• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16

Design an experiment to predict and test the output from a simple AC generator.

Extracts from this document...

Introduction

Christopher Barr                        November 2004

Design an experiment to predict and test the output from a simple AC generator

Design an experiment to predict and test the output from a simple AC generator.

Planning

A simple ac generator will produce an alternating emf of magnitude V0 and frequency f.

image00.png

The AC generator is shown below,

image01.png

A manufacturer of AC generators has asked you to help design a clockwork torch of peak power 1W. The bulb is powered by a hand driven AC generator that is rotated at 2Hz. It is assumed that the light bulb has a constant resistance once in use because it doesn't have time to cool down as the current alternates. The manufacturer has provided you with a horseshoe magnet of approximate field strength 50mT and gap 20cm, a selection of commercially available reels of copper wire of different diameters and the light bulb. The rest of the material necessary to build the generator is provided by the school. The coil ABCD is square.

Using classroom apparatus you must plan experiments to;

a] Determine the unknown properties of the magnet, copper wire and bulb.

b] Test the final constructed version.

You will need to find the relevant Physics to explain,

1] The factors affecting the peak emf in an ideal generator and how these can be measured.

2] The factors affecting the peak current in an ideal generator and how these can be measured.

You will need to carry out research into;

i] Commercially available bulbs and wire to give reasonable relevant values for calculations.

ii] Methods for determining magnetic field strengths.

Design

Emf is the electromotive force and is the ‘driving’ force behind electrical circuits. EMF or E for short is sometimes referred to as the voltage of the cell or the power supply. E is measured in volts (V)

...read more.

Middle

As can be seen the combination of wires 7, 8 and 9 and bulb 4 and wires 6, 7, 8 and 9 and bulb 5 work.

However when we take into consideration that the length of the wires in some cases is a great as 18089m then the resistance from the wires is going to be so great that it must be regarded.

Internal resistance can play major factor in the designing of the circuit and generator. Internal resistance is the resistance generated inside a power supply such as a battery or generator. It is often denoted by r as opposed to R. it is depicted in circuit diagrams with a small resistor inside the power supply.

        r

        R

The resistance of the wire can be worked out using the resistivity and information already gained.

The formula for resistance involving resistivity is

r=ρl

     A

Where r is internal resistance, ρ (rho) is resistivity, l is the total length of the wire and A is the cross sectional area of the wire.

ρ for copper wire is 1.72x10-8Ωm and therefore is merely a constant as it is a property of the material. L can be changed to be specific for an AC generator. The total length involves the length of each side multiplied by the number of coils on each side multiplied by the number of sides in other words l=4lN. A can be found from the griffin electrical accessories catalogue.

So equating the equation for r

r = ρ4lN

      A

Internal resistance removes some of the total voltage to be dissipated in the external circuit. The emf is therefore the sum of the combined voltages. As voltage is the product of resistance, whether it be internal or external, and current. The equation for the total emf including the internal resistance is

E=IR + Ir

Where E is the emf, I is the current, R is the external resistance and r is the internal resistance.

IR is equal to the voltage across the bulb and therefore can be replaced by the voltage previously calculated to give 1W of power to the bulb.

E is also equal to another formula 2BlvN so equating all 3 formulas gives an expression with only 1 variable N

2BlvN = Videal + ρ4lN

                            A

Rearranging the formula to get N as the subject the formula becomes

2BlvNA = AVideal + ρ4lN

2BlvNA – ρ4lN = AVideal

N (2BlvA – ρ4l) = AVideal

                                                                           N =        AVideal    _

                    (2BlvA – ρ4l)

All the values are now either constants or things that can be worked out.

Below are the results of finding the generator width’s using a core of 17cm

Bulb

Wire

Number of turns

Coil width (m)

Total Generator width (m)

Diagonal Length

Does it work?

4

1

279

2.67E-02

0.223493246

0.225088011

NO

4

2

314

1.59E-02

0.201878435

0.202506694

NO

4

3

352

1.33E-02

0.196637385

0.197087921

YES

4

4

438

1.17E-02

0.193432465

0.193786968

YES

4

5

678

1.17E-02

0.193429880

0.193784309

YES

4

6

1151

1.36E-02

0.197139224

0.197605688

YES

4

7

-467

ERROR

ERROR

ERROR

NO

4

8

-128

ERROR

ERROR

ERROR

NO

4

9

-35

ERROR

ERROR

ERROR

NO

5

1

208

2.31E-02

0.216162476

0.21739126

NO

5

2

270

1.48E-02

0.199585949

0.200133413

NO

5

3

367

1.36E-02

0.197219314

0.197688342

YES

5

4

873

1.65E-02

0.203092708

0.203765625

YES

5

5

-875

ERROR

ERROR

ERROR

NO

5

6

-350

ERROR

ERROR

ERROR

NO

5

7

-88

ERROR

ERROR

ERROR

NO

5

8

-37

ERROR

ERROR

ERROR

NO

5

9

-12

ERROR

ERROR

ERROR

NO

The cells with error written are such because of an incompatibility with the bulb and the wire. Also the wire/bulb combinations with diagonal lengths greater than 20 cannot work either.

As is shown above there are 6 viable options for the generator combination. Bulb 4 with wires 3, 4, 5 and 6 and bulb 5 with wires 3 and 4. However some combinations use less but the wire is more expensive whereas some combinations use more wire but it is cheaper. The only true way to choose defiantly a wire and bulb combination is to work out the total cost for each generator combination.

The prices are given in the griffin electrical accessories catalogue and are shown below with the price per meter found.

Wire

Cost per 250g

meters per kilogram

cost per meter

1

£9.30

56

£0.66

2

£9.70

177

£0.22

3

£10.05

284

£0.14

4

£10.25

456

£0.09

5

£10.35

706

£0.06

6

£10.85

893

£0.05

7

£11.50

1820

£0.03

8

£12.15

3580

£0.01

9

£12.25

9195

£0.01

No using the viable generator combinations the cost can be calculated. As the product to be designed is a prototype the costs are likely to be high as commercial materials are likely to be much cheaper than the materials from the griffin catalogue.

Bulb

Wire

Meters of wire

Cost of Bulb

Cost of wire

Total cost

4

3

239.2850

0.435

£33.87

£34.3056

4

4

297.6520

0.435

£26.76

£27.1976

4

5

460.8547

0.435

£27.02

£27.4596

4

6

782.5711

0.435

£38.03

£38.4681

5

3

249.8542

0.435

£35.37

£35.8017

5

4

593.6596

0.435

£53.38

£53.8123

...read more.

Conclusion

-1. The oscilloscope should look something like

(Not to scale)image03.png

The voltage can then be measured of the oscilloscope by reading of the peak voltage. The voltage decays like this due to the component of motion perpendicular to the magnetic field. As the generator turns the velocity changes with time. When the generator is horizontal all of the velocity is of vertical component whereas as the generator continues to turn the component of velocity is split with more of the velocity in the horizontal direction, parallel to the magnetic field and less of it in the vertical direction, perpendicular to the field. As the generator becomes vertical there is no perpendicular component of velocity so no emf is induced so we get a sinusoidal wave like that shown above. As the generator rotates 180 degrees there is a total reversal in the way the current flows due to the reversal of the poles of the sides of the generator. The poles start reversing when the generator is vertical hence there is no charge as the going from positive to zero to negative or vice versa. Then as the generator rotates the send 180 degrees the poles reverse again with no charge and hence no current when the generator is vertical in the magnetic field

The peak voltage measured should be the voltage necessary to produce 1W of power in the bulb. So in 1 complete turn the voltage peaks when the generator is horizontal decreases to zero at the vertical then increases in the opposite direction to the opposite horizontal.

Again there are no major safety issues except the generator is spinning and fingers maybe trapped, also as with all of the other experiments simple precautions to do with electricity must be observed such as don’t touch bare live wires, don’t hand wires with wet hands and always switch off the power at the mains before leaving dismantling or modifying the circuit in anyway.

Page  of

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Electricity and Magnetism essays

  1. Marked by a teacher

    To investigate how the resistance, R, of a length of wire, l, changes with ...

    4 star(s)

    When the constantan wire is being tested, the length and the diameter of the wires should be constant. Prediction Taking the content on the previous pages into account, I think that the electrical resistance of a wire would be expected to be greater for a longer wire, less for a

  2. Peer reviewed

    How does the power dissipated by a light bulb vary with voltage?

    5 star(s)

    While doing the experiment, I first set the voltage to the lowest I could get to, and then to the highest. Voltage / V Current / I Resistance / ? Lowest values 0.48 0.07 6.86 Highest values 6.00 0.29 20.69 Method Validity To make the experiment fair, I will use the same light bulb every time I repeat the experiment.

  1. An Experiment to test which combinations of metals will produce the highest voltage in ...

    If the anode was zinc (-0.76) then the difference would be 1.1 - not as large a difference as the magnesium due to it being less reactive. Other examples could be zinc and magnesium, a difference of 1.61, lead (-0.13) and iron (0.04), a difference of 17 and so on.

  2. An Experiment To Find the Resistivity of a Wire

    This error may have been encountered while measuring the length of the wire. This is because it simply was not very practical to hold a piece of wire straight, whilst holding it next to a ruler and then trying to accurately fix crocodile clips to the right part on the wire.

  1. To find out what happens to the efficiency of a motor as I change ...

    CHANGE IN VERTICAL HEIGHT (m) POTENTIAL ENERGY (J) ... So in the case of an object weighing 800N undergoing a change in height from 1000m to 3000m above ground, the GPE of the object can be worked out using the formula: CHANGE IN GRAV. POT. ENERGY = WEIGHT ? CHANGE IN VERTICAL HEIGHT = 800N ? (3000m - 1000m)

  2. An experiment to find the resistivity of nichrome

    These inaccuracies are fairly large. The error bars, however, are too small to be drawn accurately on the graph. They are at most 3% inaccurate, using the same formula as before. This suggests that the inaccuracies were not experimental, but permanent errors due to problems with the measuring equipment.

  1. Resistance of a wire. Jack has been given a second hand D.C. dynamo ...

    Type Of Wire Current (I) Voltage (v) Voltage � Current Resistance (?) Nichrome 0.1 0.7 0.1�0.7 7 Copper 0.15 0.01 0.15�0.01 0.06 Constantan 0.08 0.26 0.26�0.08 3.25 After we got these results we decided constantan would be the best wire to use as it would not get hot and affect the accuracy of our results or burn

  2. Aim: I am going to carry out an investigation to find how the resistance ...

    20k 22k 43k 56k 75k 100k Resistor 2 During the actual method resistor 2 were always be kept at 47k ohms Method Again I set up the experiment in a science lab as it is the most appropriate venue for an experiment.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work