Introduction:
The equation we use to find the resistance from the current and voltage is:

Resistance (R) = Voltage (V) ÷ Current (I)

Put more simply, it is the number of volts difference across the object when one amp of current flows. You should recall that voltage is the number of joules of energy transferred by one coulomb of charge, and that current is the number of coulombs of charge passing a place each second.

The resistance of a conductor of uniform cross-section depends on three factors:
1. The length (l ) of the conductor. The longer the length, the greater the resistance will be.
2. The cross-sectional area (A). The greater the cross-sectional area, the less the resistance will be. This is similar to pumping water through a thick pipe compared with pumping it through a thin pipe: it is easier with a thick pipe.
3. The type of material the conductor is made of. Some materials are better conductors of electricity than others. This depends on the atomic structure of the material and how many free charge carriers there are to provide the current. The property of a material, which contributes to resistance, is its resistivity (ρ — the Greek letter ‘rho’). The unit of resistivity is the ohm metre (Ω
Mathematically:

For those of you who investigated the relationship between resistance and surface area, the area of each of the wires is as follows: Wire 1: 0.156 m2 Wire 2: 0.376 m2 Wire 3: 0.560 m2 You will need to draw a graph of resistance against area.
Aim: Determine, whether surface area of a wire affects current.
Hypothesis: The larger to surface area is on the wire, the more current allows flowing through.

Variables:
Independent – The independent variable will be the surface area of the wire. This was changed by keeping the length measured between negative and positive wires the same, but changing the thickness of the wire (thicknesses present above). The thickness is changed in order to directly measure the dependent variable.
Dependent – The dependent variable will be the measure of current that flows through the wire in between a constant distance (50cm). The current is measured to test accurately whether the hypothesis is correct and therefore can be supported.
Variables kept the same: * Wire length (50cm) * Voltmeter * Power Pack
Equipment:
* Retort Stand * Stopwatch * Rubber ball with string connected (at least 1m in length) * Retort Clamp
Method:
1. Collect all equipment from the given list 2. Use a lead to connect from the negative of the power pack to one end of the wire with a crocodile clip, another lead is connected from the first lead onto the wire to one end of the voltmeter 3. Use three leads to connect the ammeter to the other end of the wire with a crocodile clip, one lead from the voltmeter connecting into the lead on the wire, then another lead from the ammeter to the positive of the power pack. 4. Method from Practical: 5. Set up the circuit first using the steps above 6. Measure the distance between the two alligator clips and keep it constant for each test, we chose 50cm 7. Turn on the power pack to 10 volts and record the numbers on the ammeter and voltmeter in a table. 8. Move the alligator clips onto the second wire with a smaller cross sectional area to the first 9. Record the numbers on the ammeter and voltmeter in the table 10. Move the alligator clips onto the thinnest wire 11. Record the number on the ammeter and the voltmeter in the table 12. Repeat steps 3-7 3 times each
The hypothesis will be supported if there is a clear increase in current as the thickness/cross section of the wire increases accordingly. The thicker the wire present, the more current is able to flow through.
Results
Thickness (m2) | Amps (x10-3) | Volts | Length (m) | (Ὼ) | | 1 | 2 | 3 | Ave | 1 | 2 | 3 | Ave | | | 0.560 | 5.86 | 6.02 | 5.99 | 5.96 |