Good conductors of electricity Essay Example for Free

Good conductors of electricity Essay observational PROCEDURES In order to record my observations, I exit use the following cases of observation tables, and I go forth display the manner in which my telegram ordain 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 solely the other factors are kept as a constant, i. e. the thickness, the material, and the temperature. Thickness = 0. 3 mm Material = nichrome Temperature = agency temperature The index is adjusted, to vary the defense, 10 alternative readings of current and voltage are taken, at uniform intervals. For any 0. 2 volts, I will be measuring the current, for each equip, and I will be observing, and written text the readings on the ammeter, in a table like this. SR. NO VOLTAGE CURRENT medium R= V / I IN VOLTS(V) increase DECREASING CURRENT (? ) 1 0. 00 thirty Xxx Xxx Xxx 80 X 11 2. 00 X TOTAL AVERAGE RESISTANCE = Xxx x My evaluate graphical records will date like this. The shorter the wire, the lesser the oppositeness there will be. ? = 1/ gradient ? 20 cm wire has the grea screen out gradient, so less apology.The immunity on should t wholey 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 totally the other factors are kept as a constant, i. e. the length, the material, and the temperature. Length = 50 cm Material = nichrome Temperature = agency 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 assure like this. The thicker the wire, the lesser the opponent 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 al l 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 fall the voltage supplied. I will also make use of serial and parallel circuits, to verify the righteousness of resistance.To investigate the justice of resistance for length. I will use the following type of board for this. The resistance for the 25 cm wire is bear witnessn 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 confabulate, 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 spot 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 spot 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(prenominal) 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 proportionate ly, 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 surrounded by 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 toget her 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.