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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound electrons are electrons with negative charge in an orbit around the nucleus; the bound electrons cannot leave their orbit. The free electrons are not in orbit but free to go where ever in the metal. The free electrons are the charge carriers, the charge is a negative charge in electricity and in conduction of heat the free electrons carry heat energy, which mean they are responsible for electrical conduction and heat conduction. As the plastics have no free electrons there can be no conduction, but as metals have free electrons this allows conduction to happen in the metals. An atom of a metal: - - + - - - An atom of a plastic: - + - There are no free electrons in the plastic so there will not be any conduction, because the free electrons are the charge carriers, without charge carriers no conduction. The free electrons are sub-atomic particles and have a negative charge so when there is a potential difference voltage the free electrons will move to the positive in one direction this is because opposite charges attract. The free electrons are moving in one direction but randomly, the free electrons don't all travel in straight lines but they are going in the same direction. This is a crystal of a metal or a space lattice: - The atoms are fixed meaning that they have a regular arrangement and that they are rigid and orderly. There will be collisions between the free electrons and the atoms. The atoms may not move and are rigid but they do vibrate. The reason why the free electrons are called charge carriers is because current is the flow of charge and free electrons carry this flow of charge. Free electrons move at high speeds as they travel large distances between collisions with the metal atoms, they can transfer energy at very quickly. Electrical conduction happens best in metals like silver, copper, aluminium, iron. Poor conductors and good resistors of electrical energy are glass, plastics, wood. Good conductors have low resistance. Bad conductors have high resistance. Resistance: - Resistance is the opposing of the flow of current. Electrical resistance is the property where electric flow or electric energy is transformed into heat energy this heat energy will then be opposing the electrical energy. Resistance involves the collisions between the free electrons and the vibrating atoms, to be more precise the collisions between free electrons and the bound electrons which makes up the structure of the conductor. The collisions with the free electrons and the vibrating atoms will slow down the flow of free electrons therefore slow down the flow of electricity. The resistance of a wire is directly proportional to its length and inversely proportional to its cross sectional area. The temperature and the number of collisions are related, if the temperature rises then the atoms will vibrate more. The more vibrations of the atoms, would decrease the space so there will be more collisions and the more collisions will slow down the charge carriers free electrons so there will be more resistance. From this diagram we can see that the charge carriers the free electrons are having more collisions because the atoms are vibrating more because of the higher temperature. P2a SAFETY Safety will always be a vital factor when carrying out any experiment. This experiment will not involve the use of electricity through the mains. This amount of electricity will be around 240volts coming through the mains, this is too much because it is dangerous enough too kill you. So the voltage should be lowered in this experiment so that is why cells are used and not the mains, this also is could because a low voltage would keep the temperature of the wire the same. It is much safer to experiment with circuits powered by low voltage source. In the experiment to make the level of voltage to be safe I would use battery cells which have low voltages, the lowest voltage the cells could be 2 volts and the highest would be 5volts so this source would be much safer. Another safety precaution that I will take is will I will not work near any desktops with water taps. The reason for this is that if water goes into the circuit or more importantly into the electrical socket it could results to hazards like fire hazards, so I will for safety do the experiment on a side desktop away from gas taps and water taps. P4a FAIR TEST For this experiment to be a fair and get reliable results I will have to undertake a fair text. The fair test will have to undertake these following things: - The non-variables or the things that I will keep the same through out the experiment will be the temperature; at room temperature cross sectional area of wire which is 0.5mm or 0.05cm, the material of the wire will be kept the same which will be Nichrome. I will also keep the same amount of cells which will be three cells throughout the experiment; keeping the same amount of cells the same will make the experiment a fair test because if the amount of cells were changed each time then this will affect the results because if the cells were changed the starting voltage of the power supply would be different so you the results would not be reliable. When the temperature of the wire is increased then the atoms of the wire will vibrate more and this will decrease the space in the space lattice so there will be more collisions and there will be more resistance. We know that if the temperature is not kept the same throughout the experiment it will not be a fair test, because of the changing temperature. If you keep the non-variables the same then you can get correct and fair results as you change the variable which in this case is the length of the wire. Each length will be repeated twice and the average taken to get fair results. P4b LIST OF EQUIPMENT AND APPROPRIATENESS Equipment/Instruments: - "¢ Ammeter "¢ Voltmeter "¢ Crocodile clips "¢ Three cells "¢ Nichrome wire "¢ Meter ruler "¢ Board to put cells in. For the measurement instruments they will need a least count for appropriateness and consistency. The least count for the meter ruler will be 1mm or 0.1cm. The voltmeter will be in volts and the least count will be 0.1volts. For the ammeter the measurement will be amperes and the least count will be 0.1amperes. P6a THEORY AND PREDICTION The factors that affect the resistance of a wire include the following: - "¢ Material "¢ Temperature "¢ Cross sectional area of wire "¢ Length of a wire The material of a wire has three things that affect the resistance: 1. The type of metal: - each different types of metal are structured differently; their space lattice or crystals are different to one another. The interatomic distance was big then there would be fewer collisions so there will be less resistance. If the interatomic distance in the structure of the metal was small there would be more collisions therefore more resistance because the free electrons flow has been slowed. 2. The amount of free electrons: - this also affects the resistance, if the amount of free electrons in one metal is more than in another metal then the metal with more free electrons will have less resistance. The more the number of free electrons then less resistance, the less the number of free electrons then the more the resistance. The reason for this is that if there are more free electrons means that more electrical charge is been carried so there is less resistance. 3. Arrangement of atoms in the metal: - if the structure of the atom or arrangement is different then this can affect the resistance. If the arrangement of the atoms has an atom in the middle there would be more collisions so there will be more resistance. If the structure had no atom in the middle there would be fewer collisions therefore less resistance. This diagram shows two differently arranged atoms one is the body-centred cubic bcc arrangement the other is faced-centred cubic FCC arrangement. The bcc arranged atom has less atoms and only one atom in the middle so it would have less collisions than the FCC arrangement because the FCC arrangement has more atoms and more in the middle, this will increase the amount of collisions and then there will be more resistance. The temperature will affect the resistance of a wire. A falling temperature will increase the conductivity. Most of the resistance to the motion of the free electrons therefore resistance of the wire is the thermal vibrations of the atoms. If the temperature is reduced to absolute zero then thermal vibrations will stop, then this will stop the resistance due to temperature. The cross-sectional area of a wire will affect the resistance. A larger cross sectional area will mean a larger current I and that makes smaller resistance. A smaller cross-sectional area has a smaller current which is a bigger resistance. The resistance of a wire is inversely proportional to its cross-sectional area. The length of a wire affects the resistance. If I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. This means that the resistance is directly proportional to the length of the wire. When the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire. L no: of collisions/sec. Resistance=R 2L 2N Resistance= 2R p8a Ohms law: - these are laws that relate to current, potential difference, and resistance. The resistance is measured in ohms, the potential difference is measured in volts and the current is measured in amperes amps. Ohms law states the amount of steady current through materials is directly proportional to the potential difference Voltage. An example that proves this is that if the potential difference voltage between two ends of a wire is tripled, the current is tripled. Ohms law also says that I=V/R, or the current in the conductor equals the potential difference voltage across the conductor divided by the resistance of the conductor. It also says that the potential difference across a conductor equals the current in the conductor times its resistance, V=IxR. Ohms law also explain that resistance of a material is its potential difference voltage divided by the current, R=V/R. The larger the resistance the greater the voltage is needed to push each ampere through it: there is a resistance of one ohm and a voltage p.d of one volt will drive a current of one ampere through it. Conductors that obey ohms law are called ohmic conductors; conductors that don't obey ohms law like semiconductors are called non-ohmic conductors. Resistivity: - is the electrical resistance of a conductor of unit cross-sectional area and unit length. A property of each material resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. The unit of resistivity is ohm-meter or ohm-centimetre. A high resistivity is in poor conductors, a poor conductor has high resistance, so high resistivity has high resistance and a conductor with low resistivity will have low resistance. Resistivity P is quantitatively equal to the resistance R of e.g. a wire times its cross-sectional area A and then divided by its length L. P =R A /L. Resistance = R= P L /A p8a Method: - 1. First I will connect the cells to the circuit board where the cells are placed. 2. Then I will then use the three pairs of connecting wire crocodile clips to connect the cells to the ammeter and the wire and back to the cells to make a complete circuit. I will also add a voltmeter opposite to the power pack/in parallel. 3. Thirdly the wire will be on a board and on that there will be a ruler so I can measure the different lengths. 4. Next, when I have measured one length I will break the circuit by taking off a connecting wire, this is so that the wire does not change temperature and affect results. 5. Then when I am ready to take the next length I will connect the circuit and move the crocodile clip to the new length. 6. I will do all the lengths three times to find reliable readings. I will also have a good range of lengths so I can find a full set of results to see how the length of a wire affects the resistance of the wire. Cells Voltmeter V Before the power is put on the ammeter is at zero. The point where the crocodile clip and the wire meet is at zero. The ammeter has two scales and the first one is 0.00amps to 1.00amps this is for a low voltage that I am using so this is the appropriate scale that I should use as I am using low voltage cells. The voltmeters least count is 0.01volts and before the experiment the voltmeter was at absolute zero. In my method I could have used digital ammeter but there was not one provided, a digital display ammeter would give me even more accurate readings. p6b NO: OF READINGS The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm 55.00cm 60.00cm 65.00cm 70.00cm 75.00cm 80.00cm 85.00cm 90.00cm 95.00cm 100.00cm This is my table where I am going to show all my results for my experiment. I will take the readings from the voltmeter to find the potential difference voltage, I will take the ampere readings to find the current I. The table has three sets of reading to put in because this means I am repeating the readings and trying to find consistent results and reliable results, any anonymous results will be marked with an asterisk and then repeated again. The average will be used to find average resistance for the three sets of results and see if they are reliable e.g. 3.1Ω, 3.2Ω, 3.1Ω average will be 3.1Ω+3.2Ω +3.1Ω=9.4Ω divided by 3= 3.1Ω. The average can let me set a graph showing how the average resistance against the length of a wire. The final column will let me work out the resistance once I have collected all the data. The table has a suitable range for the length of the wire, this is because this is my variable and I need to get a full range of results. The table has the least counts of scale units and the decimal places I am using. p8b PRELIMINARY WORK The dissipation of electric energy in the form of heat, even though small, affects the amount of electromotive force, or driving voltage, required to produce a given current through the circuit. In fact, the electromotive force V measured in volts across a circuit divided by the current I amperes through that circuit defines quantitatively the amount of electrical resistance R. Precisely, R = V/I. Thus, if a 12-volt battery steadily drives a 2-ampere current through a length of wire, the wire has a resistance of 6 volts per ampere, or 6 ohms. Ohm is the common unit of electrical resistance, equivalent to one volt per ampere and represented by the capital Greek letter omega Ω. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Resistance also depends on the material of the conductor. The Electrical resistance of a conductor is dependent on cross-sectional area and length. A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors. Most resistance is down to the motion of free electrons comes from the thermal vibration of the atoms; if the temperature is reduced to almost absolute zero, where thermal motion essentially stops, conductivity can increase several thousand times over. The value of resistivity depends also on the temperature of the material; tabulations of Resistivities usually list values at 20 C. Resistivity of metallic conductors generally increases with a rise in temperature; but resistivity of semiconductors, such as carbon and silicon, generally decreases with temperature rise. The resistivity of an exceedingly good electrical conductor, such as hard-drawn copper, at 20 C 68 F is 1.77 10-8 ohm-metre, or 1.77 10-6 ohm-centimetre. At the other extreme, electrical insulators have Resistivities in the range 1012 to 1020 ohm-metres. The unit of resistance is the ohm. In the metre-kilogram-second mks system, the ratio of area in square metres to length in metres simplifies to just metres. Thus, in the metre-kilogram-second system, the unit of resistivity is ohm-metre. If lengths are measured in centimetres, resistivity may be expressed in units of ohm-centimetre. OBSERVATIONS o4b o4a o6b o6b table of results: - The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Ω Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I=Current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p 05.00cm 1.18v 2.22 0.53 Ω 1.18v 2.22 0.53 Ω *2.90v 1.18v *2.97 2.22 0.53 Ω 0.5 Ω 10.00cm 1.64v 1.91 0.86 Ω 1.64v 1.91 0.86 Ω 1.63v 1.90 0.86 Ω 0.9 Ω 15.00cm 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.2 Ω 20.00cm 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 1.5 Ω 25.00cm 2.49v 1.34 1.86 Ω 2.49v 1.34 1.86 Ω 2.51v 1.33 1.89 Ω 1.9 Ω 30.00cm 2.67v 1.22 2.19 Ω 2.67v 1.22 2.19 Ω 2.67v 1.21 2.21 Ω 2.2 Ω 35.00cm 2.82v 1.12 2.52 Ω 2.83v 1.11 2.55 Ω 2.82v 1.12 2.52 Ω 2.5 Ω 40.00cm 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.9 Ω 45.00cm 3.06v 0.96 3.19 Ω 3.06v 0.95 3.22 Ω 3.06v 0.96 3.19 Ω 3.2 Ω 50.00cm 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.5 Ω 55.00cm 3.24v 0.84 3.86 Ω 3.22v 0.83 3.88 Ω 3.22v 0.83 3.88 Ω 3.9 Ω 60.00cm 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 4.4 Ω 65.00cm 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 4.6 Ω 70.00cm 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 4.8 Ω 75.00cm 3.49v 0.68 5.13 Ω *4.00v 3.49v *0.98 0.67 *4.08 Ω 5.21 Ω 3.49v 0.68 5.13 Ω 5.2 Ω 80.00cm 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 5.5 Ω 85.00cm 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 5.9 Ω 90.00cm 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 6.1 Ω 95.00cm 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 6.5 Ω 100.00cm 3.69v 0.54 6.83 Ω 3.69v 0.54 6.83 Ω *4.29v 3.70 *0.32 0.54 6.85 Ω 6.8 Ω o8a: - *note: the circled and crossed out results are anonymous so I repeated the reading to get a reliable reading and result, I did this because I thought that the reading was wrong and I needed to do it again, it was common sense because the anomalous results obviously didn't follow certain patterns in the table. I did the reading again and I got the same as I got in the other two sets of readings for that length. o6b: - the table is an accurate record of observations because each column of reading is to the same decimal place and is accurate. * In the planning I planned to use an analogue ammeter and voltmeter but I was able to use a digital ammeter and voltmeter as the teacher could get one in the last lesson of the practical. * I rounded the final average resistance to one decimal place because this is common sense and this will make it easier to plot the graph. ANALYSIS A.2a After completing the experiment and getting a full range of reliable results from the experiment, I have found out that in general that as the length of the wire goes up the resistance of the wire also goes up. This states that my theory and prediction is correct. I have found out that this general finding says that the resistance of the wire is proportional to the length of the wire and in many occasions the length is directly proportional to the resistance of the wire. A4.b PATTERN IN READINGS: - After collecting my results from the observations I carried out I can now analyse the results and make sense of the readings in the table of results. The table of results has a general pattern that suggests that as the length of the wire is increased the current decreases and therefore the resistance increases as my theory states through different rules. This general finding means that the resistance of the wire is proportional to the length of the wire. The readings from the table of result shows as the length of the Nichrome wire increases the current in the circuit decreases and as the theory explains the resistance should increase and this occurs in the table of readings. The readings in my results table show that the length is proportional to the resistance of a wire and in some cases nearly directly proportional and other cases exactly the length is directly proportional to the resistance of the wire. E.g. the result from 5cm to 10cm, the average resistance for 5cm is 0.5Ω and the average resistance for 10cm is 0.9Ω, this shows that the length of the wire is proportional to the resistance, if it was directly proportional then the resistance for 10cm will be 1.0Ω, this is because when the length is doubled then the resistance is doubled this then would be directly proportional, this result was not accurate by 0.1 because I could of made minor human error like not putting the exact length. The results from 30cm has the resistance of 2.2Ω and the and the doubled length that is 60cm has the resistance of 4.4Ω, this shows that as the length is doubled the resistance is doubled which means that the length is directly proportional to the resistance of the wire, these result from 30cm to 60cm proves that my theory that the length of wire is directly proportional to the resistance of a wire. The mistakes I made called experimental like not breaking the circuit when changing the length could of have affected the temperature of the wire therefore create more vibrations for the atoms in the wire and cause a bigger resistance then it should have been without the experimental error. A4a A6a A4b Graph of results: - A4b PATTERN ON GRAPH: - The pattern that appears on the graph of results tells us that the length of the wire is directly proportional to the resistance of the wire. The line of best fit provides the graph with the results. The results that prove that the resistance of a wire is directly proportional to its length include the result of the lengths 15cm to 30cm, 15cm has the resistance of 1.1Ω and the resistance of 30cm is 2.2Ω this shows as the length is doubled the resistance also is doubled which means that the resistance of the wire is directly proportional to the length of the wire, this result shows the graph of results to state my theory and to be correct as the result agrees with my theory. Other results in my graph also state that the resistance of the wire is directly proportional to its length. They include: - 26cm that had the resistance of 1.9Ω and then when the length was doubled to 52cm the resistance was 3.8Ω, this result shows that as the length is doubled the resistance is also doubled which means that the length is directly proportional to the resistance of a wire, the result again proves my prediction correct because I predicted that the length of wire is directly proportional to resistance of the wire, the results from the graph state this with these results. The other results from the graph also tell us that the length is proportional to the resistance of the wire they include: - 5cm R=3.5Ω and 10cm R=7.5Ω, 20cm R=1.5 & 40cm R=2.9Ω. These results that are not directly proportional but it is only by a few 0.1 ohms this could be down to not drawing a 100% straight line and this cause these results to be a bit off. A6b A8a: - explanation using theory From my readings in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω, so the results from the table of readings state my theory and prediction, this also proven by the diagram theory I did: - From the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. The resistivity is quantitatively equal to the resistance, from the results this would be the resistance R of the wire times the cross-sectional area A divided by the length L of the wire. From this you can find out if the resistance is directly proportional to the length and inversely proportional to the cross-sectional area and see if your results are correct. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory. Results are correct because the resistivity and the resistance are quantitatively equal. This result also shows that my main prediction that is, the length of the wire is directly proportional to the resistance of the wire; this prediction was proven by this result as the theory of resistivity and the equations P =R A /L and R= P L /A were worked out using my result that showed that the resistance and resistivity were quantitatively equal this is proof that my results are reliable and follow the theory of rules in resistance and resistivity and states my prediction. A8b: - conclusion of results and analysis. Throughout the analysis of the results I have found many results from my readings in the table of results and the results from the graph, they show the relationship of the length of the wire and to the resistance of the wire. The general conclusion I have made after analysing these results is that the length of a wire is directly proportional to the resistance of a wire. The proof of this conclusion lies within the results in the readings and the graph which shows the relationship between the length of the wire and its resistance. The results from the table of readings and the graph do agree with my theory this shows that my results are proof of the theory. This is shown e.g. in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω. Also shown in from the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. This result I worked out about resistivity also can prove that the results are proof of the theory. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory of resistivity and show results are correct so the results are the proof of the theory. EVALUATION E.2a: - This coursework that I undertook on the factors that affect the resistance of a wire was quite challenging but at the same time enjoyable to carry out. The reason for which I found it enjoyable was that the physics topic the coursework came under is one of my favourite areas as I have an interest within physics. The coursework to me was easy in parts but other parts of the coursework were challenging as I had to implicate the theory to real experiments and I also enjoyed the independent research that was needed. The coursework made me understand the factors affecting the resistance of a wire, and it also allowed me to understand certain laws in physics like the ohms law, and the theory of conduction. E.4a: - The coursework I took on was accurate this could be proven by the results and the graph. The points that lie on my graph is quite adequate out of twenty points of results ten where on the graph line of best fit the other eight were only a few decimal places on the graph below or above, they were slightly off the line of best fit, while only two were anomalous but by only small margins. The anomalous results were for 60cm and 10cm readings as the points were the furthest away from the graph or line of best fit. The accurateness was also proven by the graph passing through the origin which is the co-ordinates 0, 0, this level of accuracy suggest that my results and graph is reliable and accurate. E.4b: - evaluation of method For this coursework the method was very important because if the method was faulty and had a lot of gaps in it the results would be affected therefore most of the coursework. The method that I used was generally a good one to use as with anything it could have improved e.g. I could have used a power pack connected to the mains as the power supply, the power pack could easily change the voltage you want and keep it the same, this would also subsequently avoid complications of finding the voltage with a voltmeter and make the method quicker. A power pack would also allow me to switch of the circuit and not affect the temperature and then the results. In the method instead of taking the readings at every 5cm I could take it at every 2cm or even 1cm to give a better range and reliable results. Other improvement I could make to the method would be to make more accurate by putting the crocodile clips together without any part of the clip not connected. These changes and improvement would make my method much better and get rid of the faults that affected the results I got. E.6a: - evidence. A table to show the reliability of the results and the anomalous results present. LENGTH cm RESISTANCE From GRAPH Ω ohms RESISTANCE from EXPERIMENT Ω ohms DIFFERENCE 2.d.p % DIFFERENCE =diff÷ R graph × 100 1d.p * 5.00 0.35 0.50 0.15 *42 % 10.00 0.80 0.90 0.10 1 % 15.00 1.10 1.20 0.10 *9 % 20.00 1.50 1.50 0.00 0 % 25.00 1.82 1.90 0.08 4 % 30.00 2.20 2.20 0.00 0 % 35.00 2.52 2.50 0.02 1% 40.00 2.90 2.90 0.00 0 % 45.00 3.25 3.20 0.05 1 % 50.00 3.55 3.50 0.05 1 % 55.00 3.90 3.90 0.00 0 % 60.00 4.25 4.40 0.15 3 % 65.00 4.60 4.60 0.00 0 % 70.00 4.91 4.80 0.11 2 % 75.00 5.25 5.20 0.05 1% 80.00 5.60 5.50 0.10 1 % 85.00 5.90 5.90 0.00 0 % 90.00 6.20 6.10 0.10 1 % 95.00 6.50 6.50 0.00 0 % 100.00 6.80 6.80 0.00 0 % AVERAGE ERROR = 3.35 % Here you can see the average error in % is acceptable within experimental error, this is evidence to justify that my conclusion, because it is within the scope of experimental error so the results I got should be considered reliable because it is a low % of error, so this would prove that the length of the wire is directly proportional to the resistance of the Nichrome wire and also prove my analysis and conclusion to be reliable. In the table which shows the reliability of results also showed that my experiment had only two anomalous results, this is acceptable because within the experiment I could have made only one significant error to cause this anomalous result. The rest of the results were reliable as the table shows. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.0 Ω 0.20 ÷ 5.0 × 100% = 4 %. This is within experimental error. Length =30.00 cm, Resistance = 2.20 Ω Doubled = 60.00cm, Resistance = 4.40 Ω 0.00 Ω off 4.40 Ω 0.00 ÷ 4.40 × 100% = 0%. This is within experimental error, because there is no experimental error in this result. Length = 15.00cm, Resistance = 1.10 Ω Doubled = 30.00cm, Resistance = 2.20 Ω 0.00 Ω off 2.20 Ω 0.00 ÷ 2.20 × 100% = 0%. There is no experimental error here. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.00 Ω 0.20 ÷ 5.00 × 100% = 4%, again this is within experimental error. All these results I have showed show that the experimental error was low or there was no experimental error, so the results can be expected to be reliable and therefore the results are reliable enough to justify my conclusion. With the table showing the reliability of results with the difference and the individual evidence of results I have picked out it can be said my results are reliable. E6b: - further work. For the coursework if I were to do it again I would do further work to get more evidence. The further that I would carry out to get more evidence would consist of doing a wider range of lengths of wire and also to try other materials, like copper, constantan in the experiment, all this so I could get more evidence. Further work could also consists of choosing another factor and investigate how that factor affects the resistance of the wire. The factor could be the S.W.G or the thickness of the wire. The principle of the S.W.G gauge is that the electrical resistance of a wire changes when either stretched or compressed or dependent on the thickness. If I were to the experiment for the factor of the thickness I would need to have a new circuit, equipment and a new method for the experiment. The equipment that will be need is:- 1. Crocodile clips or connecting wire. 2. Power pack. 3. Ammeter. 4. Voltmeter. 5. Constantan or Nichrome wire depending on what you want to use. Method: - Power pack Voltmeter 1. First I would connect the power pack to the main power but not turning on the switch for safety purposes. 2. I then would set the power pack to A.C and the voltage that could be suitable a voltage between 4 volts to 10 volts could be used. 3. Then I would have to connect the circuit, using the connecting wire and crocodile clips. I will connect the voltmeter opposite to the power pack and the ammeter anywhere in the circuit. 4. then I would have to connect the circuit to the wire using the crocodile clips I will have to make sure that al of the crocodile clip is in contact with the wire at both sides. 5. I will take the readings for that specific thickness of the wire then remove the wire putting another thickness then taking the readings, I will take the readings to the appropriate range of thickness that would have to be chosen. 6. When changing the wires I will have to shut the power pack so the circuit doesn't over heat and also the wire doesn't heat up because this can affect the resistance. Table to show how I would collect the readings for the experiment: - READING NUMBER ONE READING NUMBER TWO READING NUMBER THREE AVERAGE RESISTANCE Average= 1.d.p Cross sectional area/thickness of wire A A+-0.1cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm [7,657] WORDS
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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound...
count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p Voltage

Volts

v+- 0.01v

2.d.p Current I

Amperes amps Least count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p

05.00cm

10.00cm

15.00cm

20.00cm

25.00cm

30.00cm

35.00cm

40.00cm

45.00cm

50.00cm

[7,657] WORDS

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We are trying to... We are trying to find out what factors affect frictions. Friction is the force that opposes motion, or makes it difficult an object to move across a surface. The amount of friction depends on the surface type and the force pressing two objects together. I am going to research three factors: "¢ Different surfaces "¢ Area of contact "¢ If the surface is wet For our preliminary work we used a block of wood with a hook in the end of it. We attached a Newton metre to measure the amount of friction; we did this by exerting force onto the Newton metre and recording our results. Each time we added a sandbag to our block of wood which weighed about 1kg and took the results. The sandbag is not accurately measured as it only around 1Kg Each time we added a sandbag the amount of friction became larger we know this as we recorded these results as seen below. This shows us that the more weight on the wood the more friction there is. This experiment is not completely accurate as the weights sandbags weigh only around 1 kg not exactly 1kg. No. of weights Sandbags Friction in N 0 0.7N 1 4.4N 2 10N 3 14N For more reliable results I am going to; "¢ Attach some string and a pulley to the block of wood I will do this at the edge of a table. "¢ On the block of wood I will add one different weight each time 0g, 50g, 100g, 150g, 200g. "¢ After doing this I will add as many weighs as it takes to move the block of wood onto the pulley. The weights will be100g, 50g, 10g, 5g; I am mixing the weights so that my results will be more accurate. "¢ I may not use all the different weights but I can not tell this until I do the experiment. To make it even more accurate I will make sure that the surface table I do this experiment on is the same when I repeat the experiment, I will also make sure the surface is not wet. I am going to take my results 3 times so I know they are accurate. If I find a result that is not right it is called an anomalous result I will show this by doing a table and a graph. Whilst doing this experiment I experienced a few slight problems for example I added the pulley the wood moved but these problems did not cause too many problems. Weights On Block 1st 2nd 3rd 0g 60g 90g 100g 50g 80g 110g 110g 100g 140g 130g 140g 150g 100g 140g 130g 200g 190g 160g 150g As the table above shows I came across three anomalous results whilst doing my experiments, I have plotted this in my graph. There could be many different reasons why this occurred the surface area could have become wet or when the pulley with the weights could have slightly moved therefore there would be a different surface contact. My graph shows that the force needed to move the block of wood increases when there is a heavier weight is on it. The shape of my graph curves as it goes up which shoes an increase in the force. By doing this investigation I have learned many things about friction and what affects friction. When I did this experiment I learnt that when there is a big weight on an object or the object itself weighs a lot then a higher force is needed to move the object. My results all seemed fairly accurate as each time I did the experiment it followed a similar pattern each time. I found three anomalous results two of which were in my final experiment something could have affected this as I have talked about earlier. I think that my results back up my conclusion quite well. My method worked very well I did not experience any real problems and found working by step by step method straight forward. To develop my experiment more I could; "¢ Do this experiment several more times. "¢ Do the same experiment but use several different surfaces, smooth table, rough sandpaper or a shiny black bag. This experiment has tested many theory's and has provide me with a basic level of understanding.   

We are trying to find out what factors affect frictions. Friction is the force that opposes motion, or makes it difficult an object to move across a surface. The amount of friction depends on the surface type and the force pressing two objects together. I am going to...

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INTRODUCTION: In this coursework... INTRODUCTION: In this coursework investigation, I will be doing a series of experiments to see how changing the length of a piece of wire affects the resistance of the wire. PREDICTION & BACKGROUND THEORY: I predict that as the length of wire becomes longer, the resistance will become gradually bigger. A piece of wire is made up of several hundred atoms. An electrical flow is able to run through the wire, these are e- electrons. As the current increases the velocity of the e- electrons increases which increases the amount of collisions between the electrons and the atoms. This, in turn, creates a higher resistance. We know that if the current increases, the resistance does too. There are three factors which can affect the resistance of a wire, these are the length of the wire, the material it is made of and the area of the wire. The structure of the wire is show below, it is similar to a lattice structure with atoms in rows, packed quite closely together. = Atoms = Electrons As we can see from the diagram, the electrons will have to work harder to make their way through the piece of wire. PRELIMINARY WORK: To make sure I have done my research and preparation work correctly, I will do a preliminary experiment. This is like a "test run" to make sure I don't run into any major problems. Should a problem become apparent, I will be able to change it before I start to record my results of the experiment for real. I will record my results, however, to see if they look correct and in the right range. METHOD: a Set up apparatus in the following way b Set up wire in the appropriate way. i.e. choose correct length and put crocodile clips either side of chosen length c Ensure ammeter is set to "0". d Switch on power supply e Record results f Switch off power supply g Alter length of wire h Repeat steps "b" to "g" for remaining length of wire SAFETY PROCEDURES: a Switch off power supply between readings b Ensure wires are connected t power supply correctly c Ensure wires are safe "“ i.e. not bare or broken ACCURACY PROCEDURES: a Measure a certain length of wire in more than one place to see if you get a similar reading. b Always start experiment with Ammeter at "0" c Switch of power supply to allow wire to go cool for all readings PROCEDURE FOR FAIRNESS: a Use same equipment for all experiments i. Ammeter ii. Volt Meter iii. Wire b Allow wire to go cold before taking next reading. Temperature could cause changes. RESULTS: Length in CM Voltage Average Resistance V1 V2 V3 In W 5 0.5 0.2 0.1 0.3 3 10 0.7 0.3 0.2 0.4 4 15 0.9 0.4 0.3 0.5 5 20 1.1 0.5 0.4 0.6 6 25 1.2 0.6 0.5 0.9 9 30 1.4 0.7 0.6 1.2 12 35 1.5 0.8 0.7 1.0 10 40 1.6 1.0 0.8 1.1 11 45 1.7 1.1 0.9 1.2 12 50 1.8 1.2 1.0 1.3 13 Current = 0.1amp Wire = Nickel Chrome ANALYSIS AND EVALUATION From the graph overleaf, we can deduce that our prediction was correct. As the length of the wire increases, the resistance does too. We have noticed that two points don not follow this trend. This could be due to a flaw in the experiment, such as "we left the power on too long and caused overheating" or "we simply took a mis-reading". However, this does not happen for the majority of the readings so we have an adequate set of results to draw this conclusion. The results could be changed if we did things differently a second time around: - Changed current - Used different wire - Left the power on for longer Our prediction has been proved correct. As the length of wire increases the resistance gets bigger. They are proportional to each other in this case.   

INTRODUCTION: In this coursework investigation, I will be doing a series of experiments to see how changing the length of a piece of wire affects the resistance of the wire. PREDICTION & BACKGROUND THEORY: I predict that as the length of wire becomes longer, the resistance...

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Introduction What is electrical resistance?...Introduction What is electrical resistance? Electrical resistance is what limits the current inside a circuit. What effects resistance? "¢ Length: The longer the length of a wire the higher the resistance level will be. "¢ Diameter: The thicker the diameter of a wire the lower the resistance level will be. "¢ Material: Insulators have a very high resistance level and conductors have a very low resistance level. Metals are normally good conductors and plastics are very good insulators. In this assignment I am going to investigate how the length of a wire affects the level of resistance. Background Theory What is electrical current? Electrical current is the electron flow through a wire. Why are metals good conductors? Metals have free electrons inside a lattice of positive ions. This makes them very good conductors where insulators don't have free moving electrons. The electrons are very closely packed. Electrical current is electron flow. When electrons flow through a wire they collide with positive ions. The more collisions that take place between the electrons and the positive ions, the smaller the current is that flows the higher the resistance level is. Prediction I predict that if the length of the wire is increased then the resistance will also increase. I also predict that if the length of the wire is doubled the resistance will also double. When the length doubles the number of ions will double and due to this there will be double the amount of collisions. This will then cause the electron flow to halve and when the electron flow halves the current will also halve. This then causes the resistance level to double. If the length of the wire is then the resistance will also increase. If the length doubles then I believe the resistance will double and this shows the resistance is proportional to the length. Pre-test I am going to investigate the resistance change when the length of the wire is changed. The diameter of the wire will also affect the resistance so I am going to perform a pre-test to decide which diameter of wire I am going to use. I am going to use three different diameters and they are 0.55mm, 0.37mm, 0.27mm of constantan wire. Diameter = 0.27mm Length cm Voltage V Current A Resistance 20 2 1.4 1.43 20 1 0.75 1.33 50 2 0.6 3.33 50 3 0.8 3.75 Diameter = 0.37mm Length cm Voltage V Current A Resistance 20 1.6 1.8 1.56 20 2.0 2.3 1.57 50 2.6 1.3 3.70 50 3.3 1.7 3.88 Diameter = 0.55mm Length cm Voltage V Current A Resistance 20 0.8 2.3 0.35 20 0.4 1.2 0.33 50 1.8 1.9 0.95 50 1.2 1.7 0.71 From my pre-test results I can see that the wire that would be best to use would be the 0.55mm wire. The 0.27mm and 0.37mm wires got too hot after a short while and the 0.55mm wire did not increase in temperature. The rise in temperature, of the wire caused the resistance to increase because all the particles gained energy which caused them to move more which meant more collisions where taking place. Due to this I have chosen the 0.55mm diameter constantan wire. Plan Apparatus 0.55mm diameter constantan wire Ammeter Connecting wires Crocodile clips Meter ruler Power pack Variable resistor Voltmeter Diagram Method First I am going to set up my apparatus and then I am going to mark out nine different measurements and them being 10cm, 20cm, 30cm, 40cm, 50cm, 60cm, 70cm, 80cm and 90cm. I will move the crocodile clips, which are connected to the ammeter, voltmeter and constantan wire up every 10cm. I am then going to take the voltage for each measurement and also repeat the method three times. I am then going to divide the voltage by the current and my answer will be the resistance. I will then take an average of my three resistance results. My results charts will look like these. Length cm Voltage V Current A Voltage V Current A Voltage V Current A 10 20 30 40 50 60 70 80 90 Length cm Resistance 1 Resistance 2 Resistance 3 Average Resistance 10 20 30 40 50 60 70 80 90 Results Length cm Voltage V Current A Voltage V Current A Voltage V Current A 10 0.14 0.6 0.15 0.6 0.16 0.6 20 0.26 0.6 0.26 0.6 0.27 0.6 30 0.39 0.6 0.39 0.6 0.39 0.6 40 0.51 0.6 0.50 0.6 0.50 0.6 50 0.59 0.6 0.61 0.6 0.61 0.6 60 0.70 0.6 0.73 0.6 0.73 0.6 70 0.80 0.6 0.83 0.6 0.83 0.6 80 0.90 0.6 0.94 0.6 0.94 0.6 90 1.01 0.6 1.03 0.6 1.03 0.6 Length cm Resistance 1 Resistance 2 Resistance 3 Average Resistance 10 0.24 0.25 0.27 0.25 20 0.42 0.42 0.45 0.43 30 0.65 0.65 0.65 0.65 40 0.85 0.83 0.83 0.84 50 0.98 1.02 1.02 1.01 60 1.67 1.23 1.27 1.38 70 1.34 1.86/1.36 1.38 1.52 80 1.50 1.57 1.57 1.75 90 1.86 1.72 1.71 1.76 Analysis My graph shows that the resistance is proportional to the length up to 70cm and then it curves slightly after that. Up to 70 cm on my graph shows that my prediction was correct. I predicted that if the length of the wire doubled then so would the resistance. I took three readings off my graph: 15, 30 and 45. This is shown in the table below. Length cm Average Resistance 15 33 30 66 45 98 My table of results above shows that when the length doubles so does the resistance and when the length triples so should the resistance triple. This proves what I said in my prediction was correct that the resistance is proportional to the length. This can be explained because metals have free moving electrons in a bed of positive ions. All the free electrons will collide with the ions inside the wire and these ions will gain more energy and cause many more collisions to take place. The more collisions that take place the smaller the current will be and this causes the level of resistance to increase. After the 70cm mark the resistance was lower than expected. The resistance stopped being proportional to the length as it begun to curve at the end. It could not be due to the wire overheating because from that I would have expected the resistance to rise, and this is because the warmer the wire the more collisions that would have taken place. The more collisions that take place, the smaller the current and the higher the resistance would be. Since my graph begun to curve at the end I'm not totally sure that the resistance is proportional to the length. Maybe it is up to a point. Another reason why my graph curved toward the end could be human error. The current may not have been 0.6A for all the lengths. Evaluation I think that my investigation was successful but it could have been more accurate. Most of the points on my graph are near the line of best fit. I had one anomalous result which was the second reading for 70 cm. My reading was 1.83 which was way out so I did another test and now my new result for that reading is 1.36. This is most likely due to human error. I could have moved the slider on the resistor which would have changed the current and also my graph could not have been accurate from 70 "“ 90 cm. I may also have taken the length measurements wrong. Improvements I used a digital voltmeter which is accurate to +-0.01V and I used an analogue ammeter which is accurate to +-0.1A. If I had used a digital ammeter , which is has an accuracy of 0.01A then that would have given me more3 accurate results. I took three measurements where I could have taken five. If I had taken five measurements then I would have been able to have more points to plot on my graph. This would have helped me to see if I should have had a straight or curved line. The crocodile clips I used were all thick in diameter and all had different diameters. If I had to same slim diameter for each crocodile clip then I would have had a more accurate set of results. Further Work My investigation showed that up to 70cm my graph agreed with my prediction but after 70cm I'm not sure if the resistance is proportional to the length. I could have taken more readings on my graph after 70 cm and I could have taken measurements for every 2 cm after the 70cm point. I could also take my measurements up to 10 m to see if the resistance was proportional to the length for a longer length.   

Introduction What is electrical resistance? Electrical resistance is what limits the current inside a circuit. What effects resistance? • Length: The longer the length of a wire the higher the resistance level will be. • Diameter: The thicker the diameter of a wire the lower...

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I plan to see... I plan to see if the body size of an animal affects its heat loss. I am going to do this by using three glass beakers with various amounts of boiling water in them. I will then see which beaker looses its heat the fastest or the slowest. Prediction I predict that the smaller the animal the faster that it will loose its body heat so I predict the smaller beaker will loose its heat the fastest and the largest one will loose its heat the slowest. The science I predict this because for example if you have a drop of boiling water and a jug of boiling water the drop of boiling water will cool down the quickest. I have also predicted this because the wild animals in the cold adapt to their environment by the size of their bodies and shape. I will be focusing on the size. E.g. polar bears live in cold weather so they are big to reduce heat loss. Apparatus "¢ 3 glass beakers "¢ kettle "¢ thermometer "¢ 3 timers Safe test I will make sure that my experiment is safe by "“ "¢ Making sure that I handle the boiling water carefully. "¢ Making sure that the water doesn't spill near any electricity or on the floor "¢ Make sure that I do not smash anything Fair test I will make sure that my experiment is fair by "“ "¢ Having the water at the same temperature at the start of the experiment. "¢ Don't not insulate them "¢ Do the same to each one. E.g. if I stir one I will have to stir the others. Range and extent "¢ 20 minutes timing "¢ Beakers "“ 100ml, 200ml, 300,ml Variables There will be 100ml in one beaker, 200ml in another and 300ml in another. This will not affect my investigation because these variables are significant to my experiment. Method Firstly I will set up my equipment data logger, water and computer. The beakers will be set out with the right amount in each one and the temperature readers will be placed in each beaker. And then I will press start on the computer and the data loggers will record my results on a graph and also on a table so I can spot exactly where any anomalies occur. This will be done for 20 minutes and when it is finished reading I will print out the required information that I need. Conclusion From doing this experiment have found out that the smaller the animal the quicker that it looses its heat and the larger the animal the slower it looses its heat. My prediction was proved correct. The water temperatures started near enough at the same temperature and they have all dropped in Celsius but the 100ml dropped even more. Here I how they dropped: Start Finish Range 100ml 89.4 52.5 39.9 200ml 91.2 59.2 32 300ml 91.2 61.9 29.9 In the table you can clearly see how my prediction was correct however I would have expected more of a difference between the 200ml and the 300ml. The bigger the animal the slower it looses its heat and the smaller the animal the quicker it looses its heat. Animals in colder conditions are mostly large because the condition they live in is cold. If you look at the graph you can see how my results dropped. Evaluation I don't think that my experiment was very successful even though my prediction was correct. I don't think it was very successful because of the actual experiment. Here are my reasons: "¢ The 100ml beaker didn't start off at the same temperature of the other two. "¢ There are two anomalies in my results where the temperature had dropped dramatically I think that this has happened because maybe the temperature reader came out of the water for a few seconds. That is the only explanation that it could possibly be. Apart from the anomalies, my results do support my conclusion. If I was to repeat this investigation then I would stick the data loggers down to the beaker so that they do not come out of the water. I would also repeat the experiment 3 times to be more accurate and I would also include a 400ml and a 50ml amount in the experiment. By doing this I would have found out more information and made my results more accurate. If I had more time then I would research the sizes of animals in different climates. This will have made me surer about my conclusion.   

I plan to see if the body size of an animal affects its heat loss. I am going to do this by using three glass beakers with various amounts of boiling water in them. I will then see which beaker looses its heat the fastest or the slowest....

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