Electromagnetism Topic 12 2 Alternating Current Rotating Coils

Electromagnetism Topic 12. 2 Alternating Current

Rotating Coils Most of our electricity comes from huge generators in power stations. l There are smaller generators in cars and on some bicycles. l These generators, or dynamos, all use electromagnetic induction. l When turned, they induce an EMF (voltage) which can make a current flow. l

l l l 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.

AC Generator

l l l 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.

The graph shows how the current varies through one cycle (rotation). l It is a maximum when the coil is horizontal and cutting field lines at the fastest rate. l It is zero when the coil is vertical and cutting no field lines. l

AC Generator Output

The Sinusoidal Shape l As the emf can be calculated from l ε = - N Δ (Φ/ Δt) and Φ = AB cos θ It can be clearly seen that the shape of the curve must be sinusoidal. l l

The following all increase the maximum EMF (and the current): l increasing the number of turns on the coil l increasing the area of the coil l using a stronger magnet l rotating the coil faster. l (rotating the coil faster increases the frequency too!) l

Alternating Current The graph shows the values of V and I plotted against time l Can you see that the graphs for both V and I are sine curves? l They both vary sinusoidally with time. l Can you see that the p. d. and the current rise and fall together? l We say that V and I are in phase. l

l The time period T of an alternating p. d. or current is the time for one complete cycle. This is shown on the graph l The frequency f of an alternating pd or current is the number of cycles in one second. l The peak values V 0 and I 0 of the alternating p. d. and current are also shown on the graph

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? l We could use the peak value, but this occurs only for a moment. l What about the average value? l This is zero over a complete cycle and so is not very helpful! l

In fact, we use the root‑mean‑square (r. m. s. ) value. l This is also called the effective value. l The r. m. s. value is chosen, because it is the value which is equivalent to a steady direct current. l

You can investigate this using the apparatus in the diagram l Place two identical lamps side by side. l Connect one lamp to a battery; the other to an a. c. supply. l The p. d. across each lamp must be displayed on the screen of a double‑beam oscilloscope. l

Adjust the a. c. supply, so that both lamps are equally bright l The graph shows a typical trace from the oscilloscope We can use it to compare the voltage across each lamp. l

l l 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.

l In fact:

Why √ 2 Why The power dissipated in a lamp varies as the p. d. across it, and the current passing through it, alternate. l Remember power, P = current, (I) x p. d. , (V) l l If we multiply the values of I and V at any instant, we get the power at that moment in time, as the graph shows

The power varies between I 0 V 0 and zero. l Therefore average power = I 0 V 0 / 2 l l Or P = (I 0 / √ 2) x (V 0 / √ 2) l Or P = Irms x Vrms

Root Mean Square Voltage

Root Mean Square Current

Calculations Use the rms values in the normal equations} l Vrms = Irms R l P = Irms Vrms l P = Irms 2 R l P = Vrms 2 / R l

Transformers A transformer changes the value of an alternating voltage. l It consists of two coils, wound around a soft‑iron core, as shown l

l In this transformer, when an input p. d. of 2 V is applied to the primary coil, the output pd. of the secondary coil is 8 V

How does the transformer work? l l 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.

An Ideal Transformer This is 100% efficient l Therefore the power in the primary is equal to the power in the secondary l Pp = Ps l i. e. Ip Vp = Is Vs l

Step-up Step-down A step‑up transformer increases the a. c. voltage, because the secondary coil has more turns than the primary coil. l In a step‑down transformer, the voltage is reduced and the secondary coil has fewer turns than the primary coil. l

The Equation

Note: l • In the transformer equations, the voltages and currents that you use must all be peak values or all r. m. s. values. l Do not mix the two. l Strictly, the equations apply only to an ideal transformer, which is 100 % efficient. l