1. 1 AC Electricity

2. Time variation of a DC voltage or current 2 I V Current Voltage time t

3. Sinusoidal Waveform 3 voltage or current Sinusoidal Wave - + 0 T t The relationship between frequency, f, in hertz and period, T, in seconds, is given by the expression: f = 1/T Mathematically, we can represent the sinusoid as a function of time using the equation is the peak value of the waveform and f is its frequency in Hz. This is simply the sine function of an angle  measured in radians and where the angle varies with time in accordance with the angular frequency  = 2  f, ie  is measured in radians per second

4. 4

5. 5 a b c At point a the magnetic flux linking the coil is maximum, the rate of change of magnetic flux is zero thus the induced voltage is zero. At point c the magnetic flux linking the coil is zero, the rate of change of magnetic flux being cut is maximum thus the induced voltage is maximum At point b the magnetic flux linking the coil is NØ = N A B cos θ, θ = ωt = 2πft = 45 o, thus the induced voltage v = -NdØ/dt = (ωNAB) sin θ, θ = ωt = 2πft = 45 o. where N = the number of turns on the coil, A is its cross-sectional area, and B is the magnetic flux density between the poles of the magnet.

6. 6 0

7. 7 45 0

8. 8 90 0

9. 9 135 0

10. 10 180 0

11. 11 225 0

12. 12 270 0

13. 13 315 0

14. 14 360 0 Or 0 0

15. 15 0 360 0

16. 16 Definitions 1 Waveform 2 Instantaneous value 3 Peak value 4 Peak to peak (p/p) value 5 Periodic waveform 6 Period (T) 7 Cycle 8 Frequency (f)

17. 17 Important parameters for a sinusoidal voltage.

18. 18 Defining the cycle and period of a sinusoidal waveform.

19. 19 Demonstrating the effect of a changing frequency on the period of a sinusoidal waveform. The unit of measurement for frequency is the hertz (Hz) where 1 hertz (Hz) = 1 cycle per second (c/s)

20. 20 Determine 1 Period 2 frequency 3 Amplitude

21. 21 FIGURE 13.18 Basic sinusoidal function.

22. 22

23. 23

24. Power dissipation in a DC circuit 24 I (I 2 R) Power Current t t (a) Current variation through a resistance R in a DC circuit (b) Power dissipation in a resistance R in a DC circuit

25. Power dissipation in an AC circuit 25 i 2 (t) I rms 2 Average of squared current = 1/2 I p 2 I p 2 Time, t Current i (t) Power dissipation is proportional to voltage (or current) squared)

26. 26 0 -2 +2 +4 -4 Imagine the sinewave shown below A = 2.sin α is squared.

27. 27 The waveform shown in red below is obtained Note that it is all positive And twice the frequency of the blue waveform 0 -2 +2 +4 -4

28. 28 0 -2 +2 +4 -4 Peak Value = 2 Peak Squared Value = 4 Mean Square Value = 2 Root Mean Square Value

29. 29 The root mean square value of an ac voltage or current is the equivalent d.c.voltage or current that will produce the same power (heating effect). ROOT MEAN SQUARE (r.m.s.) value

30. 30 AC Voltage & Current In a Resistor For a purely resistive element (i.e. one that does not contain any capacitive or inductive elements) the voltage across and the current through the element are in phase, with the ratio of voltage across to the current through giving the resistance. i.e. the same as for a d.c. circuit

31. 31

32. For the circuit shown, calculate: (a)The rms current supplied by the generator. (b)The rms voltage across the 2.2 kΩ resistor. (c)The average power dissipated in the 6.8kΩ resistor. V 18 V rms 1 kHz R 1 = 6.8 k  R 2 = 2.2 k  Solution (a)The total circuit resistance, R T = 6.8 + 2.2 = 9k  The rms current, I = v/R = 18/(9 × 10 3 ) = 2mA (b)Let v 2 = rms voltage across R 2 = 2.2k  = 27.2mW = 2 × 10 -3 ×  2.2 × 10 3 = 4.4V (c)Let P= average power dissipation in R 1 = 6.8k  = (2 × 10 -3 ) 2 × 6.8 × 10 3 = 27.2 × 10 -3