EXPERIMENTAL PROCEDURES In order to record my observations, I will use the following types of observation tables, and I will display the manner in which my wire will be setup, in order that I will be able to experiment with them. 1. LENGTH For length, we have to make sure that only the length is changed, and that all the other factors are kept as a constant, i. e. the thickness, the material, and the temperature. Thickness = 0. 3 mm Material = nichrome Temperature = room temperature The index is adjusted, to vary the resistance, 10 alternative readings of current and voltage are taken, at uniform intervals.

For every 0. 2 volts, I will be measuring the current, for each wire, and I will be observing, and recording the readings on the ammeter, in a table like this. SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (? ) 1 0. 00 Xxx Xxx Xxx Xxx 80 X 11 2. 00 X TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. The shorter the wire, the lesser the resistance there will be. ? = 1/ gradient ? 20 cm wire has the greatest gradient, so less resistance.

The resistance on should tally with my table readings; otherwise, it will mean that there is an error somewhere. 2. Thickness For thickness, we have to make sure that only the thickness is changed, and that all the other factors are kept as a constant, i. e. the length, the material, and the temperature. Length = 50 cm Material = nichrome Temperature = room temperature SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (? ).

11 2. 00 X TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. The thicker the wire, the lesser the resistance there will be. ? = 1/ gradient ? 3 mm wire has the greatest gradient, so it has the least resistance. 3. Material For material, we have to make sure that only the material is changed, and that all the other factors are kept as a constant, i. e. the length, the material, and the temperature. Length = 50 cm Thickness = 0. 4 mm Temperature = room temperature SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (?)

TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. Different conductors have different resistances, thus, the copper wire has the greatest gradient, and so it has the least resistance. In order to increase the reliability of my resulting readings, I am going to record the readings while increasing and decreasing the voltage supplied. I will also make use of series and parallel circuits, to verify the law of resistance.

To investigate the law of resistance for length. I will use the following type of board for this. The resistance for the 25 cm wire is shown by : The resistance for the 50 cm wire is shown by : This the type of graph I would be expecting to get. As you can see, the line for the 20+30 cm graph falls just a little short of the 50 cm. R = R1+ R2 In addition, to verify the law of resistivity for thickness, we use parallel circuit, which are connected in this manner: Here we will test to see if the resistance of two . 4 mm wires connected in a parallel, is equal to the resistance of a .56 mm wire.

This should get me a graph like the one that follows: In order to plot this type of graph, I will have to record my results in a table like this: The resistance for the 0. 4 mm + 0. 4 mm wire is shown by : The resistance for the 0. 56 mm wire is shown by : I did a prior test, or an introductory pre experiment test, to get me used to how to know to work the rheostat, and connect the circuit, and the results I got, are on the next page.

Analyzing evidence As you can see from my graphs, which are more or less like the graphs, I had expected to get, in my planning,In order to show that when the length of the wire was changed, the resistance changed proportionately, I created this bar graph. Thus as you can see, when the lengths in crease, the resistance of the wire increases, as there are more collisions between the electron, (which is moving from the negative end to the positive), and between the atom. When length is doubled, resistance doubles. Therefore length is directly proportional to resistance. In addition, I compared the resistance obtained from the tables, when I changed the thickness of the wire, and this is the resulting pie chart.

Here too, it is plain to see that when the thickness doubles, the resistance is halved. This is due to, when the thickness increases, there is more space for the electron to pass through, without colliding, and thus resistance decreases. Thus resistance is inversely proportional to resistance. Where as in my series and parallel graphs, the gradient achieved for both the graphs is almost the same, thus I state that the resistance of a longer wire, is the same as two shorter wires connected together in a series circuit. In addition, the resistance of a thicker wire is the same as that of two thinner wires connected in a parallel.

For every 0. 2 volts, I will be measuring the current, for each wire, and I will be observing, and recording the readings on the ammeter, in a table like this. SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (? ) 1 0. 00 Xxx Xxx Xxx Xxx 80 X 11 2. 00 X TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. The shorter the wire, the lesser the resistance there will be. ? = 1/ gradient ? 20 cm wire has the greatest gradient, so less resistance.

The resistance on should tally with my table readings; otherwise, it will mean that there is an error somewhere. 2. Thickness For thickness, we have to make sure that only the thickness is changed, and that all the other factors are kept as a constant, i. e. the length, the material, and the temperature. Length = 50 cm Material = nichrome Temperature = room temperature SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (? ).

11 2. 00 X TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. The thicker the wire, the lesser the resistance there will be. ? = 1/ gradient ? 3 mm wire has the greatest gradient, so it has the least resistance. 3. Material For material, we have to make sure that only the material is changed, and that all the other factors are kept as a constant, i. e. the length, the material, and the temperature. Length = 50 cm Thickness = 0. 4 mm Temperature = room temperature SR. NO VOLTAGE CURRENT AVERAGE R= V / I IN VOLTS(V) INCREASING DECREASING CURRENT (?)

TOTAL AVERAGE RESISTANCE = Xxxx My expected graphs will look like this. Different conductors have different resistances, thus, the copper wire has the greatest gradient, and so it has the least resistance. In order to increase the reliability of my resulting readings, I am going to record the readings while increasing and decreasing the voltage supplied. I will also make use of series and parallel circuits, to verify the law of resistance.

To investigate the law of resistance for length. I will use the following type of board for this. The resistance for the 25 cm wire is shown by : The resistance for the 50 cm wire is shown by : This the type of graph I would be expecting to get. As you can see, the line for the 20+30 cm graph falls just a little short of the 50 cm. R = R1+ R2 In addition, to verify the law of resistivity for thickness, we use parallel circuit, which are connected in this manner: Here we will test to see if the resistance of two . 4 mm wires connected in a parallel, is equal to the resistance of a .56 mm wire.

This should get me a graph like the one that follows: In order to plot this type of graph, I will have to record my results in a table like this: The resistance for the 0. 4 mm + 0. 4 mm wire is shown by : The resistance for the 0. 56 mm wire is shown by : I did a prior test, or an introductory pre experiment test, to get me used to how to know to work the rheostat, and connect the circuit, and the results I got, are on the next page.

Analyzing evidence As you can see from my graphs, which are more or less like the graphs, I had expected to get, in my planning,In order to show that when the length of the wire was changed, the resistance changed proportionately, I created this bar graph. Thus as you can see, when the lengths in crease, the resistance of the wire increases, as there are more collisions between the electron, (which is moving from the negative end to the positive), and between the atom. When length is doubled, resistance doubles. Therefore length is directly proportional to resistance. In addition, I compared the resistance obtained from the tables, when I changed the thickness of the wire, and this is the resulting pie chart.

Here too, it is plain to see that when the thickness doubles, the resistance is halved. This is due to, when the thickness increases, there is more space for the electron to pass through, without colliding, and thus resistance decreases. Thus resistance is inversely proportional to resistance. Where as in my series and parallel graphs, the gradient achieved for both the graphs is almost the same, thus I state that the resistance of a longer wire, is the same as two shorter wires connected together in a series circuit. In addition, the resistance of a thicker wire is the same as that of two thinner wires connected in a parallel.