This will consequently increase the resistance of the wire. However if the wire was shorter, the free electrons will not collide with the atoms as much as it did when it was longer. So now I can conclude that resistance is directly proportional to length and that if the length of the wire doubles its resistance doubles. If I draw a graph representing current against voltage for different lengths of wire, then it will look like this: The shortest wire of these will be the steepest in the graph; this is because it has the least resistance and therefore supplied the least voltage.

The longest wire will have the greatest resistance and will be represented by the lowest line on the graph. Material of the wire: Here I use two different materials for the wires and calculate their resistance. I have chosen copper and nichrome. I will use the same procedure as mentioned before to calculate the resistance and I will set up the apparatus as first mentioned while keeping the thickness and the length of the wires same. If I were to draw a graph representing current against voltage, for this experiment it will look sort of like this: Thickness of the wire: Resistance: Resistance and length:

If I plot a graph to show how resistance varies with length, it will show me something similar to this predicted graph: The graph passes through the origin, which concludes that , however much I increase the length, the resistance will increase by the same amount. Resistance decreases if the cross-section area is increased. An example can help to get a clearer picture; a narrow wire has fewer paths existing for the electrons to move through. While a larger wire has many more paths they could take. This makes conduction easier. It can be shown that the relationship between the cross-section area, A and resistance is R ? 1/A OR R ?

K/A Where k is a constant that depends on the length and type of material. If I plot a graph to show how resistance varies with area, this will be shown: Unlike the graph for length, the line doesnt pass through the origin. However: R ? 1/A means that a graph of R against 1/A will show direct proportion. If I plot a graph to show how resistance varies with 1/Area, I will get this: Free electron conducting in metal Conductors: Conductors ( e. g. copper, aluminum ) are those substances which easily allow the passage of electric current through them. It is because there are a large number of free electrons available in a conductor.

In terms of energy band, the valence and conduction bands overlap each other as shown below. Due to this overlapping, a slight potential difference across a conductor causes the free electrons to form electric current. Thus the electrical behaviour of conductors can be effectively explained by the band energy theory of materials. Prior test In order to rely on my results, I take the readings of the current for the increasing and the decreasing currents; giving a prior test. Material: Length: Thickness: Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Experiment This first trial is to test the accuracy and the realism of the experiment itself. It also shows us that as the temperature has an effect on resistance. I will use a 100cm long strip of Nichrome wire and attach it to the circuit and the current will be raised and recordings will be taken at different levels. 1. Attach 100cm Nichrome wire. 2. Turn on the power supply and raise the current. 3. Take reading from the voltmeter. 4. Continue raising the power recording voltmeter readings. This above procedure will require the following equipment given below:

1. 100cm Nichrome wire 2. Ammeter 0 to 200 mA 3. Voltmeter 0 to 20 volts 4. Rheostat 5. Crocodile clips 6. Battery 7. Switch with key 8. Connecting wires By adjusting the rheostat the voltage are increasing in steps of 0. 15 V to 0. 50 V. Each time noting down the corresponding ammeter readings. I do this to make sure the readings of the ammeter while increasing and decreasing the voltage are the same almost with a slight variation, therefore making sure that no heating has taken place. I will then note down the ammeter readings while decreasing the voltage in steps of 0. 50 V to 0.

15 V, and Ill take the average current readings as this will improve the reliability of my experiment. I will use a table similarly to the one drawn and record my readings and calculate the resistance. Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V).

Sly. No Average = 5. 70 And now I will draw a graph representing current against voltage, and then I will take the line of best fit from which I will take the gradient and check whether the results I have obtained graphically matches the results in the table. Obtaining Evidence In this part of my task, I will show all my graphs and results that I have obtained after carrying out the experiments. I have done the same process as I have planned earlier. My experiment will be based on these: Length (same thickness and different lengths) Thickness (same length and different thickness).

Resistors in series Resistors in parallel Different Lengths Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Material: Nichrome Wire Length: 75cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA).

Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Average = 6. 49 Material: Nichrome Wire Length: 50cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Material: Nichrome Wire Length: 25cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Resistance = 1/440 = 0. 00227 x 1000 = 2. 27 Different Thickness Material: Nichrome Wire Length: 100cm Thickness: 0. 90mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage)Voltage (V) Sly. No Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Average = 8. 40 Material: Nichrome Wire Length: 100cm Thickness: 0. 56mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Average = 4. 86 Material: Nichrome Wire Length: 100cm Thickness: 0. 32mm mA Resistance (? )Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. NAverage = 8. 73 Resistance from graph Thickness:

93 Material: Nichrome Wire Length: 100cm and 50cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. Average = 12. 79 Resistance from graph 100cm and 75cm Gradient = y/x = 25. 15/0. 40 =62. 87 Resistance = 1/62. 87 = 0. 01590 x 1000 = 15. 90 100cm and 50cm.

Gradient = y/x = 23. 55/0. 30 = 78. 5 Resistance = 1/78. 5 = 0. 012738 x 100 = 12. 73 Parallel Material: Nichrome Wire Length: 100cm and 75cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. Average = 3. 69 Material: Nichrome Wire Length: 100cm and 50cm Thickness: 0. 45mm mA.

Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Resistance Analyzing Evidence.

Here in this part of my experiment I will prove that my hypothesis and my obtained results obey the ohms law. This is where the results from my graphs and the results I have already obtained earlier will be compared; talking about its proportionality. Variation in length: When the length of the wire boosts, the amount of atoms present in it also boosts. The variation of these atoms block the passage of electrons passing through the wire. A slower flow of electrons will therefore lead to less current passing through the wire hence the longer the wire, the longer the electrons have to travel, so they come across more collision.

From this statements I predict that the resistance increases with the length of the increasing wire. Hypothesis: It is expected that the resistance should increase in proportion to the length. The resistance should be considerably higher for the 100cm length than it is for the 50cm length. Theoretically the resistance for the 100cm length should be 2 times that of the 50cm length.

The reason for this was explained earlier. Resistance will increase with length. Resistance is proportional to length. In this table below I will show the results I have obtained from the graph and the table: ? R/L = constant Average R (graph)R (table) Length of wire (cm).

From the above table I have concluded that resistance increases with length and as the length doubles, the resistance doubles about with it. The column R/L is roughly constant. This supports my hypothesis that resistance of a wire is directly proportional to its length. The even increase of resistance with length can be explained by the clashes that take place in a wire as current flows through it. When the current flows through a wire, the free electrons collide with the atoms of the wire.

The longer the wire the more collisions occur. And this will result in an increase in the resistance. However, the shorter the wire, less the collision, hence less resistance. I will show the relationship between length of a wire and its resistance on a graph, from the values of the above table. And this will confirm my conclusion that length is directly proportional to its resistance. Results The resistance is clearly increasing as the length of the wire increases. And when the length of the wire doubles, its resistance also doubles. The results shown in graph is exactly what is anticipated to happen as stated in the hypothesis.

From the table above we can see that as the length doubles, the resistance also approximately doubles. In the last column R/L if found to be constant somewhat, therefore making it obey the ohms law R ? L The predicted graph drawn between R and L looks like this:

Variation of thickness This experiment is needed to confirm that the resistance of a wire is inversely proportional to its diameter. If the cross-section of a wire is enlarged this means that the area on which the electrons move will be enlarged. Thus suggesting that there will be no clouds of electrons and atoms. The current can travel easily with nothing increasing the resistance.

Likewise if the wire is narrower the obstruction will be crossed by the electrons will be lesser and therefore the current will decrease. Hypothesis It is expected that the thinnest wire will have the highest resistance because a thicker wire offers less resistance to current than a thinner one of the same material. This is because current consists of electrons flowing through the metal of the wire. The electrons hop from atom to atom in the metal in reaction to the electric field in the circuit. A conductor with a larger cross-section allows more electrons to intermingle with the fields.

The longest wire will have the greatest resistance and will be represented by the lowest line on the graph. Material of the wire: Here I use two different materials for the wires and calculate their resistance. I have chosen copper and nichrome. I will use the same procedure as mentioned before to calculate the resistance and I will set up the apparatus as first mentioned while keeping the thickness and the length of the wires same. If I were to draw a graph representing current against voltage, for this experiment it will look sort of like this: Thickness of the wire: Resistance: Resistance and length:

If I plot a graph to show how resistance varies with length, it will show me something similar to this predicted graph: The graph passes through the origin, which concludes that , however much I increase the length, the resistance will increase by the same amount. Resistance decreases if the cross-section area is increased. An example can help to get a clearer picture; a narrow wire has fewer paths existing for the electrons to move through. While a larger wire has many more paths they could take. This makes conduction easier. It can be shown that the relationship between the cross-section area, A and resistance is R ? 1/A OR R ?

K/A Where k is a constant that depends on the length and type of material. If I plot a graph to show how resistance varies with area, this will be shown: Unlike the graph for length, the line doesnt pass through the origin. However: R ? 1/A means that a graph of R against 1/A will show direct proportion. If I plot a graph to show how resistance varies with 1/Area, I will get this: Free electron conducting in metal Conductors: Conductors ( e. g. copper, aluminum ) are those substances which easily allow the passage of electric current through them. It is because there are a large number of free electrons available in a conductor.

In terms of energy band, the valence and conduction bands overlap each other as shown below. Due to this overlapping, a slight potential difference across a conductor causes the free electrons to form electric current. Thus the electrical behaviour of conductors can be effectively explained by the band energy theory of materials. Prior test In order to rely on my results, I take the readings of the current for the increasing and the decreasing currents; giving a prior test. Material: Length: Thickness: Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Experiment This first trial is to test the accuracy and the realism of the experiment itself. It also shows us that as the temperature has an effect on resistance. I will use a 100cm long strip of Nichrome wire and attach it to the circuit and the current will be raised and recordings will be taken at different levels. 1. Attach 100cm Nichrome wire. 2. Turn on the power supply and raise the current. 3. Take reading from the voltmeter. 4. Continue raising the power recording voltmeter readings. This above procedure will require the following equipment given below:

1. 100cm Nichrome wire 2. Ammeter 0 to 200 mA 3. Voltmeter 0 to 20 volts 4. Rheostat 5. Crocodile clips 6. Battery 7. Switch with key 8. Connecting wires By adjusting the rheostat the voltage are increasing in steps of 0. 15 V to 0. 50 V. Each time noting down the corresponding ammeter readings. I do this to make sure the readings of the ammeter while increasing and decreasing the voltage are the same almost with a slight variation, therefore making sure that no heating has taken place. I will then note down the ammeter readings while decreasing the voltage in steps of 0. 50 V to 0.

15 V, and Ill take the average current readings as this will improve the reliability of my experiment. I will use a table similarly to the one drawn and record my readings and calculate the resistance. Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V).

Sly. No Average = 5. 70 And now I will draw a graph representing current against voltage, and then I will take the line of best fit from which I will take the gradient and check whether the results I have obtained graphically matches the results in the table. Obtaining Evidence In this part of my task, I will show all my graphs and results that I have obtained after carrying out the experiments. I have done the same process as I have planned earlier. My experiment will be based on these: Length (same thickness and different lengths) Thickness (same length and different thickness).

Resistors in series Resistors in parallel Different Lengths Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Material: Nichrome Wire Length: 75cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA).

Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Average = 6. 49 Material: Nichrome Wire Length: 50cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Material: Nichrome Wire Length: 25cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Resistance = 1/440 = 0. 00227 x 1000 = 2. 27 Different Thickness Material: Nichrome Wire Length: 100cm Thickness: 0. 90mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage)Voltage (V) Sly. No Material: Nichrome Wire Length: 100cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Average = 8. 40 Material: Nichrome Wire Length: 100cm Thickness: 0. 56mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly.

Average = 4. 86 Material: Nichrome Wire Length: 100cm Thickness: 0. 32mm mA Resistance (? )Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. NAverage = 8. 73 Resistance from graph Thickness:

93 Material: Nichrome Wire Length: 100cm and 50cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. Average = 12. 79 Resistance from graph 100cm and 75cm Gradient = y/x = 25. 15/0. 40 =62. 87 Resistance = 1/62. 87 = 0. 01590 x 1000 = 15. 90 100cm and 50cm.

Gradient = y/x = 23. 55/0. 30 = 78. 5 Resistance = 1/78. 5 = 0. 012738 x 100 = 12. 73 Parallel Material: Nichrome Wire Length: 100cm and 75cm Thickness: 0. 45mm mA Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. Average = 3. 69 Material: Nichrome Wire Length: 100cm and 50cm Thickness: 0. 45mm mA.

Resistance (? ) Average Current (mA) Current (decreasing in voltage) Current (increasing voltage) Voltage (V) Sly. No Resistance Analyzing Evidence.

Here in this part of my experiment I will prove that my hypothesis and my obtained results obey the ohms law. This is where the results from my graphs and the results I have already obtained earlier will be compared; talking about its proportionality. Variation in length: When the length of the wire boosts, the amount of atoms present in it also boosts. The variation of these atoms block the passage of electrons passing through the wire. A slower flow of electrons will therefore lead to less current passing through the wire hence the longer the wire, the longer the electrons have to travel, so they come across more collision.

From this statements I predict that the resistance increases with the length of the increasing wire. Hypothesis: It is expected that the resistance should increase in proportion to the length. The resistance should be considerably higher for the 100cm length than it is for the 50cm length. Theoretically the resistance for the 100cm length should be 2 times that of the 50cm length.

The reason for this was explained earlier. Resistance will increase with length. Resistance is proportional to length. In this table below I will show the results I have obtained from the graph and the table: ? R/L = constant Average R (graph)R (table) Length of wire (cm).

From the above table I have concluded that resistance increases with length and as the length doubles, the resistance doubles about with it. The column R/L is roughly constant. This supports my hypothesis that resistance of a wire is directly proportional to its length. The even increase of resistance with length can be explained by the clashes that take place in a wire as current flows through it. When the current flows through a wire, the free electrons collide with the atoms of the wire.

The longer the wire the more collisions occur. And this will result in an increase in the resistance. However, the shorter the wire, less the collision, hence less resistance. I will show the relationship between length of a wire and its resistance on a graph, from the values of the above table. And this will confirm my conclusion that length is directly proportional to its resistance. Results The resistance is clearly increasing as the length of the wire increases. And when the length of the wire doubles, its resistance also doubles. The results shown in graph is exactly what is anticipated to happen as stated in the hypothesis.

From the table above we can see that as the length doubles, the resistance also approximately doubles. In the last column R/L if found to be constant somewhat, therefore making it obey the ohms law R ? L The predicted graph drawn between R and L looks like this:

Variation of thickness This experiment is needed to confirm that the resistance of a wire is inversely proportional to its diameter. If the cross-section of a wire is enlarged this means that the area on which the electrons move will be enlarged. Thus suggesting that there will be no clouds of electrons and atoms. The current can travel easily with nothing increasing the resistance.

Likewise if the wire is narrower the obstruction will be crossed by the electrons will be lesser and therefore the current will decrease. Hypothesis It is expected that the thinnest wire will have the highest resistance because a thicker wire offers less resistance to current than a thinner one of the same material. This is because current consists of electrons flowing through the metal of the wire. The electrons hop from atom to atom in the metal in reaction to the electric field in the circuit. A conductor with a larger cross-section allows more electrons to intermingle with the fields.