Plot Resistance Versus Cross Section example essay topic

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 oh msr = resistivity in ohm- = 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 x 10 Wm.

Different materials have different; 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 0.25 mm. We used the following formula to work out the cross sectional area: Cross Sectional Area = 'r^2 r = the radius of the wire. The radius is half the diameter so it should therefore be 0.125 mm. 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 (mm^2) 10.0520. 1030.1540. 2050.25 ApparatusPowerpackConstantan wire Crocodile clipsLeadsTest Rig We set up the apparatus as shown belowFairtest: 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 15 cm 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: 1. 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 (mm^2) Experiment 1 Experiment 2 Experiment 3 Volt-age (V) Cur -rent (I) Resist-a nce (W) Volt-age (V) Cur -rent (I) Resist-a nce (W) Volt-age (V) Cur -rent (I) Resist-a nce (W) 0.053. 000.704. 293.000. 704.293. 000.704.

290.103. 001.502. 003.001. 601.883.

001.402. 140.153. 002.201. 363.002.

101.433. 002.101. 430.203. 002.901. 033.002. 901.033.

002.801. 070.253. 003.500. 863.003.

600.833. 003.500. 86 Cross Sectional Area (mm^2) Experiment 1 Experiment 2 Experiment 3 Average Resistance (W) Resistance (W) Resistance (W) Resistance (W) 0.054. 294.294. 294.290. 102.001.

882.142. 010.151. 361.431. 431.410. 201.031. 031.071.

040.250. 860.830. 860.85 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 1/Cross Sectional Area 0.05200. 10100.156. 670.2050. 254 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..