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Outline Plan

In my investigation I will be doing two experiments. The first one will be to test the resistance of a wire when the lengths are altered making sure the area of the wire is kept constant. The second one will be to test the resistance of a wire when the area (diameter) is altered making sure the length of the wire is kept constant. I will have two methods, direct and indirect. The first one (altering the length) I will be using the direct method, where I will be using a multimeter.

The second experiment (altering the diameter) I will be using the indirect method, where I will be using an ammeter and voltmeter.

Aim

-Experiment one – To test the resistance of a wire, (nicrome) measured in SWG/mm, to find out if resistance changes as we change the length of the wire.

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-Experiment two – To test the resistance of a wire (nicrome) to see if the resistance changes as we change the area of the wire.

Resistance – to test the resistance of wire – Apparatus:

Ohms law:

V=IR

Voltage = current X resistance

R=V/I (resistance = voltage/current)

Ammeter always wired in series

Voltmeter always wired in parallel.

This is the indirect method. I will be using this to measure the resistance in the different diameters of the wire.

The direct method will be using a multimeter to find the resistance in the different lengths of the wire.

Prediction

Experiment one -As the length of the wire increases the resistance also increases.

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This is because you are increasing the number of particles. Therefore if you double the length of wire, you double the amount of particles, which is doubling the number of collisions, therefore the resistance is double.

Metallic Bonding.

As current travels through a wire it gets really hot. This is because atoms vibrate more, meaning more collisions, causing the electrons to slow down.

Keep current down because the wire gets hot. Try to keep it low.

Experiment two – As the diameter of the wire increases the resistance decreases. This is because there is more space for the electrons to flow. Therefore if you double the diameter of the wire, you double the amount of space for the electrons to flow through, which is halving the number of collisions, therefore halving the resistance

A1 = TTr

A2 = TT(2r )= 4TTr =4A

Variables

Independent variables – the length and the diameter of the wire.

Dependant variable – Resistance

Control

-All equipment the same

-Voltage the same

-No. of cells the same

-Thickness/type of wire the same

-Temp of room

-Length of time before taking reading

-Zero Error

Safety

-Exposed conductors

-Electricity and Wet hands

-Potentially Hot Conductors.

Trial Investigation

To prove that my prediction is correct and that the experiment works, I will be doing a trial experiment.

Detailed Plan

Experiment one – Set up apparatus as shown. In this experiment, I will be varying the lengths of the wire from 5cm to 55cm going up in 10cm steps making sure I keep the area of the wire at a constant. I will repeat this three times to make sure I get accurate results. I will then record the results, and average the three recording of the resistance out and draw up a graph of my results. I have to take into account the zero error. For this I experiment I will be using a multimeter.

Experiment two – Set up apparatus as shown. In this experiment, I will be varying the diameter of 5 wires making sure I keep the length constant. I will be using an ammeter and voltmeter and using Ohm’s Law. I will again repeat this three times making sure I get accurate results. I will measure the voltage off the voltmeter and the current off the Ammeter. I will then calculate the resistance using R=V/I. I will calculate the average resistance. I will record the results in a table and create a graph of average Resistance over average area

Resistivity

Resistance length

} Resistance Length/area

Resistance area

Resistance = constant x length/area

Resistivity (p) = a property of the material.

R=pl/a …therefore… p=r x a/l (ohm meter)

Calculate p and compare to Published resistivity (1.01 x 10 m)

O – Observing

* At least three repeats for each.

LENGTH (with a constant area of 0.00108 cm )

Length of wire (cm)

Resistance

1 (ohms)

Resistance 2 (ohms)

Resistance 3 (ohms)

Average Resistance (ohms)

5.0 cm

00.7

00.7

00.7

00.7

15.0cm

01.7

01.6

01.6

01.6

25.0cm

02.7

02.7

02.7

02.7

35.0cm

03.8

03.6

03.6

03.7

45.0cm

04.6

04.6

04.6

04.6

55.0cm

05.7

05.7

05.6

05.7

AREA (with constant length of 20cm)

Diameter

(cm)

V1

(volts)

V2 (volts)

V3

(volts)

C1 (amps)

C2 (amps)

C3 (amps)

0.027

0.23

0.37

0.88

0.05

0.09

0.21

0.031

0.18

0.34

0.80

0.06

0.11

0.27

0.037

0.12

0.22

0.68

0.06

0.11

0.34

0.045

0.09

0.15

0.58

0.06

0.10

0.40

0.055

0.06

0.11

0.46

0.06

0.11

0.48

*V=Voltage

*C=Current

Calculate resistance…

Diameter (cm)

Voltage (volts)

Current

(Amps)

Resistance

(Ohms)

0.027

0.23

0.05

4.6

0.027

0.37

0.09

4.1

0.027

0.88

0.21

4.2 =4.3*

0.031

0.18

0.06

3.0

0.031

0.34

0.0.11

3.1

0.031

0.80

0.27

3.0 =3.0*

0.037

0.12

0.06

2.0

0.037

0.22

0.11

1.8

0.037

0.68

0.34

2.0=1.9*

0.045

0.09

0.06

1.5

0.045

0.15

0.10

1.5

0.045

0.58

0.40

1.5 = 1.5*

0.55

0.06

0.06

1

0.55

0.11

0.11

1

0.55

0.46

0.48

1 =1*

* “… = 4.3” -Average of the three resistances

Calculate by using R=V/I

Diameter (cm)

Area (cm )

Area (to the power of minus 4, cm )

Resistance (ohms)

1/ area

(1/cm )

0.027

0.000573

5.73 x 10

4.3

1747

0.031

0.000756

7.56 x 10

3.0

1325

0.037

0.00108

10.8 x 10

1.9

930

0.045

0.00159

15.9 x 10

1.5

629

0.055

0.00238

23.8 x 10

1

421

A – Analysis

Experiment one – Graph one

From my graph you can see that as you increase the length, you also increase the resistance

For example: Length – 5.0cm, Resistance – 0.5 ohms

If you double the length…

Length – 10.0 cm

…The resistance also doubles…

Resistance – 1.0 ohms

Therefore my prediction was correct. This is because you are increasing the number of particles in the wire. So if you double the length of the wire, you double the amount of particles, which is doubling the number of collisions, therefore doubling the resistance.

From this I can conclude Length is proportional to Resistance.

Experiment Two – Graph Two

From my graph you can see that as you increase the length, the resistance decreases

For example: Area – 10.5 x 10 cm , Resistance – 2 ohms

If you double the area…

Length – 21 x 10 cm

…The resistance also doubles…

Resistance – 1.0 ohms

Therefore my prediction was correct. This is because there is more space for the electrons to flow. So if you double the area of the wire, you double the amount for the current to flow, which is halving the number of collisions, therefore halving the resistance.

Graph Three – Resistance over 1/area

As the area increases, the resistance goes down, but not at a steady pace. The graph is curved and in a result is not proportional.

I then plotted a graph of resistance over the inverse of the area to make it proportional. It turned out a straight line therefore proving resistance is inversely proportional to the area.

During my experiment, I did not need to repeat any anomalous results because they were between an acceptable margin of error.

E – Evaluation

My experiment went not to badly yet I did struggle with the resistivity. My results came out fairly accurately when I did repeats and came to look at my graphs and final figures.

When averaging out my final results to compare them to the published resistivity, I found one of my final results were very far out. This may hav been because it was the smallest length therefore the current may have been high causing the wire to heat up. When the wire heats up the particles vibrates more. Next time, I will not take readings at 5.0 cm and I will keep the voltage low. I left ths result out when averaging it out

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Outline Plan. (2020, Jun 02). Retrieved from http://studymoose.com/outline-plan-new-essay

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