Example research essay topic: Inversely Proportional Cross Sectional - 1,082 words

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Aim I am going to be studying the resistance of wire. The purpose of this investigation is to see how length and thickness of wire affect the dependent variable, resistance. Prediction I predict that, as the length of the wire doubles, the resistance will also double, but as the cross-sectional area of the wire doubles, the resistance halves. This means that the length will affect the resistance more than the thickness will. Hypothesis Resistance is caused by electrons bumping into ions. If the length of the wire is doubled, the electrons bump into twice as many ions so there will be twice as much resistance.

So If the cross-sectional area of the wire doubles there will be twice as many ions and twice as many electrons bumping into them, but also twice as many electrons getting through twice as many gaps. If there are twice as many electrons getting through, as there is twice the current, the resistance must have halved. This means that I am assuming that the temperature is kept constant and that the material is kept constant. We can include this in our equations by adding a constant Method Equipment needed: 1 x Power Pack (to give varied voltage) 1 x Voltmeter 1 x Ammeter 5 x wires (with crocodile clips) wire of varied length and thickness Controlled variables: Temperature (room temperature) Wire material Dependent variable: Resistance Independent variables: Thickness of wire Length of wire Circuit diagram First, set up the experiment as shown above.

Turn on the power and set the power pack so that the voltmeter reads 0. 1 volts. Take the reading from the ammeter recording both the current and the voltage. Then do exactly the same again but use voltages of 0. 2 volts, 0. 3 volts, 0. 4 volts and 0. 5 volts. This is so that when we work out the resistance (V/I) we will have five readings and can then take an average resistance. Then carry out the whole thing again, varying the length of the wire in intervals of 10 cm from 10 cm to 100 cm.

To do the thickness experiment, set up the equipment again as shown. Turn on the power and set the power pack to read 0. 2 volts. Take the current reading then turn off the power and start again. Take four readings like this so that an average resistance can be found. Next, change the thickness of the wire and do the experiment again.

Use the diameters 0. 71 mm, 0. 56 mm, 0. 28 mm and 0. 20 mm. Although the diameters havent the same interval between them, once we have worked out the resistance, we can draw a graph to discover any relationship between the thickness and the resistance of wire. The equation for resistance = V/I Results SWG (thickness / mm ) Voltage / volts Current / amps V/I = R/ohms Average R/ohms 22 (0. 71) 0. 2 0. 29 0. 28 0. 29 0. 28 0. 69 0. 71 0. 69 0. 71 0. 70 24 (0. 56) 0. 2 0. 23 0. 24 0. 24 0. 24 0. 87 0. 83 0. 83 0. 83 0. 84 32 (0. 28) 0. 2 0. 09 0. 08 0. 08 0. 08 2. 22 2. 543 36 (0. 20) 0. 2 0. 04 0. 04 0. 04 0. 05 5 5 4 4. 75 Thickness investigation (Length kept constant at 15 cms) Graph 1 relationship between the wires thickness and its resistance Length / cm Voltage / volts Current / amps V/I = R/ohms Average R/ohms 100 0. 1 0. 3 0. 5 0. 04 0. 08 0. 13 0. 17 0. 22 2. 50 2. 50 2. 31 2. 35 2. 27 2. 386 90 0. 1 0. 3 0. 5 0. 04 0. 09 0. 14 0. 19 0. 24 2. 50 2. 22 2. 14 2. 11 2. 08 2. 21 80 0. 1 0. 3 0. 5 0. 05 0. 10 0. 15 0. 21 0. 26 2. 00 2. 00 2. 00 1. 90 1. 92 1. 964 70 0. 1 0. 3 0. 5 0. 05 0. 11 0. 17 0. 23 0. 29 2. 00 1. 82 1. 76 1. 74 1. 72 1. 808 60 0. 1 0. 3 0. 5 0. 06 0. 13 0. 20 0. 26 0. 33 1. 67 1. 54 1. 50 1. 54 1. 52 1. 554 50 0. 1 0. 3 0. 5 0. 07 0. 15 0. 22 0. 30 0. 38 1. 43 1. 33 1. 36 1. 33 1. 32 1. 354 40 0. 1 0. 3 0. 5 0. 09 0. 18 0. 26 0. 36 0. 44 1. 11 1. 11 1. 15 1. 11 1. 14 1. 124 30 0. 1 0. 3 0. 21 0. 32 0. 43 0. 54 1. 00 0. 95 0. 94 0. 93 0. 93 0. 95 20 0. 1 0. 3 0. 5 0. 15 0. 26 0. 40 0. 54 0. 67 0. 67 0. 77 0. 75 0. 74 0. 75 0. 736 10 0. 1 0. 3 0. 5 0. 17 0. 34 0. 50 0. 68 0. 85 0. 59 0. 59 0. 60 0. 59 0. 59 0. 592 Length investigation (Thickness kept constant at SWG 24) Graph 2 relationship between the wires length and its resistance After doing the two graphs I have decided to do a graph of 1 /thickness 2, to see if thickness is inversely proportional to resistance. Thickness / mm 1 /Thickness 2 Average resistance / ohms 0. 71 1. 98 0. 7 0. 56 3. 19 0. 84 0. 28 12. 76 2. 43 0. 20 25 4. 75 Investigating thickness 2 and resistance Graph 3 relationship between the wires thickness 2 and its resistance Conclusion I conclude that, as the length of a wire doubles, the resistance also doubles (provided the thickness of the wire is kept constant).

I also conclude that, as the cross-sectional area of the wire doubles, the resistance halves (provided the length of the wire stays constant). I can conclude this because my graph shows that resistance is inversely proportional to 1 / (thickness 2). So The theory behind these conclusions are: As the length doubles the resistance doubles. Resistance is caused by electrons bumping into ions. If the length of the wire doubles, the electrons bump into the ions twice as much so the resistance will double.

Research essay sample on Inversely Proportional Cross Sectional