Essay,
Pages 6 (1347 words)

Views

10

Research

The equation for finding resistance is

R= V/I

Where V = potential difference (volts) and I = current (amps)

Current is the rate of flow of charge. An amp = 1 coulomb/second. The coulomb is the standard unit of charge. Potential difference is the amount of electrical energy transferred per unit of charge between two points. It is measured in joules per coulomb, or volts.

The opposition to the flow of charge is resistance, measured in ohms (?). The larger a materials resistance, the greater the amount of potential difference needed to make a current flow through it.

A material such as steel will have a relatively high resistance compared to that of gold. Some materials, known as superconductors have no resistance whatsoever!

When a potential difference is applied to a conductor, all the free electrons in it move in the same direction. When the electrons ‘move’ through the conductor they collide with the atoms in the material, so they are continually accelerating and decelerating.

Because of this the electrons do not have a constant velocity, so we give them an average velocity, known as the drift velocity. If the length of the conductor is increased, there will be more atoms for the electrons to collide with, therefore the electrons will have a slower drift velocity, therefore there is more resistance.

Resistivity is a quantity, which is used to compare the resistance of material r with the same direction.

R=? x L/A

Therefore: ?= RA/L

R= resistance, ?= Resistivity, L= length, A= area.

L

This is a diagram of a wire in a circuit. The number of electrons in this circuit is given as

nAL

where:

* n= number of electrons per unit volume

* A= cross-sectional area of the wire

* L= length of the wire

Therefore the total amount of charge in the circuit is given as

nALe

where

* e= charge of one electron

Preliminary Data

Before I performed my main experiment, I felt it was necessary to do some preliminary research into the most appropriate wire to use and the length of wire to use. I decided to measure the resistance of both copper and nichrome wire for a number of different lengths. I used a multimeter to do this. This is what I found:

Nichrome

5 cm

10 cm

15 cm

20 cm

25 cm

30 cm

35 cm

40 cm

Swg 22

0.4

0.6

0.8

0.9

1.2

1.5

1.8

2.0

Swg 30

1.0

1.8

2.5

3.2

4.0

4.7

5.5

6.2

Swg 36

2.4

4.6

7.2

8.1

10.4

12.0

13.6

15.5

Copper

Swg 22

0.6

0.9

1.2

1.4

1.5

1.6

2.2

2.5

Swg 30

0.3

0.4

0.7

1.0

1.2

1.4

1.6

1.9

Swg 36

0.3

0.5

0.6

0.8

0.9

1.1

1.2

1.5

V is directly proportional to I: If there is an increase in V there will be an increase in I. The best wire for this experiment is one with a relatively high resistance per unit length but also one which will not overheat with higher resistance (a wire with good temperature stability.) In my experiment, I will use nichrome wire, because it displays both of the above qualities. I will use 0.5 amps through 30cm of wire.

I will carry out my experiment as fairly as possible, but there are bound to be errors and inaccuracies. Many of these will be due to the measuring equipment. The most accurate ammeters and voltmeters available to me are accurate to ï¿½0.01Amp (or Volt). This will affect the accuracy of my readings for p.d and current and more importantly, my calculated resistance, where the associated percentage errors will be added together, increasing the overall %error. The metre rule used to measure the wire is accurate to ï¿½0.001m. The micrometer used to measure the diameter of the wire is accurate to ï¿½0.01mm. Although a hundredth of a millimetre is a very small inaccuracy, we must remember that we are using this measurement to calculate the cross-sectional area of the wire – and because this process involves a square, the percentage error will be doubled. One way to reduce the overall effect of this type of inaccuracy is to take the mean value of a number of readings and I have done this throughout my experiment.

All other physical conditions should ideally be kept constant. The effect of heating due to current has been minimised by using a relatively small current. The wire chosen (nichrome) was also selected because its physical properties would help keep my results accurate and reliable.

Apparatus

* Power Pack 0-12V DC

* Leads

* Metre ruler

* Ammeter 0-10a ï¿½0.01a

* Voltmeter 0-70V ï¿½0.01V

* Rheostat 0-15?

* Crocodile clips

* Nichrome wire Swg 22 – 36

Diagram

DC current 0 – 12V

Method

1. Wire up the circuit as shown

2. Tape a piece of wire (Swg 22) tightly to a metre ruler

3. Measure 30cm into the wire and attatch the crocodile clips

4. Switch on power pack

5. Adjust the rheostat so the ammeter reads 0.5 amps

6. Take a reading from the voltmeter

7. Move the crocodile clips to a different place on the wire, but still with a separation distance of 30cm

8. Repeat steps 5 – 7

9. Repeat step 8

10. Repeat steps 2 – 9 with Swg 24, 28, 30, 32, 34, 36.

Having collected and written down all of my data, I fed it into a spreadsheet, so I could manipulate and calculate the values easily and with minimal risk of error.

Calculation of errors

1. Length measured = 0.1m

Accuracy of measuring equipment = ï¿½0.001m

Percentage error = accuracy/recorded length x100 = 0.001/0.1 x100 = ï¿½1%

2. Current measured = 0.5A

Accuracy of measuring equipment = ï¿½0.01A

Percentage error = accuracy/recorded value x100 = 0.01/0.5 x100 =ï¿½2%

3. Voltage measured =

Accuracy of measuring equipment = ï¿½0.01V

Percentage error = accuracy/recorded value x100

= 0.01/voltage reading x100

4. Resistance measured =

%error in V + %error in I

Error bars were used when drawing the graph. I calculated these by working out the total %errors in both r and L/A and then drawing them onto the graph in an ‘I’ shape.

Analysis

Once all the data for my experiment had been collected, averages could be taken and percentage errors found. For accuracy, and also to save time, I put all the raw data into a spreadsheet, entered the correct formulas and hence calculated all the necessary values (see results table). Having done this I plotted a graph of resistance against L/A, but instead of just plotting points I drew error bars. By doing this I was able to draw a number of gradients, and therefore gain higher accuracy. Having taken the gradients, I was able to use the mathematical equations shown at the start to get a value for resistivity.

The gradient is the resistivity of the wire because:

R= ? L/A

So

?= RA/L = R/L/A = gradient of graph.

To find the gradient I divided resistance by L/A:

* Minimum gradient: 0.9636 x10-6 ?m

* Best gradient: 1.0363 x10-6 ?m

* Maximum gradient: 1.0909 x10-6 ?m

The graph I drew was a straight-line graph of resistance against L/A. L, the length was constant. Therefore

? 1

A

The greater the L/A value, the greater the resistivity.

I now had my values for the resistivity of the wire. However, my graph did feature one anomaly, which lay quite a way out. I chose to ignore it when drawing the gradient because it was obvious it was inaccurate. This was probably due to human error somewhere along the process. I found that because there were so many equations to get to the actual resistivity, there was more chance for human error.

Evaluation

When doing my experiment, I tried to make my work as accurate as possible by using the best equipment available to me. However there were still small discrepancies in my work. I took an average value for resistance by measuring three different lengths of wire within a metre of wire taped to a metre ruler. To improve the accuracy of my work, I could have chosen three separate lengths of wire, maybe from different coils, which would have given me a better average. Also, when I recorded the data, I did it in two sessions; a number of conditions that could have affected my experiment could have changed, for example the heat.

The actual value for resistivity, given by the catalogue the wire was obtained from is 1.13 x10-6?m. All my values for resistivity lay slightly below this value, but my closest value, taken from my maximum gradient was 1.0909 x10-6?m, which is only 0.0391 x10-6?m.

I believe that my experiment was of a good level of accuracy and I regard it a success.

👋 Hi! I’m your smart assistant Amy!

Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.

get help with your assignment