Aim: The resistance of a wire depends on certain factors. Investigate the effect of two of these factors ¡V
             Some variables that will be relevant to this investigation are:
             Of these the variables will be input and output voltages in experiment one, and length and resistance in experiment two. The other variables (temperature, material and voltage) will have to be kept constant in both experiments to make sure that only length, thickness and resistance are investigated. In experiment 1 the same bit of wire and the same thickness need to be kept constant. In experiment 2, the length will need to be kept constant to make sure only the variables indicated are investigated to ensure a fair test.
             Metals conduct electricity because the atoms in them do not hold on to their electrons very well, and so creating free electrons, carrying a negative charge to jump along the line of atoms in a wire. Resistance is caused when these electrons flowing towards the positive terminal have to 'jumps¡ atoms. So if we double the length of a wire, the number of atoms in the wire doubles, so the number of jumps double, so twice the amount of energy is required: There are twice as many jumps if the wire is twice as long.
             The thinner the wire is the less channels of electrons in the wire for current to flow, so the energy is not spread out as much, so the resistance will be higher: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve,
             This can be explained using the formula
             where there is 2X the current, and the voltage is the same, therefore R will halve. I did some research and in a book called 'Ordinary Level Physics¡ By A. F. Abbott, it says 'that doubling the area will therefore halve the resistance¡- in other words the resistance of a wire is inversely proportional to its area, or R 1/A , but we are measuri...