1. ALTERNATING VOLTAGE AND CURRENT

2. Example of instantaneous value of i or v in electrical circuits i,v t v i i t 0 -Im-Im ImIm v t v t 0 VmVm Direct current Saw wave Unipolar binary waveform bipolar binary waveform

3. x (  ) y 1 0 90 270 360 The instantaneous value varies with time following the sine or cosine waveform. This is the common waveform for alternating current (AC). Graphically can be represented by the following equations i(t) = I m sin  tataui(t) = I m kos  t v(t) = V m sin  t orv(t) = V m cos  t where  = 2  fand f is a frequency in Hertz (Hz ), V m and I m are the maximum amplitude of voltage and current respectively Sinusoidal waveform

4. t (s) v (V) VmVm -V m 0 T/4 3T/4 T/2 T V P-P 1.Above figure represents one cylcle of voltage waveform which is methamatically represented by v = V m sin  t. V m to –V m is called V P-P (peak to peak value). 2.One cycle is equivalent to one wavelength or 360 o or 2  in degree. 3.One cycle also is said to have a periodic time T (sec). 4.A number of cycle per sec is said to have a frequency f (Hz) 5.The relationship between T and f is Features of the voltage waveform or

5. i 0 ImIm -I m T/4 T/2 3T/4 T t 1.The following figure is a current waveform represented mathematically as i = I m cos  t 2.It starts maximum at t=0 which equivalent to cos (0)=1 and ends maximum at t=T or 360 o or 2 .

6. t (ms) i (mA) 170 -170 0 20 10 From the graph I m = 170 mA;T = 20 ms = 0.02 s  f = 1/T = 1/0.02 = 50 Hz  i(t) = I m sin  t = 170sin2  ft = 170sin100  t mA The following is a one cycle sinusoidal current waveform. Obtain the equation for the current in the function of time.

7. t (ms) v (V) 15 -15 0 0.4 0.2 A sinusoidal AC voltage has a frequency of 2500 Hz and a peak voltage value is 15 V. Draw a one cycle of the voltage. V m = 15 VT =1/f= 1/2500= 0.4 ms  Thus the diagram as follows

8. v (V) 0 156 -156 0.625 1.25 1.875 2.5 t (ms) A sinusoidal AC voltage is given by equation : v(t) = 156 cos( 800  t )V Draw a one cycle of voltage. From equation v(t) = V m cos(  t) = 156 cos( 800  t ) V We have V m = 156;  = 2  f = 800   f = 400 and thus T = 1/f = 1/400 = 2.5 ms

9. x (  ) 90 180270 360 0 YmYm -Y m aa Waveform which is not begin at t=0 In this case, the waveform is given by y = Y m sin(x + a  ) x =angle a o = phase difference refer to sine wave begins at t=0 For current and voltage, the equations are given by i(t) = I m sin(  t +  ) v(t) = V m sin(  t +  )  =phase difference

10. Draw one cycle of sinusoidal current wave given by the equation i(t) = 70sin(8000  t + 0.943 rad) mA From the equation i(t) = I m sin(  t +  ) = 70sin(8000  t + 0.943 rad)  I m = 70;  = 2  f = 8000  ;f = 4000 Hz = 4 kHz; T = 1/f = 1/4000 = 0.25 ms;  = 0.943 rad = 54 o i (mA) t (ms)0 70 -70 57 54  0.25 0.125

11. From waveform: T = 20 ms  f=1/T = 1/0.02 = 50 Hz  = 2  f = 100  3 ms = 3 x 360/20 = 54   = 90 – 54 = 36  V m = 339V Equation for voltage: v(t) = V m sin(  t +  ) = 339sin(100  t + 36  )  Obtain the equation of the following waveform

12. v, i t  VmVm ImIm 0 -V m -I m T v i v(t) = V m cos  t;i(t) = I m cos(  t +  ) The current i(t) is leading the voltage by  (the minimum or maximum comes first. The voltage v(t) is lagging the current by 

13. t (  s) i (mA) 50 2518.1 i2i2 i1i1 From the waveform: I m1 = 60 mA; I m2 = 80 mA T = 50  s  f = 1/(50 x 10 -6 ) = 20 kHz  i 1 (t) = 60sin(4 x 10 4  t)  = 25 – 18.1 = 6.9  s 6.9  s  6.9 x 360/50 = 50   i 2 (t) = 80sin(4 x 10 4  t + 50  ) Following is a sinusoidal waveform for current i 1 (t) and i 2 (t). Obtain the equation for those current.

14. Average value for one cycle of waveform is zero For half-wave can be calculated as follows i(t) = I m sin  t Area under the curve But

15. Power P = I 2 R (i.e. P  I 2 ) The area under the I m 2 is A. Equate this to the rectangular of the same area A=h 2 x 2  i(t) = I m sin  t  i 2 (t) = I m 2 sin 2  t= ½I m 2 (1 - cos2  t)

16. Area under the I m 2 Height of the rectangular r.m.s value

17. An alternating current is given by an equation i(t)=0.4sin 100  t A; flowing into a resistor R=384  for 48 hours. Calculate the energy in kWh consumed by the resistor. I m = 0.4  I = 0.707 x 0.4 = 0.283 A P = I 2 R = 0.283 2 x 384 = 30.7 W W = Pt = 30.7 x 48 = 1.474 kWh

18. A sinusoidal voltage as in figure is applied to a resistor 56 . Calculate the power absorbed by the resistor V m = 339 V  V = 0.707 x 339 = 240 V P = V 2 /R= 240 2 /56= 1029 W Power absorbed

19. A purely resistive resistor of 17 W dissipates 3.4kW when a sinusoidal voltage of frequency 50Hz apply across it.Give an equation for the current passing through the resistor in a function of time. P = I 2 RorI =  (P/R)=  (3400/17)= 14.14 A I m = I/0.707= 14.14/0.707= 20 A  = 2  f= 2  x 50= 100  i(t) = I m sin  t= 20 sin(100  t)

20. A moving –coil ammeter, a thermal ammeter and a rectifier are connected in series with a resistor across a 110V sinusoidal a.c. supply. The circuit has a resistance of 50  to current in one direction and, due to the rectifier, an infinity resistance to current in the reverse direction. Calculate : (1)The readings on the ammeters; (2)The form and peak factors of the current wave Initially the moving coil-ammeter will read the I av for the first half of a cycle. The second half, the value of current will be zero (due to rectifier- reverse ). For the whole cycle, it will read 1.98/2=0.99A

21. Thermal ammeter only response to the heat. This heat effect is corresponding to power dissipated in the resistor and given by the equation Full power is Since only half cycle give the heating effect, the other half is no current ( due to rectifier-reverse). Therefore the heating power will be Thus equivalent I rms read by the meter is thus

22. Form factor Peak factor The actual value for full cycle is But only half a cycle then the reading is

23. e = E m sin  t = E m sin  EmEm 

24. e 1 = E m sin  t e 2 = E m sin (  t +  ) Here the magnitudes are same but the phases are different A B

25. e = E m sin  t i = I m sin  t Here the phases are same but the magnitudes are different

26. i 1 = I m1 sin  t i 2 = I m2 sin (  t +  )  I m1 I m2  i 2 is leading the i 1 by  or  i 1 is lagging the i 2 by 

27. AC B O x y   AXAX AYAY BYBY CYCY CXCX BXBX Vertical components A y = OA sin  B y = OB sin  C y = A y + B y = OA sin  + OB sin  Horizontal components A x = OA cos  B x = OB cos  C x = A x + B x = OA cos  + OB cos  Resultant = OC =  (C y 2 + C x 2 )

28. A C B O -B   AxAx x -B y CxCx -B x AyAy CyCy y Vertical components A y = OA sin  -B y = -OB sin  C y = A y - B y = OA sin  - OB sin  Horizontal components A x = OA cos  -B x = -OB cos  C x = A x - B x = OA cos  - OB cos  Resultant = OC =  (C y 2 + C x 2 )

29. v1v1 v2v2 v Given v 1 = 180 sin 314t volt ;and v 2 = 120 sin (314t +  /3) volt. Find 1.The supply voltage v in trigonometry form; 2.r.m.s voltage of supply 3.Supply frequency

30. V m2 V m1 VmVm  /3  V m2 sin  /3 V m2 cos  /3 O A B OA = V m1 + V m2 cos  /3 = 180 + 120 x 0.5 = 240 OB = V m2 sin  /3 = 120 x 0.866 = 104 V m =  ((OA) 2 + (OB) 2 ) =  (240 2 + 104 2 ) = 262  = tan -1 (OB/OA) = tan -1 (104/240) = 0.41 rad v = 262 sin (  t + 0.41)volt

31. v2v2 v1v1 v (b)Rms value. = V = 0.707 V m = 0.707 x 262 = 185 V (c)Frequency = f = 314/2  = 50 Hz Graph showing the three components

32. Solution e 1 = 25 sin  t [ V ]E m1 = 25 volt e 2 = 30 sin (  t +  /6) [ V ]E m2 = 30 volt e 3 = 30 cos  t [ V ]E m3 = 30 volt e 4 = 20 sin (  t -  /4) [ V ]E m4 = 20 volt Find graphically or otherwise the resultants of the following voltages e 1 = 25 sin  t,e 2 = 30 sin (  t +  /6), e 3 = 30 kos  t,e 4 = 20 sin (  t -  /4) Express in the same form

33. E m1 E m2 E m3 E m4 E m2 sin(  /6) E m4 sin(  /4) E m4 cos(  /4 ) E m2 cos(  /6) EyEy ExEx EmEm Horizontal components: E x = E m1 + E m2 cos(  /6) + E m4 cos(-  /4) = 25 + (30 x 0.866) + (20 x 0.707) = 65.1 Vertical components: E y = E m3 + E m2 sin(  /6) + E m4 sin(-  /4) = 30 + (30 x 0.5) + (20 x -0.707) = 30.9

34.  = tan-1(E y /E x ) = tan -1 (30.9/65.1) = 25  = 5  /36  e = e 1 + e 2 + e 3 + e 4 = 72 sin(  t + 5  /36) Peak value for e: E m = (E x 2 + E y 2 ) ½ = (65.1 2 +30.9 2 ) ½ = 72 [V] Phase angle for e :

35. Show graphically the waveform and phasor diagrams of the resultant of the following voltages e = 339cos100  t + 339cos(100  t + 120  ) + 339cos(100  t + 240  ) What is the value of e? Say: e 1 = 339cos100  t, e 2 = 339cos(100  t + 120  ), e 3 = 339cos(100  t + 240  )

36. e 1, e 2, e 3 (V) t (ms) 20 ms (360  ) e1e1 e2e2 e3e3 15 ms (270  ) 10 ms (180  ) 5 ms (90  )

37. E 1 = 240 V E 2 = 240 V E 3 = 240 V 120  e = 0 for all instants

38. The instantaneous values of two alternating voltages are represented respectively by v 1 =60 sin  volts and v 2 =40 sin(  /3) volts. Derive an expression for the instantaneous values of (a)The sum (b)The difference of these voltages. First we consider  =0 or t=0 as reference in order to simplified the phasor diagram. Thus v 1 will be in the x-axis and v 2 will be –  /3 or -60 o behind (lagging) v 1. Magnitude for v 1 is 60V and v 2 is 40V.

39. OA=60V ; OB= 40V Horizontal components OA+OD=60 + 40 cos 60 o = 60 + 20 = 80V Vertical components OY= - 40 sin 60 o = -34.64V ResultantOC= Equation for the voltage V= 87.2 sin (  -23.4 o ) V and (a) V m1 V m2 VmVm

40. OA=60V ; OB= 40V Horizontal components OA-OE=60 - 40 cos 60 o =OD = 60 - 20 = 40V Vertical components OY= 40 sin 60 o = 34.64V ResultantOC= Equation for the voltage V= 52.9 sin (  +40.9 o ) V and (b) V m1 V m2 -V m2 VmVm