1. AC POWER ANALYSIS Tunku Muhammad Nizar Bin Tunku Mansur
Pegawai Latihan Vokasional
Pusat Pengajian Kejuruteraan Sistem Elektrik

2. Content Average Power Maximum Average Power Transfer Complex Power
Power Factor Correction

3. AVERAGE POWER

4. Average Power
Average Power, in watts (W), is the average of instantaneous power over one period

5. Average Power Resistive load (R) absorbs power all the time.
For a purely resistive circuit, the voltage and the current are in phase (v = i).

6. Average Power Reactive load (L or C) absorbs zero average power.
For a purely reactive circuit, the voltage and the current are out of phase by 90o (v - i = ±90).

7. Exercise 11.3 Find the average power supplied by the source and the average power absorb by the resistor

8. Solution The current I is given by
The average power supplied by the voltage source is

9. Solution The current through the resistor is
The voltage across resistor is
The average power absorbed by the resistor is
Notice that the average power supplied by the voltage source is same as the power absorbed by the resistor.
This result shows the capacitor absorbed zero average power.

10. Practice Problem 11.3 Calculate the average power absorbed by the resistor and the inductor. Then find the average power supplied by the voltage source

11. MAXIMUM AVERAGE POWER TRANSFER

12. Maximum Power Transfer
For maximum power transfer, the load impedance ZL must equal to the complex conjugate of the Thevenin impedance Zth

13. Maximum Average Power The current through the load is
The Maximum Average Power delivered to the load is

14. Maximum Average Power
By setting RL = Rth and XL = -Xth , the maximum average power is
In a situation in which the load is purely real, the load resistance must equal to the magnitude of the Thevenin impedance.

15. Exercise 11.5 Determine the load impedance ZL that maximize the power drawn and the maximum average power.

16. Solution First we obtain the Thevenin equivalent
To find Zth, consider circuit (a)
To find Vth, consider circuit (b)

17. Solution
From the result obtained, the load impedance draws the maximum power from the circuit when
The maximum average power is

18. Practice Problem 11.5 Determine the load impedance ZL that absorbs the maximum average power. Calculate the maximum average power.

19. Example 11.6 Find the value of RL that will absorbs maximum average power. Then calculate that power.

20. Solution First we obtain the Thevenin equivalent Find Zth Find Vth
By using voltage divider

21. Solution The value of RL that will absorb the maximum average power is
The current through the load is
The maximum average power is

22. Practice Problem 11.6 Find the value of RL that will absorbs maximize average power, Then calculate the power.

23. COMPLEX POWER

24. Complex Power Apparent Power, S (VA) Real Power, P (Watts)
Reactive Power, Q (VAR)
Power Factor, cos 

25. Complex Power
Complex power is the product of the rms voltage phasor and the complex conjugate of the rms current phasor.
Measured in volt-amperes or VA
As a complex quantity
Its real part is real power, P
Its imaginary part is reactive power, Q

26. Complex Power (Derivation)

27. Complex Power (Derivation)

28. Complex Power (Derivation)
From derivation, we notice that the real power is or
and also the reactive power or

29. Real or Average Power
The real power is the average power delivered to a load.
Measured in watts (W)
The only useful power
The actual power dissipated by the load

30. Reactive Power
The reactive power, Q is the imaginary parts of complex power.
The unit of Q is volt-ampere reactive (VAR).
It represents a lossless interchange between the load and the source
Q = 0 for resistive load (unity pf)
Q < 0 for capacitive load (leading pf)
Q > 0 for inductive load (lagging pf)

31. Apparent Power
The apparent power is the product of rms values of voltage and current
Measured in volt-amperes or VA
Magnitude of the complex power

32. Power Factor
Power factor is the cosine of the phase difference between voltage and current.
It is also cosine of the angle of the load impedance.

33. Power Factor The range of pf is between zero and unity.
For a purely resistive load, the voltage and current are in phase so that v- i = 0 and pf = 1, the apparent power is equal to average power.
For a purely reactive load, v- i = 90 and pf = 0, the average power is zero.

34. Power Triangular
Comparison between the power triangular (a) and the impedance triangular (b).

35. Problem 11.46
For the following voltage and current phasors, calculate the complex power, apparent power, real power and reactive power. Specify whether the pf is leading or lagging.
V = 22030o Vrms, I = 0.560o Arms.
V = 250-10o Vrms, I = 6.2-25o Arms.
V = 1200o Vrms, I = 2.4-15o Arms.
V = 16045o Vrms, I = 8.590o Arms.

36. Problem 11.48 Determine the complex power for the following cases:
P = 269 W, Q = 150 VAR (capacitive)
Q = 2000 VAR, pf = 0.9 (leading)
S = 600 VA, Q = 450 VAR (inductive)
Vrms = 220 V, P = 1 kW, |Z| = 40  (inductive)

37. Problem 11.42
A 110Vrms, 60Hz source is applied to a load impedance Z. The apparent power entering the load is 120VA at a power factor of lagging. Calculate
The complex power
The rms current supplied to the load.
Determine Z
Assuming that Z = R + j L, find the value of R and L.

38. Problem 11.83
Oscilloscope measurement indicate that the voltage across a load and the current through is are 21060o V and 825o A respectively. Determine
The real power
The apparent power
The reactive power
The power factor

39. POWER FACTOR CORRECTION

40. Power Factor Correction
The process of increasing the power factor without altering the voltage or current to the original load.
It may be viewed as the addition of a reactive element (usually capacitor) in parallel with the load in order to make the power factor closer to unity.

41. Power Factor Correction
Normally, most loads are inductive. Thus power factor is improved or corrected by installing a capacitor in parallel with the load.
In circuit analysis, an inductive load is modeled as a series combination of an inductor and a resistor.

42. Implementation of Power Factor Correction

43. Calculation If the original inductive load has apparent power S1, then
P = S1 cos 1 and Q1 = S1 sin 1 = P tan 1
If we desired to increased the power factor from cos1 to cos2 without altering the real power, then the new reactive power is
Q2 = P tan 2
The reduction in the reactive power is caused by the shunt capacitor is given by
QC = Q1 – Q2 = P (tan 1 - tan 2)

44. Calculation
The value of the required shunt capacitance is determined by the formula
Notice that the real power, P dissipated by the load is not affected by the power factor correction because the average power due to the capacitor is zero

45. Example 11.15
When connected to a 120V (rms), 60Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

46. Solution If the pf = 0.8 then, cos1 = 0.8  1 = 36.87o
where 1 is the phase difference between the voltage and current.
We obtained the apparent power from the real power and the pf as shown below.
The reactive power is

47. Solution When the pf raised to 0.95, cos2 = 0.95  2 = 18.19o
The real power P has not changed. But the apparent power has changed. The new value is
The new reactive power is

48. Solution
The difference between the new and the old reactive power is due to the parallel addition of the capacitor to the load.
The reactive power due to the capacitor is
The value of capacitance added is

49. Practice Problem 11.15
Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume the load is supplied by a 110V (rms) 60Hz power line.

50. Problem 11.82
A 240Vrms, 60Hz source supplies a parallel combination of a 5 kW heater and a 30 kVA induction motor whose power factor is Determine
The system apparent power
The system reactive power
The kVA rating of a capacitor required to adjust the system power factor to 0.9 lagging
The value of capacitance required