1. The instantaneous power
real (or average) power (Watts)
Which is the actual power absorb by the element
Examples Electric Heater , Electric Stove , oven Toasters, Iron …etc
reactive power
Which is the reactive power absorb or deliver by the element
Reactive power represents energy stored in reactive elements
(inductors and capacitors). Its unit is Volt Ampere Reactive (VAR)

2. The instantaneous power
real (or average) power (Watts)
reactive power (Volt Ampere Reactive (VAR) )

3. Complex Power
Previously, we found it convenient to introduce sinusoidal voltage and current in terms of the complex number the phasor
Definition
Let the complex power be the complex sum of real power and reactive power

4. Advantages of using complex power
- We can compute the average and reactive power from the complex power S
- complex power provide a geometric interpretation
were
The geometric relations for a right triangle mean the four power triangle dimensions
( , P, Q,  ) can be determined if any two of the four are known

5. Power Calculations
were
Is the conjugate of the current phasor
Circuit

6. This implies that in any circuit, conservation of average power and
In any circuit, conservation of complex power is achieved
This implies that in any circuit, conservation of average power and
Conservation of reactive power are achieved
However, the apparent power (the magnitude of the complex power)
is not conserved

7. Ex:6.13 Determine the average and reactive power delivered by the source.
The phasor current leaving the source is
The average power delivered by the source is:

8. The reactive power delivered by the source is:
And the complex power delivered by the source is

9. The voltage across the elements are:
Determine the average power and reactive power delivered to each element
The voltage across the elements are:
Thus the complex power delivered to each element is

10. show that conservation of complex power, average power, and reactive power is achieved.

11. 6.6.1 Power Relations for the Resistor
The voltage and current are in phase so
Average power is:
Reactive power is zero for resistor

12. 6.6.1 Power Relations for the Inductor
The voltage leads the current by 90 so that

13. 6.6.1 Power Relations for the Capacitor
The current leads the voltage by 90 so that

14. Recall the Instantaneous power p(t)
The power factor
Recall the Instantaneous power p(t)
The angle qv - qi plays a role in the computation of both average and reactive power
The angle qv - qi is referred to as the power factor angle
We now define the following :
The power factor

15. The power factor
Knowing the power factor pf does not tell you the power factor angle , because
To completely describe this angle, we use the descriptive phrases lagging power factor
and leading power factor
Lagging power factor implies that current lags voltage hence an inductive load
Leading power factor implies that current leads voltage hence a capacitive load

16. 6.6.2 Power Factor
Since

17. Circuit

18. Circuit

19. EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load
Average power

20. EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load OR

21. EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load OR

22. This could also be calculated from the complex power
delivered to the load
The power factor of the load is:
The load is lagging because the current lags the voltage

23. A typical power distribution circuit
The consumer is charged for the average power consumed by the load
The load requires a certain total apparent power

24. The powers consumed in the line losses
Ex Suppose that the load
voltage figure is 170V
The line resistance is 0.1 ohm
The load requires 10KW of average power.
Examine the line losses for a load power factor of unity and for a power factor of 0.7 lagging.
The load current is obtained from
For unity power factor this is
For power factor of 0.7
The powers consumed in the line losses
720 W extra power to be generated if pf is 0.7 to supply the load

25. Power Factor Correction
Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct the power factor to unity if power frequency is 60Hz.
(Negative because lagging)
The angle of the current is:
Thus the current into the load is:
The current through the added capacitor is:
Hence the total current
Solving this to cancel out the reactive component gives:
C=120.02/(2*3.14*60*170)= F

26. Power Factor Correction
Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct
the power factor from 0.7 to unity if power frequency is 60Hz.
From Ex 6.16
For power factor of 0.7
power factor 0.7 lagging
The current through the added capacitor is:
Hence the total current
Unity power factor
Imaginary component of the line current is zero

28. The average power delivered to the load is:

29. 6.6.3 Maximum Power Transfer
Source-load Configuration
Determine the load impedance so that maximum average power is delivered to
that load.
Represent the source and the load impedances with real and imaginary parts:
The load current is:

30. The average power delivered to the load is:
Since the reactance can be negative and to max value, we choose
leaving
Differentiate with respect to RL and set to zero to determine required RL which is RL= RS
Hence:
In this case the load is matched to the source.
The max power delivered to the load becomes:

31. 6.6.4 Superposition of Average Power
Average power computation when circuit contains more than one source

33. The instantaneous power delivered to the element is
Substituting
Using the identity

34. Using the identity

35. The instantaneous power becomes
Average powers delivered individually by the sources
Suppose that the two frequencies are integer multiples of some frequency as
The instantaneous power becomes

36. Averaging the instantaneous over the common period
THUS: we may superimpose the average powers delivered by sources of
different frequencies, but we may not, in general, apply superposition to
average power if the sources are of the same frequency.
where

37. Ex 6.18: Determine the average power delivered by the two sources of the circuit
Hence the average power delivered by the voltage source is
This can be confirmed from average powers delivered to the two resistors

38. of average powers delivered individually by each source
By current division:
The voltage across the current source is
Hence the average power delivered by the current source is
This may be again confirmed by computing the average power delivered to the
Two resistors:
Since frequencies are not the same, total average power delivered is the sum
of average powers delivered individually by each source

39. EX 6.19: Determine the average power delivered by the two sources
Since both sources have the same frequency, we can’t use superposition.
So we include both sources in one phasor circuit. The total average power
delivered by the sources is equal to the average power delivered to the resistor
We use superposition on the phasor circuit to find the current across the resistor

40. The phasor current is:
Hence the average power delivered to the resistor is
Note that we may not superimpose average powers delivered to the resistors by the
individual sources
We can compute this total average power by directly computing the average power
delivered by the sources from the phasor circuit
The voltage across the current source is
The average power delivered by voltage source is
The average power delivered by the current source is
The total average power delivered by the sources is

41. 6.6.5 Effective (RMS) Values of Periodic Waveforms
Sinusoidal waveform is one of more general periodic waveforms
Apply a periodic current source with period T on resistor R
The instantaneous power delivered to the resistor is
The average power delivered to the resistor is
Hence the average power delivered to the resistor by this periodic waveform
can be viewed as equivalent to that produced by a DC waveform whose value is
This is called the effective value of the waveform or the root-mean-square
RMS value of the waveform

42. Ex 6.20 Determine the RMS value of the current waveform and the average power this would deliver to resistor
The RMS value of the waveform is
Hence the average power delivered to the resistor is

43. RMS voltages and currents in phasor circuits
The sinusoid
has a RMS value of
Hence the average power delivered to a resistor by a sinusoidal voltage or current
waveform is
In general, the average power delivered to an element is
Therefore, if sinusoidal voltages and currents are specified in their RMS values
rather than their peak values, the factor ½ is removed from all average-power
expressions. However, the time-domain expressions require a magnitude multiplied
by square root of 2
Since X is the peak value of the waveform. Common household voltage are specified
as 120V. This is the RMS value of the peak of 170V.

44. Ex 6.21 Determine the average power delivered by the source and the time-domain current i(t)
Phasor circuit with rms rather than peak
The phasor current is
Hence the average power delivered by the source is
The time-domain current is

45. Commercial Power Distribution

46. Time domain representation

50. Using Vectors

54. We are going to investigate the transmission of power from
6.9.1 Wye-Connected Loads
We are going to investigate the transmission of power from
three phase generator to two types of load:
I – WYE connected load
Hence the current returning through the neutral wire is zero,
and the neutral may be removed

55. Power calculation
For a balanced wye-connected load, the voltages across the individual
loads are the respective phase voltages whether the neutral is connected
or not.
The power delivered to the individual loads is three times the power
delivered to an individual load, because the individual loads are identical.
No ½ factor in power expression because values are rms

56. Since the line voltage is more accessible than the phase voltage
Power in terms of line-to-line voltages and load current gives

57. Hence, the average power delivered to each load is
Example 6.24 Consider a balanced, wye-connected load
where each load impedance is
the phase voltages are 120 V.
Determine the total average power delivered
to the load.
The line currents are
Hence, the average power delivered to each load is
The total average power delivered to the load is

58. Example 6.25 If the line voltage of a balanced, wye-connected load is 208 V and the total average power delivered to the load is 900 W, determine each load if their power factors are 0.8 leading.
Thus
The phase voltage is
Thus the magnitude of the individual load impedance is
Since the power factor is 0.8 leading (current leads voltage; voltage lags current)
Thus the individual loads are

59. Example 6.25
If the line voltage of a balanced, wye-connected load is 208 V and the total average power delivered to the load is 900 W, determine each load if their power factors are 0.8 leading.
Thus
The phase voltage is
Thus the magnitude of the individual load impedance is
Since the power factor is 0.8 leading (current leads voltage; voltage lags current)
Thus the individual loads are