperature, material and voltage) will have to be kept constant in both experiments to make sure that only length, thickness and resistance are investigated. In experiment 1 the same bit of wire and the same thickness need to be kept constant. In experiment 2, the length will need to be kept constant to make sure only the variables indicated are investigated to ensure a fair test.
Metals conduct electricity because the atoms in them do not hold on to their electrons very well, and so creating free electrons, carrying a negative charge to jump along the line of atoms in a wire. Resistance is caused when these electrons flowing towards the positive terminal have to 'jumps' atoms. So if we double the length of a wire, the number of atoms in the wire doubles, so the number of jumps double, so twice the amount of energy is required: There are twice as many jumps if the wire is twice as long.
The thinner the wire is the less channels of electrons in the wire for current to flow, so the energy is not spread out as much, so the resistance will be higher: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve,
i.e. Resistance= 1/Area.
This can be explained using the formula
R = V/I
where there is 2X the current, and the voltage is the same, therefore R will halve. I did some research and in a book called 'Ordinary Level Physics' By A. F. Abbott, it says 'that doubling the area will therefore halve the resistance'- in other words the resistance of a wire is inversely proportional to its area, or R ? 1/A , but we are measuring diameter, so if the area is: ?r2 = ?(d?2) 2 A= ?d2 ? 4 Where A is area and d is diameter.
Method
Experiment One - First a length of wire over a metre long is sellotaped to a metre rule. The positive crocodile clip is attached at 0cm. And the negative is moved up and down the wire, stopping at 20, 40, 60, 80 and 100cm. Each time reading the ammeter and voltmeter to work out resistance R = V/I. This is using 30 SWG wire. Other variables, voltage, thickness, and temperature will be kept constant, although the temperature will rise once current is passing through it, which will cause the atoms in the wire to vibrate, and so obstruct the flow of electrons, so the resistance will increase, creating an error. In both experiments constantan wire is used because it does not heat up as much as copper, so the resistance is not effected as much.
Experiment Two - The circuit isset up is the same, as is the method apart from the length is constant at 50cm, and the thickness is changed between 28, 30, 32, 34, 36, 38 and 40 swg. For both experiments the voltage will be kept the same at 2V dc from a power pack. Both experiments will be done twice with different ammeters in case of any damaged or old equipment to gain more accurate results.
Results
Experiment 1
Length (cm)
V1 (volts)
V2 (volts)
A1 (amps)
A2 (amps)
Average resistance (Ohms)
00
.00
.00
0.20
0.20
5.00
80
.00
.00
0.30
0.28
4.00
60
0.90
0.90
0.40
0.30
2.80
40
0.90
0.85
0.50
0.40
.94
20
0.70
0.80
0.80
0.75
0.94
Experiment 2
Thickness (mm)
Area (mm2)
V1 (volts)
V2 (volts)
A1 (amps)
A2 (amps)
Resistance
28
0.36
0.107
0.8
0.8
0.61
0.59
.3
0.29
0.066
0.9
0.9
0.49
0.51
.8
0.32
0.25
0.049
0.9
0.9
0.35
0.4
0.24
0.34
0.18
0.025
.0
.0
0.25
0.25
.0
.0
0.25
0.4
0.36
0.16
0.020
.0
.0
0.16
0.17
0.61
0.38
0.12
0.011
.1
.1
0.09
.00
0.09
.00
1.6
40
...
This is a preview of the whole essay
0.8
0.61
0.59
.3
0.29
0.066
0.9
0.9
0.49
0.51
.8
0.32
0.25
0.049
0.9
0.9
0.35
0.4
0.24
0.34
0.18
0.025
.0
.0
0.25
0.25
.0
.0
0.25
0.4
0.36
0.16
0.020
.0
.0
0.16
0.17
0.61
0.38
0.12
0.011
.1
.1
0.09
.00
0.09
.00
1.6
40
0.05
0.0020
.2
Evaluation
Evaluation Experiment one: This experiment was quite accurate, as when it is compared to the manufactures line which is on the same graph, we can see that this line is at most only 0.4? different form the manufactures line. This is a percentage difference of approximately 8%, using the formula: Difference ? original X 100 This shows that the results were good, as 8% is a very small margin of error. The error bars on the graph show that the most inaccurate result was the 60cm result. This could be down to an error in the measurement of the wire or a temperature rise. The two results for 100cm are exactly the same, and it is near to the manufacture's line, so this is the most accurate point.
The other three readings have almost the same inaccuracy, an average of 10%, which again, is fairly accurate. The inaccuracy could have been because of the wire coming from a different manufacturer to the predicted results, as there is some discrepancy between the amount of copper and nickel in different brand's wire. The ammeters and voltmeters could have been damaged and reading falsely on both the meters used.
Measuring the lengths of the wire is also a inaccuracy as the rulers used are not exact, and it is difficult to get an accurate reading of length by eye, as the wire might not be completely straight, it may be of different thicknesses throughout the length. These would have contributed as well to the error. These results would be difficult to improve on as they are reasonably accurate, and there were no anomalous results. But if I were to do this experiment again, I would use newer, more accurate ammeters and voltmeters, a more accurate method of measurement, and take a much wider range of readings, and more readings so that a more accurate average can be taken.
I would also investigate other factors, such as temperature, voltage and current, and see how these effect the resistance. I would also do the experiments under different conditions such as temperature and pressure to see if it makes any difference to resistance. As these results had a range of only 5 readings, from 0-100cm, and were only repeated twice, and that the results are not 100%, accurate due to the errors discussed earlier, then I would say that these results are not strong enough to base a firm conclusion on because there are so many sources of error, which are explained earlier.
Experiment two - These results were not as accurate as experiment one. I had predicted that the resistance should halve as area doubles, which it does, however not to the predicted curve. When the resistance is 24ohms, the % inaccuracy is 6%, and when the resistance is 6 ohms, the inaccuracy is 8%. These inaccuracies are fairly large. The error bars, however, are too small to be drawn accurately on the graph. They are at most 3% inaccurate, using the same formula as before. This suggests that the inaccuracies were not experimental, but permanent errors due to problems with the measuring equipment.
These results were this inaccurate as the tool used for measuring the diameter of the wire were very inaccurate due to a zero error on the screw reading, i.e. the mark given for zero mm was not the real mark, hence throwing all the results off by the same amount. The ammeters and voltmeters could have been damaged and reading falsely on both the meters used. Measuring the lengths of the wire is also a inaccuracy as the rulers used are not exact, and it is difficult to get an accurate reading of length by eye, as the wire might not be completely straight, it may be of different thicknesses throughout the length. These would have contributed as well to the error.
There was one slightly anomalous result, at 0.25mm2. This could have been due to a unique error in the measuring and or reading of the meters, or a temperature change. These results could be done better. If I were to do this experiment again, I would use newer, more accurate ammeters and voltmeters, a more accurate method of measurement, and take a much wider range of readings, and more readings so that a more accurate average can be taken. I would also investigate other factors, such as temperature, voltage and current, and see how these effect the resistance. I would also do the experiments under different conditions such as temperature and pressure to see if it makes any difference to resistance.
As these results had a range of only 7 readings, from 0.1mm2, and were only repeated twice, and that the results are not 100% accurate, due to the errors discussed earlier, then I would say that these results are not strong enough to base a firm conclusion on because there are so many sources of error, which have been explained earlier.
Analysis
The graph of experiment 1 is a straight line through the origin, which means R is directly proportional to L. This means that if the length is 40cm, and resistance is 2?, then if length is doubled to 80cm, resistance also doubles to 4.
This is because of the scientific idea, stated in the planning that if you double length, you double the number of atoms in it, so doubling the number of electron 'jumps', which causes resistance: The results support my predictions well, the results turned out the way I had expected, they match the predicted line well. I had predicted a straight line through the origin, which means R is directly proportional to L.
The graph of experiment 2 is an inversely proportional curve. This is because R is directly proportional 1/A, this means when A doubles, R halves. for example when the Area is 0.025mm2 the resistance is 4.8. When A doubles to 0.05, R halves to 2.4?. When A doubles again, R halves again to 1.2. This is because, as stated earlier: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve, i.e. Resistance is directly proportional 1/Area
Hypothesis
I predict that as the temperature increases, the speed of the reaction will increase therefore the gas will be produced faster. I believe this because most chemical reactions happen faster when the temperature is higher. At higher temperatures molecules mover around faster, which makes it easier for them to react together. Usually, rises of 100C will double the rate of reaction.
Chemical reactions take place by chance. Particles need to collide with enough velocity so that they react. As the temperature is increased the particles move faster since they have more energy. This means that they are colliding more often and more of the collisions have enough velocity to cause a reaction. Since there are more collisions the chemical reaction takes place faster.
Pilot Experiment
To decide on the best volume and concentration of hydrochloric acid and best mass of magnesium a number of calculations were done and a pilot experiment conducted.
The equation for the reaction is:
Magnesium(s) + Hydrochloric Acid(l) Magnesium Chloride(l) + Hydrogen(g)
Mg(s) + 2HCL(l) MgCl2(l) + H2(g)
We were advised to use 0.1g of magnesium ribbon (found to be 10.9 cm long). The Relative Molecular Mass (RMM) of magnesium is 24, therefore the moles of magnesium to be used was:
Moles= 0.1
24
Moles= 0.00416
In the reaction above, 1 mole of magnesium reacts with 2 moles of hydrochloric acid. The concentration of acid was 1mol/dm3. Therefore the volume of hydrochloric acid used was:
Volume = 0.00416 * 2
Volume = 0.0083dm3
Volume = 8.3cm3
It was decided to use an excess of hydrochloric acid to ensure all the magnesium reacted, therefore 10cm3 of acid was used in the pilot experiment. At room temperature 10cm3 of hydrochloric acid was added to 0.1g of magnesium and the gas was collected (see fig 1). The volume of gas produced was measured every 15 seconds. It was found that the reaction was too rapid to be effectively measured, therefore 10cm3 of water was added to halve the concentration of the acid.
Investigation Experiment
I am going to investigate how temperature affects the rate of reaction between magnesium and hydrochloric acid. The procedure for the experiment is as follows;
* Using a measuring cylinder, measure 10cm3 of water and pour it into the side arm tube.
* Measure 10cm3 of hydrochloric acid (1 mol/ dm3) and add it to the water.
* Place the side arm tube in a water bath at 20OC, set up the apparatus below.
* Measure 10.9 cm of magnesium ribbon and check on the balance that it weighs 0.1g.
* Coil the ribbon around a pencil and then drop it into the side arm tube and quickly put a bung on the side arm tube (this must be done quickly to prevent gas escaping).
* Every 15 seconds measure the volume of gas produced until less than 1cm3 of gas is produced every 15 seconds.
* Repeat experiment two more times (for accuracy) and record all results in a table.
* Repeat the experiment for temperatures of 0.5M, 1.0M, 1.5M and 2.0M (1mole/dm)
It is important that only the temperature is changed since this is what is being investigated.
Method
* Apparatus
I have chosen to use a 10cm3 measuring cylinder to measure the volumes of substances used since it is more accurate than a pipette. I will use an electronic water bath for maintaining the mixture at a temperature since the temperature is more accurate than a water bath above a Bunsen burner.
A 100cm3 gas syringe should be appropriately accurate for measuring the gas produced since it is accurate to 1cm3 of gas. I will use a three figure balance to measure the mass of magnesium to be used since it is vital that as close to 0.1g of magnesium is used as possible.
* Variables
After deciding how to approach all of the variables in the experiment I decided how to carry out my experiment. I decided to use different concentrations of HCl in 0.5M, 1.0M, 1.5M and 2.0M (1mole/dm ). To obtain these different concentrations I used a solution of 1.0M which was already prepared and the same with a 2.0M solution. However in order to create a 0.5M and 1.5M solution I had to mix different solutions. The 0.5M solution was made using 10cm of deionised water and 10cm of 1.0M solution, this was then stirred with a glass rod. To make 1.5M solution the same concept was used, I mixed 10cm of 2.0M solution with 10cm of 1.0M solution.
I then measured and recorded how much volume of gas (cm ) was given off each 10sec using a stopwatch. Finally the gas syringe is an excellent piece of apparatus as it is very accurate, however I will have to make sure that the clamp does not hold on to it too tightly as this could effect the results.
To help obtain the best possible results I will repeat each experiment twice and then find an average set of data.
I predict that the high molarity concentration solution will have a much faster rate of reaction than the weaker solution. I have predicted this from my knowledge of a previous experiment and scientific knowledge. I think that this happened because the more HCl in the solution the greater the concentration and there will be more particles colliding and more energy and therefore a faster rate of reaction. The reason I think that the rate of reaction will increase is that the experiment is exothermic, this means that it will give itself more energy and therefore more collisions and faster collisions. Also in the 2.0M solution compared with the 0.5M solution there are much more HCl particles in the same volume so the are going to collide with each other more often
Therefore I predict that the rate of reaction will increase with the concentration of the HCl solutions.
* Rates of Reaction
Increasing the temperature increases the speed of the particles. The faster the particles move, the greater the number of collisions, and therefore the rate of the reaction increases. A 10OC rise in temperature almost doubles the rate of most reactions.
Chemical reactions take place by chance. Particles need to collide with enough velocity so that they react. As the temperature is increased the particles move faster since they have more energy. This means that they are colliding more often and more of the collisions have enough velocity to cause a reaction. Since there are more collisions the chemical reaction takes place faster.
Aim: To investigate the factors that affect the resistance in a conductor. The main factors that affect the resistance in a conductor are:
· Length
· Temperature
· Cross sectional area
· Material
· Magnetism
The factor that we are going to change is the cross sectional area.
Hypothesis: I think that the higher the cross sectional area, the lower the resistance in the conductor will be. This is because the Resistance in a metal conductor happens because as the electrons move through the material (once a voltage has been applied) they collide with the atoms in the material and as a result lose some of their energy. The idea of resistance is simply how difficult it is for the electrons to move through a material. The more difficult it is, the more energy they lose in the material on their travels.
We define electrical resistance as the ratio of voltage to current.
The equation we use to find the resistance from the current and voltage is:
Resistance (R) = Voltage (V) ÷ Current (I)
Put more simply, it is the number of volts difference across the object when one amp of current flows. You should recall that voltage is the number of joules of energy transferred by one coulomb of charge, and that current is the number of coulombs of charge passing a place each second.
What the object is made of will have an effect on its resistance. Not all metals even are equally as good at conducting electricity. A longer length will also make it more difficult for current to flow, as there is more material to travel through.
The temperature of a metallic conductor will also affect the resistance. A hot metal has a larger resistance than a cooler one, but this is tricky to test reliably in the laboratory because the temperature has to be a lot higher to get a decent change in resistance.
Current is nothing but the rate of flow:
But when the temperature rise takes place, the lattice atoms also vibrate in their own equilibrium more vigorously impeding the flow of electric charges due to more frequent collisions.
More electrons are available to conduct the current in the wire. Collisions with lattice ions are less frequent. The Current increases and resistance decreases.
However, the cross-sectional area will also have an effect, as the larger this is, the more charge can travel simultaneously through a given length. Therefore, a larger area of cross-section actually reduces the resistance. It is like having identical lengths sat side by side to be in parallel.
The cross sectional area has a continuous variable, i.e. one that is measured and can have any value. You can put your results onto a point graph and get meaningful conclusions. That is why I have chosen to change the cross sectional area of the conductor.
The main equation that describes the resistive behavior of a piece of uniform metallic wire is
R = resistance in ohms
r = resistivity in ohm-metres
l = length in metres
A = area of cross-section in square-metres
The resistivity is simply a constant number for the particular material that makes the numbers work out in S.I. units. The resistivity of Constantan wire is 47× 10 Wm. Different materials have different resistivities; the higher the resistivity the larger the resistance for a given length and cross-section.
We can see from this equation that if the material and area are kept constant, then the equation shows that resistance is directly proportional to the length assuming the wire is uniform. Hence, if you double the length, you double the resistance. As I have said already, it is like having two identical resistors in series.
To verify this you will need to take many values of resistance and length, and then plot resistance against length on a graph. If the graph is a straight line through the origin of the graph then you have verified the equation.
Now, taking cross-section you can either measure the resistance for wire from different reels which have different cross sections, or you can lay the wire from the same reel side by side; the overall effect is the same. We decided to lay them side by side. The equation implies that the resistance is inversely proportional to cross-section, so doubling the cross section should halve the resistance.
Prediction Graph:
Preliminary work:
The circuit (above) shows how we set up the circuit for our preliminary work and our experiments. In our preliminary work, we used some constantan wire in a circuit like above and used it to see what voltage is best to keep the same to find the current and resistance. We found that 3 volts was a good voltage because the current not too high or too low to get a good resistance.
We also did some preliminary work to see if there is a difference weather the two wires are apart from each other or constantly touching each other at all parts, all the time. Our preliminary findings were that it makes no difference. Since it made no difference, we decided to keep them separated so we don't have to twist the wires around each other.
We also had to work out the cross sectional area of the wires. We knew that the diameter for one wire was 0.25mm. We used the following formula to work out the cross sectional area:
Cross Sectional Area = Õr²
r = the radius of the wire. The radius is half the diameter so it should therefore be 0.125mm.
Using the formula, I have worked out the Cross Sectional Area for all the wires I will use. To get the Cross Sectional Area for two wires together you multiply what you found for the first wire by two, for three wires you multiply by 3 and so on until you find the Cross Sectional Area for all five wires. The results I got are shown in the table on the following page.
Number of wires
Cross Sectional Area(mm²)
0.05
2
0.10
3
0.15
4
0.20
5
0.25
Apparatus
Powerpack
Constantan wire
Crocodile clips
Leads
Test Rig
We set up the apparatus as shown below
Fairtest: To make it a fair test, I made sure that the same apparatus was used (i.e. crocodile clips and leads) as they can affect the resistance in the circuit. Also to make it a fair test, I will make sure only one of the factors listed in my aim will be changed. I will measure the length of the wire to 15cm using a ruler and making sure that it is correct to the nearest millimetre. I have no control over the temperature in the laboratory, so I will just have to take it into account. I will use the same material throughout the experiment. The material I will use is constantan wire. I will also make sure that no magnetic materials are placed anywhere near the experiment as it too can affect the resistance.
Method:
. Set up the apparatus as shown on the previous page using only one wire across the test rig.
2. On the power supply at 3 volts.
3. Record the voltage and current in the circuit by reading the appropriate meters.
4. Repeat steps 1-3 with two, three, four and five pieces of wire side by side across the test rig.
5. Record your results in a table and work out the resistance by dividing the voltage by the current.
Results:
Cross Sectional Area (mm²)
Experiment 1
Experiment 2
Experiment 3
Volt-age (V)
Cur -rent (I)
Resist-ance (W)
Volt-age (V)
Cur -rent (I)
Resist-ance (W)
Volt-age (V)
Cur -rent (I)
Resist-ance (W)
0.05
3.00
0.70
4.29
3.00
0.70
4.29
3.00
0.70
4.29
0.10
3.00
.50
2.00
3.00
.60
.88
3.00
.40
2.14
0.15
3.00
2.20
.36
3.00
2.10
.43
3.00
2.10
.43
0.20
3.00
2.90
.03
3.00
2.90
.03
3.00
2.80
.07
0.25
3.00
3.50
0.86
3.00
3.60
0.83
3.00
3.50
0.86
Cross Sectional Area (mm²)
Experiment 1
Experiment 2
Experiment 3
Average Resistamce (W)
Resistance (W)
Resistance (W)
Resistance (W)
0.05
4.29
4.29
4.29
4.29
0.10
2.00
.88
2.14
2.01
0.15
.36
.43
.43
.41
0.20
.03
.03
.07
.04
0.25
0.86
0.83
0.86
0.85
The graph (below) shows the average resistance of the wires plotted against the cross sectional area of the wire. As you can see it shows a curve, just like one I drew in my hypothesis. This curve is nothing but a visual expression of the relationship between the resistance and cross sectional area.
We could just plot resistance versus cross-section, but that just gives us a curve. Is this the curve we want? We always want a straight line if possible, since it is easy to see if points lie genuinely on a straight line. If we plot resistance versus 1/cross sectional area then this should give a straight line through the origin of the graph. If this happens, the equation is verified in this respect because in my hypothesis I stated that the resistance is inversely proportional to cross-section. The graph on the following page shows the resistance plotted against 1/cross sectional area.
Cross Sectional Area
/Cross Sectional Area
0.05
20
0.10
0
0.15
6.67
0.20
5
0.25
4
The graph (above) shows the resistance plotted over 1/Cross sectional area. As you can see, it is a straight line but two of the points do not lie exactly on the line. This means that there was an error while doing the experiment.
Evaluation: As I have already stated, I had two inaccurate results, although they weren't far off. These inaccurate results could have been parallax errors, or the equipment could have been faulty.
To ensure that my results were more accurate, I could have carried out the experiments a few more times, so the average would have been closer to the readings we should have had. We could have also tried out more cross sectional areas so the range is greater and it would be easier to find inaccurate results.
We could have also checked all the equipment we used, to see if they gave accurate readings.
We read directly over the meters to see where the needle was pointing so there is less chance of parallax error. If we had read it looking from the left or right side, we may have read the wrong current and therefore we will work out the resistant incorrectly.
We could have used digital meters as these will record the resistance to two decimal places and there would be no chance of making a parallax error
