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# Investigating resistance when altering thickness of wire / length of wire

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Introduction

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. ...read more.

Middle

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 1 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) ...read more.

Conclusion

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 0.05 20 0.10 10 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 ...read more.

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