Diese Präsentation wurde erfolgreich gemeldet.

# Electromagnetic Induction 12.2

Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Nächste SlideShare
A.c. generator
×

1 von 36 Anzeige

# Electromagnetic Induction 12.2

IBO HL Topic 12

IBO HL Topic 12

Anzeige
Anzeige

## Weitere Verwandte Inhalte

Anzeige

Anzeige

### Electromagnetic Induction 12.2

1. 1. Electromagnetism Topic 12.2 Alternating Current
2. 2. Rotating Coils  Most of our electricity comes from huge generators in power stations.  There are smaller generators in cars and on some bicycles.  These generators, or dynamos, all use electromagnetic induction.  When turned, they induce an EMF (voltage) which can make a current flow.
3. 3.  The next diagram shows a simple AC generator.  It is providing the current for a small bulb.  The coil is made of insulated copper wire and is rotated by turning the shaft.  The slip rings are fixed to the coil and rotate with it.  The brushes are two contacts which rub against the slip rings and keep the coil connected to the outside part of the circuit.  They are usually made of carbon.
4. 4. AC Generator
5. 5.  When the coil is rotated, it cuts magnetic field lines, so an EMF is generated.  This makes a current flow.  As the coil rotates, each side travels upwards, downwards, upwards, downwards... and so on, through the magnetic field.  So the current flows backwards, forwards... and so on.  In other words, it is AC.
6. 6.  The graph shows how the current varies through one cycle (rotation).  It is a maximum when the coil is horizontal and cutting field lines at the fastest rate.  It is zero when the coil is vertical and cutting no field lines.
7. 7. AC Generator Output
8. 8. The Sinusoidal Shape  As the emf can be calculated from ε = - N Δ (Φ/ Δt)  and Φ = AB cos θ  It can be clearly seen that the shape of the curve must be sinusoidal.
9. 9.  The following all increase the maximum EMF (and the current):  increasing the number of turns on the coil  increasing the area of the coil  using a stronger magnet  rotating the coil faster.  (rotating the coil faster increases the frequency too!)
10. 10. Alternating Current  The graph shows the values of V and I plotted against time  Can you see that the graphs for both V and I are sine curves?  They both vary sinusoidally with time.  Can you see that the p.d. and the current rise and fall together?  We say that V and I are in phase.
11. 11.  The time period T of an alternating p.d. or current is the time for one complete cycle. This is shown on the graph  The frequency f of an alternating pd or current is the number of cycles in one second.  Thepeak values V0 and I0 of the alternating p.d. and current are also shown on the graph
12. 12. Root Mean Square Values  How do we measure the size of an alternating p.d. (or current) when its value changes from one instant to the next?  We could use the peak value, but this occurs only for a moment.  What about the average value?  This is zero over a complete cycle and so is not very helpful!
13. 13.  In fact, we use the root‑ mean‑ square (r.m.s.) value.  This is also called the effective value.  The r.m.s. value is chosen, because it is the value which is equivalent to a steady direct current.
14. 14.  You can investigate this using the apparatus in the diagram  Place two identical lamps side by side.  Connect one lamp to a battery; the other to an a.c. supply.  The p.d. across each lamp must be displayed on the screen of a double ‑beam oscilloscope.
15. 15.  Adjust the a.c. supply, so that both lamps are equally bright  The graph shows a typical trace from the oscilloscope We can use it to compare the voltage across each lamp.
16. 16.  Since both lamps are equally bright, the d.c. and a.c. supplies are transferring energy to the bulbs at the same rate.  Therefore, the d.c. voltage is equivalent to the a.c. voltage.  The d.c. voltage equals the r.m.s. value of the a.c. voltage.  Notice that the r.m.s. value is about 70% (1/ √2) of the peak value.
17. 17.  In fact:
18. 18. Why √2  Why The power dissipated in a lamp varies as the p.d. across it, and the current passing through it, alternate.  Remember power,P = current,(I) x p.d., (V)  Ifwe multiply the values of I and V at any instant, we get the power at that moment in time, as the graph shows
19. 19.  The power varies between I0V0 and zero.  Therefore average power = I0V0 / 2  Or P = (I0 / √ 2) x (V0 / √ 2)  Or P = Irms x Vrms
20. 20. Root Mean Square Voltage
21. 21. Root Mean Square Current
22. 22. Calculations  Use the rms values in the normal equations}  Vrms = Irms R P = Irms Vrms P = Irms2 R P = Vrms2 / R
23. 23. Transformers A transformer changes the value of an alternating voltage.  It consists of two coils, wound around a soft‑iron core, as shown
24. 24.  Inthis transformer, when an input p.d. of 2 V is applied to the primary coil, the output pd. of the secondary coil is 8V
25. 25.  http://www.allaboutcircuits.com/worksheets/tra  Transformer simulation
26. 26. How does the transformer work?  An alternating current flows in the primary coil.  This produces an alternating magnetic field in the soft iron core.  This means that the flux linkage of the secondary coil is constantly changing and so an alternating potential difference is induced across it.  A transformer cannot work on d.c.
27. 27. An Ideal Transformer  This is 100% efficient  Therefore the power in the primary is equal to the power in the secondary  Pp = Ps  i.e. I p V p = Is V s
28. 28. Step-up Step-down A step‑up transformer increases the a.c. voltage, because the secondary coil has more turns than the primary coil.  In a step‑down transformer, the voltage is reduced and the secondary coil has fewer turns than the primary coil.
29. 29. The Equation
30. 30.  Note: • In the transformer equations, the voltages and currents that you use must all be peak values or all r.m.s. values.  Do not mix the two.  Strictly, the equations apply only to an ideal transformer, which is 100 % efficient.