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ALTERNATING VOLTAGE AND CURRENT

ALTERNATING VOLTAGE AND CURRENT. v. i,v. v. V m. i. 0. t. t. v. t. i. I m. t. 0. -I m. Example of instantaneous value of i or v in electrical circuits. Direct current. Unipolar binary waveform. bipolar binary waveform. Saw wave. y. 1. 360. 270. 0. x (  ). 90. 270.

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ALTERNATING VOLTAGE AND CURRENT

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  1. ALTERNATING VOLTAGE AND CURRENT

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

  3. y 1 360 270 0 x () 90 270 -1 Sinusoidal waveform 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 v(t) = Vmsint or v(t) = Vmcost i(t) = Imsint atau i(t) = Imkost where  = 2f and f is a frequency in Hertz (Hz ), Vm and Im are the maximum amplitude of voltage and current respectively

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

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

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

  7. v (V) 15 0.4 0 t (ms) 0.2 -15 Example 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. Vm = 15 V T =1/f= 1/2500= 0.4 ms  Thus the diagram as follows

  8. v (V) 156 1.25 0 t (ms) 2.5 0.625 -156 1.875 Example 3 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) = Vmcos(t) = 156 cos( 800  t ) V We have Vm = 156;  = 2f = 800 f = 400 and thus T = 1/f = 1/400 = 2.5 ms

  9. a Ym 180 270 x () 0 360 90 -Ym Waveform which is not begin at t=0 In this case, the waveform is given by y = Ymsin(x + a) x =angle ao= phase difference refer to sine wave begins at t=0 For current and voltage, the equations are given by i(t) = Imsin(t + ) v(t) = Vmsin(t + ) q=phase difference

  10. i (mA) 54 70 57 0.125 0 t (ms) 0.25 -70 Example 4 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) = Imsin(t + ) = 70sin(8000t + 0.943 rad) Im = 70;  = 2f = 8000; f = 4000 Hz = 4 kHz; T = 1/f = 1/4000 = 0.25 ms;  = 0.943 rad = 54o

  11. Example 5 Obtain the equation of the following waveform q 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 Vm = 339V Equation for voltage: v(t) = Vmsin(t + ) = 339sin(100t + 36)

  12. v, i v Vm  i Im t 0 T -Im -Vm Lagging and leading of phase v(t) = Vmcost; i(t) = Imcos(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. i (mA) i2 i1 t (s) 50 18.1 25 Example 6 Following is a sinusoidal waveform for current i1(t) and i2(t). Obtain the equation for those current. From the waveform: Im1 = 60 mA; Im2 = 80 mA T = 50 sf = 1/(50 x 10-6) = 20 kHz i1(t) = 60sin(4 x 104t)  = 25 – 18.1 = 6.9 s 6.9 s  6.9 x 360/50 = 50 i2(t) = 80sin(4 x 104 t + 50)

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

  15. r.m.s value Power P = I2R (i.e. P  I2) i(t) = Imsint i2(t) = Im2sin2t = ½Im2(1 - cos2t) The area under the Im2 is A. Equate this to the rectangular of the same area A=h2 x 2p/w

  16. Area under the Im2 Height of the rectangular r.m.s value

  17. Example 7 An alternating current is given by an equation i(t)=0.4sin 100pt A; flowing into a resistor R=384 W for 48 hours. Calculate the energy in kWh consumed by the resistor. Im = 0.4 I = 0.707 x 0.4 = 0.283 A P = I2R = 0.2832 x 384 = 30.7 W W = Pt = 30.7 x 48 = 1.474 kWh

  18. Example 8 A sinusoidal voltage as in figure is applied to a resistor 56 W. Calculate the power absorbed by the resistor Vm = 339 V V = 0.707 x 339 = 240 V Power absorbed P = V2/R = 2402/56 = 1029 W

  19. Example 9 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 = I2R or I = (P/R) = (3400/17) = 14.14 A Im = I/0.707 = 14.14/0.707 = 20 A  = 2f = 2 x 50 = 100 i(t) = Imsint = 20 sin(100t)

  20. Example 10 • 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 W to current in one direction and , due to the rectifier, an infinity resistance to current in the reverse direction . Calculate : • The readings on the ammeters; • The form and peak factors of the current wave Initiallythe moving coil-ammeter will read the Iav 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 Thus equivalent Irms read by the meter is

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

  23. Em q Phasor representation e = Em sin t = Em sinq

  24. B A e1 = Em sin t e2 = Em sin (t + ) Here the magnitudes are same but the phases are different

  25. e = Em sin t i = Im sin t Here the phases are same but the magnitudes are different

  26. j Im1 Im2 j i1 = Im1 sin t i2 = Im2 sin (t + ) i2 is leading the i1 by j ori1is lagging the i2 by j

  27. y CY A C AY q B BY j x BX AX CX O ADDITION AND SUBSTRACTION ADDING Vertical components Ay = OA sin  By = OB sin  Cy = Ay + By = OA sin  + OB sin  Horizontal components Ax = OA cos  Bx = OB cos  Cx = Ax + Bx = OA cos  + OB cos  Resultant = OC = (Cy2 + Cx2)

  28. y Ay Cy q -Bx j x Ax Cx -By A C B O -B SUBSTRACTING Vertical components Ay = OA sin  -By = -OB sin  Cy = Ay - By = OA sin  - OB sin  Horizontal components Ax = OA cos  -Bx = -OB cos  Cx = Ax - Bx = OA cos  - OB cos  Resultant = OC = (Cy2 + Cx2)

  29. v1 v2 v Example 11 • Given v1 = 180 sin 314t volt ;andv2 = 120 sin (314t + /3) volt. • Find • The supply voltage v in trigonometry form; • r.m.s voltage of supply • Supply frequency

  30. Vm2 Vm B /3 Vm2 sin /3  O Vm1 A Vm2 cos /3 OA = Vm1 + Vm2 cos /3 = 180 + 120 x 0.5 = 240 OB = Vm2 sin /3 = 120 x 0.866 = 104 Vm = ((OA)2 + (OB)2 ) = (2402 + 1042) = 262 = tan-1 (OB/OA) = tan-1(104/240) = 0.41 rad v = 262 sin (t + 0.41) volt

  31. v v2 v1 Graph showing the three components • Rms value. = V = 0.707 Vm = 0.707 x 262 = 185 V • (c) Frequency = f = 314/2= 50 Hz

  32. Example 12 Find graphically or otherwise the resultants of the following voltages e1 = 25 sin t, e2 = 30 sin (t + /6), e3 = 30 kos t, e4 = 20 sin (t - /4) Express in the same form Solution e1 = 25 sin t[ V ] Em1 = 25 volt e2 = 30 sin (t + /6)[ V ] Em2 = 30 volt e3 = 30 cos t[ V ] Em3 = 30 volt e4 = 20 sin (t - /4)[ V ] Em4 = 20 volt

  33. Em Ey Em3 Em2 Em2sin(/6) Ex Em1 Em4sin(/4) Em2cos(/6) Em4 Em4cos(/4) Horizontal components: Ex = Em1 + Em2cos(/6) + Em4cos(-/4) = 25 + (30 x 0.866) + (20 x 0.707) = 65.1 Vertical components: Ey = Em3 + Em2sin(/6) + Em4sin(-/4) = 30 + (30 x 0.5) + (20 x -0.707) = 30.9

  34. Peak value for e: Em = (Ex2 + Ey2)½ = (65.12+30.92) ½ = 72 [V] Phase angle for e: • = tan-1(Ey/Ex) = tan-1(30.9/65.1) = 25= 5/36 • e = e1 + e2 + e3 + e4 = 72 sin(t + 5/36)

  35. Example 13 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: e1 = 339cos100t, e2 = 339cos(100t + 120), e3 = 339cos(100t + 240)

  36. e1, e2, e3 (V) e1 e2 e3 t (ms) 5 ms (90) 10 ms (180) 15 ms (270) 20 ms (360)

  37. E3 = 240 V 120 E1 = 240 V 120 E2 = 240 V e=0 for all instants

  38. Example 14 • The instantaneous values of two alternating voltages are represented respectively by v1=60 sinq volts and v2=40 sin(q-p/3) volts. Derive an expression for the instantaneous values of • The sum • The difference of these voltages. First we consider q =0 or t=0 as reference in order to simplified the phasor diagram. Thus v1 will be in the x-axis and v2 will be –p/3 or -60o behind (lagging) v1. Magnitude for v1 is 60V and v2 is 40V.

  39. Vm1 Vm2 Vm (a) OA=60V ; OB= 40V Horizontal components OA+OD=60 + 40 cos 60o = 60 + 20 = 80V Vertical components OY= - 40 sin 60o= -34.64V Resultant OC= and Equation for the voltage V= 87.2 sin (q -23.4o) V

  40. -Vm2 Vm Vm1 Vm2 (b) OA=60V ; OB= 40V Horizontal components OA-OE=60 - 40 cos 60o=OD = 60 - 20 = 40V Vertical components OY= 40 sin 60o= 34.64V Resultant OC= and Equation for the voltage V= 52.9 sin (q +40.9o) V

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