Single Phase System

1. Single Phase System

2. i v R Pure Resistive Circuit in Series Instantaneous voltage is given by v = Vm sin t Instantaneous current is given by The maximum value for current Im and maximum value for voltage Vm can be related as Im = Vm/R The rms value for current Irms (simply I) and rms value for voltage Vrms (simply V) can be related as I = V/R

3. + Vm Im current time voltage - V I In circuit contains resistor , the V and I are in phase as in diagram below waveform phasor

4. Power dissipated in the resistor p = i2R = (Im2/R)sin2t p = v2/R = (Vm2/R) sin2t p = vi = VmIm sin2t Average value for power

5. power p current i voltage v Wave in pure resistance circuit

6. i v Pure Inductive Circuit in Series If the current is i = Im sin t; Thus Vm= w L Im The voltage is leading the current by p/2 rad (90o) OR current is lagging behind the voltage by p/2

7. v, i Vm v i Im t t /2 Voltage and current waveform in a purely inductive circuit Maximum voltage: Vm = LIm Voltage r.m.s value .: V = LI V/I = Vm/Im = L = XL XL is measured in ohm and called as inductivereactance=wL where  = 2f

8. V 90 I 90 E I, XL XL I f Changes of I and XL with frequency Phasor for purely inductive circuit

9. Power dissipated in purely inductive circuit P = vi = (Im sin t)(Vm sin (t + /2) = VmIm sin t sin (t + /2) = VmIm sin t [sin t cos /2 + cos t sin /2 ] = VmIm sin t cos t = ½VmIm sin 2t Average power Half cycle has cancelled the other half cycle that is why the average power is zero.

10. v, i, p p v i + + t - - CURRENT, VOLTAGE AND AVERAGE POWER WAVEFORM IN A PURELY INDUCTIVE CIRCUIT.

11. i v Pure capacitive Circuit in Series If the voltage is given as v = Vm sin t Then Therefore Im= w C Vm In this case the current is leading a voltage by p/2 ( 90o) OR voltage is lagging behind the current by p/2.

12. v, i Vm v i Im t /2 AC VOLTAGE AND CURRENT WAVEFORM IN PURELY CAPACITIVE CIRCUIT Maximum current value Im = CVm r.m.s value .: I = CV ratio V/I = Vm/Im = 1/C = XC XC = is measured in ohm and called as capacitive reactance =1/wC

13. I /2 V I, Xc I Xc f Changes of I and XC with frequency Phasor diagram

14. Power dissipated by capacitor p = vi = (Vm sin t)(Im sin (t + /2)) = VmIm sin t sin (t + /2) = VmIm sin t [sin t cos /2 + cos t sin /2 ] = VmIm sin t cos t = ½VmIm sin 2t Average power Half cycle has cancelled the other half cycle that is why the average power is zero.

15. voltan v arus i Kuasa p CURRENT, VOLTAGE AND POWER IN PURELY CAPACITIVE CIRCUIT

16. i vR v vL Resistor and inductor in series If i = Im sin t Then vR = iR = ImR sin tin phase with i vL = iXL = ImXLsin (t + /2) = LIm sin (t + /2) leading the i by /2 v = vR + vL = ImR sin t + LIm sin (t + /2) = ImR sin t + LIm cos t v = Vm sin (t + ) (1) Where (2) and = tan-1(L/R) (3)

17. Current and Voltage waveform in L-R serial circuit v, i v i vR t vL VR in phase with I VL is leading in phase with I by p/2

18. VL V /2  VR I Phasor diagram for I, VR, VL & V in R & L serial circuit • I and VR overlap to represent the in phase • VL is vertical to represent the 90o leading out of phase • Resultant between VL and VR give the value of V and The representation of voltage can be written as V / q

19. XL Z /2  I R impedance triangle Phasor diagram for I, R, XL & Z in R & L serial circuit Impedance is represented as

20. Power p = vi = Vm sin (t + ) Im sin t = ½VmIm [cos - cos (2t - )] = ½VmIm cos - ½VmIm cos (2t - ) component ½VmIm cos (2t - ) is zero Therefore the average value is only given by

21. V,I P p i t v i v R, XL, Z () Z XL R f (Hz) CURRENT, VOLTAGE AND POWER WAVEFORMS FOR L-R SERIAL CIRCUIT

22. q VR = V cos  V I cos  I I sin  Phasor diagram We can also calculate the power from P = I2R or P = VR2/R or VRI All in r.m.s values I= Irms , VR=VRrms From phasor diagram P=VRI=VIcosq (active power) Cos q is a power factor Reactive power (VAR) = VLI = VI sin  cos  = VR/V = R/Z.

23. Example 1 A resistance of 7.0W is connected in series with a pure inductance of 31.8mH and the circuit is connected to a 100V, 50Hz, sinusoidal supply. Calculate (a) the circuit current. (b) the phase angle

24. Example 2 A pure inductance of 318mH is connected in series with a pure resistance of 75W. The circuit is supplied from a 50Hz sinusoidal source and the voltage across the 75W resistor is found to be 150V. Calculate the supply voltage check

25. Example 3 A coil, having both resistance and inductance, has a total effective impedance of 50W and the phase angle of the current through it with respect to the voltage across it is 45o lag. The coil is connected in series with a 40W resistor across a sinusoidal supply. The circuit current is 3A; by constructing a phasor diagram, estimate the supply voltage and the circuit phase angle

26. VLr cosfLr

27. Example 4 • A coil having a resistance of 12 W and a inductance of 0.1H is connected across a 100V, 50Hz supply. Calculate: • The reactance and the impedance of the coil; • The current; • The phase difference between the current • and the applied voltage:

28. B A D C Example 5 • A serial ac circuit ,ABCD, contains a resistor AB, an inductor BC with resistance R and inductance L and a resistor CD. When a current 6.5A flow in the circuit, voltage drops across those points are: VAB = 65 V; VBC = 124 V; VAC = 149 V. The supply voltage is 220 V at 50 Hz.Calculate • Draw the circuit • Voltage drop VBC and its phase compared to the current • Resistance and inductance of the inductor Current= I = 6.5 A Supply voltage = V = 220 V Z = V/I = 220/6.5 = 33.8 (1) VAB = 65 V; R1 = 65/6.5 = 10 (2) VBC = 124 V; ZL = 124/6.5 = 19.1 (3) VAC = 149 V; ZAC = 149/6.5 = 22.9 (4)

29. From Equation (3):- ZL2 = RL2 + XL2 19.12= 364.8 =RL2 + XL2 (5) From Equations (2) and (4):- ZAC2 = (R1 + RL)2 + XL2 22.92=524.4= (10 + RL)2 + XL2 (6) From Equations (5) and (6):- RL = 3.0 (7) Substitute RL in (5) XL = 18.9 (8) Dari (1) Z2 = (R1 + RL + R2)2 + XL2 33.82=(10+3.0 + R2)2 + 18.92  R2 = 15 

30. VR1 = 6.5 x 10 = 65 V VR2 = 6.5 x 15 = 97.5 V VRL = 6.5 x 3.1 = 20.2 V VXL = 6.5 x 18.8 = 122.2 V (b) (b) Resistance of inductor = RL = 3.0  XL = 18.9 = 2fL L = 18.9/2f = 18.9/2 x 50 = 60.1 mH

31. i vR vC Resistor and capacitor in series If i = Imsint vR = iR = ImR sin t --- in phase with i v = vR + vC = (Im/C) sin (t - /2) + ImR sin t v = Vm sin (t - ) Vc is lagging by 90o refer to I where and ; ;

32. /2 / 3/2 2/ vR v vC t i CURRENT AND VOLTAGE WAVEFORMS IN R-C SERIAL CIRCUIT

33. I R - I /2 VR - Z /2 XC V VC Phasor diagram Impedance triangle  = tan-1 (VC/VR)  = tan-1 (XC/R) Can be written as Z- = 2f where

34. p t i v POWER p = vi= Vm sin (t - ) Im sin t = ½VmIm [cos - cos (2t - )] = ½VmIm cos  - ½VmIm cos (2t - )] The average of component [½VmIm cos (2t - )] is zero, therefore CURRENT, VOLTAGE AND POWER WAVEFORMS FOR R-L SERIAL CIRCUIT

35. q Z R XC I sin  I I cos  V Phasor diagram We can also calculate the power from P = I2R or P = VR2/R or VRI All in r.m.s values I= Irms , VR=VRrms From phasor diagram P=VRI=VIcosq (active power) Cos q is a power factor Reactive power (VAR) = VCI = VI sin  Cos  = VR/V = R/Z. f

36. Example 6 • A capacitor of 8.0mF takes a current of 1.0 a when the alternating voltage applied across it is 230 V. Calculate: • The frequency of the applied voltage; • The resistance to be connected in series with the capacitor to reduce the current in the circuit to 0.5A at the same frequency; • The phase angle of the resultants circuit. (a)

37. (b) (c) Leading by 30o

38. Example 7 • A metal-filament lamp, rated at 750W, 100V, is to be connected in series with a capacitor across a 230V, 60Hz supply. Calculate: • The capacitance required • The phase angle between the current and the supply voltage

39. (b)