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# resistance on a wire

Extracts from this document...

Introduction

-  -

Background Knowledge:

An electrical current is a movement of charged particles. In metals, these particles are electrons, they carry a negative charge.

Metals contain lots of electrons which are free to move through the metal carrying electrical energy.

This is why metals are good electrical conductors.

Resistance of a metal to electrical current is due to the conducting electrons colliding with the particles of the metal. The more frequent the collisions the greater the resistance.

In a simple circuit where there are thick connecting wires there is low resistance and more current flows easier. All of this is because the electrons have much more space as they move so there are fewer collisions between them and the metal particles so the resistance is low and more current can pass for a given voltage.

In a simple circuit with thin connecting wires there is high resistance and less current flows easily. This is because there is

Middle

R=V/I(ohms)

Results:

I repeated the experiment 3 times and here are my results.

To work out the resistance I divided the voltage by the current.

I also took the average resistance.

 Length (cm) I (A) V1 (V) R1 (ohms)

Conclusion

Evaluation:

Because of the similarity between all my experiment repeats I think I can make a firm conclusion about the effect of length on resistance.

My method worked well because my results came out right and the way I had predicted they would.

To improve my experiment I would have more thoroughly checked the wire for kinks as kinks in the wire affect the length.

Possible errors.

There are some errors which could have occurred during my experiment.

Length

Error=+/-1mm

Total error=2mm

% error= Total Area____ x 100= 2_ x100

Average length             500

=4%

Voltage

Error in voltage=+/-0.005

Total Error in current = 0.01

Average Voltage measured =2V

% Error in voltage=0.01x100=0.5%

2

Current

Error in current = +/-0.0005

Total Error in Current = 0.01

Average current = 1.OA

%Error in current = 0.01 x 100

1

=1%

Although there are no faults or errors in my trend line, meaning that these errors are probably not present in my data, they must still be accounted for when I am analyzing results.  In my repeat I could have divided by half a percent because there is a point like 0.5% above the line

Clodagh Mc Henry 12T,

Physics Coursework,

Dr. J. McKiernan

This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.

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