Electromagnetic Induction 12.2

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IBO HL Topic 12

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Electromagnetic Induction 12.2

  1. 1. ElectromagnetismTopic 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.

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