1. Sinusoidal Excitation of Circuits
Chapter 6
Sinusoidal Excitation of Circuits

2. 6.1 The Sinusoidal Source
Adding a phase angle to argument shift it backward

4. Example 6.1 Convert the cosine function
to a sine function and give its representation in terms
of sine and cosine function that have no phase angle.

5. 6.1.1 Representation of General Waveforms via the Fourier Series
A general periodic function satisfies the property that
Where T is the period of the waveform

6. It is possible to represent any periodic waveform as an
Infinite sum of sinusoids: a Fourier series
where

7. For example:
The sum of the
First seven terms

8. Once we replace the waveform that
has arbitrary time variation with the
sum of sinusoidal sources representing
the Fourier components of the waveform
we may apply superposition to determine
the response to the original waveform
as the sum of the responses to the individual
sinusoidal components of that waveform

9. 6.1.2 Response of Circuits to Sinusoidal Sources
Will be discussed with
6.3 The Phasor (Frequency Domain) Circuit

10. 6.2 Complex Numbers, Complex Algebra, and Euler’s Identity
The complex number can be represented as
A vector in two-dimensional complex plan
The length of the vector is the magnitude of the complex number
The angle is measured counterclockwise

11. The polar representation of complex numbers in terms of its
magnitude and angle

12. Euler’s Identity
From Euler’s identity we can prove that

13. Addition and Subtraction of Complex Numbers
For two numbers

14. Multiplication and Division of Complex Numbers
Multiplication and division is best accomplished in polar form
Multiplication and division can also be accomplished in
rectangular form

15. Define conjugate of a complex number
In polar form

16. Division in rectangular form
Multiply numerator and denominator by conjugate of
denominator
Other identities that can be proven

17. 6.3 The Phasor (Frequency- Domain) Circuit
We will represent inductors and capacitors as complex valued resistors and use all previous analysis techniques to solve problems

18. Response of Linear Circuit to Sinusoidal Source
The response will be the same trigonometric functions as the source with the same frequency
Differs from the source in magnitude and phase
Our task is to determine the magnitude and phase angle of currents and voltages

19. 6.3.1 Representation of Sinusoidal Sources with Euler’s Identity

20. The key to the Phasor method:
To determine the response to either a sine or
a cosine source, replace the source with the
complex exponential source and use superposition

21. The response to source is the real part to the source
Similarly, the response to source is the imaginary part to the source

22. 6.3.2 The Phasor Circuit (Frequency-domain circuit)
Replace the source or
with exponential source
Represent all response voltages and currents as having the same form
Cancel the factor in source and all response variables
Replace the resistors, inductors, and capacitors with their impedances

23. Resistor Impedance
For resistor, the impedance is

24. Inductor Impedance
The pahsor current and voltage of the inductor are related by
Impedance of inductor is

25. Capacitor Impedance

26. 6.4 Applications of Resistive-Circuit Analysis Techniques in the Phasor Circuit
Ex 6.4:Determine the voltage v(t) in the circuit
Replace: source with
desired voltage v(t) with
Common factor is removed from all
Impedance of capacitor is

27. A single-node pair circuit
Hence time-domain voltage becomes
Since the source was a cosine we take the real part

28. Ex 6.5 Determine the current i(t) and voltage v(t)
Single loop phasor circuit
The current
By voltage division
The time-domain

29. Ex 6.6 Determine the current i(t)
The phasor circuit is
Combine resistor and inductor

30. Use current division to obtain capacitor current
Hence time-domain current is:

31. Ex 6.7 Determine i(t) using source transformation
Phasor circuit
Transformed source
Voltage of source:
Hence the current
In time-domain

32. Ex 6.9 Find voltage v(t) by reducing the phasor circuit at terminals a and b to a Thevenin equivalent
Phasor circuit

34. The Thevenin impedance can be modeled as 1.19 resistor in series
with a capacitor with value or

35. 6.5 Circuits Containing More than One Sinusoidal Source
Apply superposition in the time domain to yield time-domain circuits with one sinusoidal source. Then solve each of these one-source circuits by reducing them to phasor circuit.

36. Ex 6.10 Determine i(t)

37. Cannot simplify further because radian frequencies are different

38. 6.5.1 Sources of the Same Frequency
Sinusoidal sources of the same radian frequency can be combined into one phasor circuit. However, all sources must be either sine or cosine form

39. Ex 6.11 Determine i(t) by placing both sinusoidal sources in the same phasor circuit
Superposition in frequency domain
In time-domain:

40. If use superposition in time-domain

41. 6.6 Power The instantaneous power delivered to an element is
The average power is

42. Average power is real power dissipated by the element usually in the
form of heat or converted to useful work as in an electric motor.
It can be computed also through the complex power

43. The complex power
The magnitude of the complex power 1/2VI is referred to as
apparent power and its units are VA.
The real part of the complex power is the average power:
Watts
The imaginary part of the complex power is the reactive power:
Reactive power represents energy stored in reactive elements
(inductors and capacitors). Its unit is Volt-Amperes Reactive VAR

44. 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

45. 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:

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

47. 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

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

49. 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

50. 6.6.1 Power Relations for the Inductor
The voltage leads the current by 90 so that

51. 6.6.1 Power Relations for the Capacitor
The current leads the voltage by 90 so that

52. 6.6.2 Power Factor
Since

53. EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load
Only through the resistor

54. 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

55. 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

56. The load current is obtained from
Ex 6.16 Suppose that the load voltage in the previous figure is 170V, the line resistance is 0.1 ohm and 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
The power needed to be generated
720 W extra power to be generated if pf is 0.7 to supply the load

57. 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

58. 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:

59. 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:

60. 6.6.4 Superposition of Average Power
Average power computation when circuit contains more than one source

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

62. 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

63. 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

64. 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

65. 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

66. 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

67. 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

68. 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

69. 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

70. 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.

71. 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

72. 6.9 Commercial Power Distribution
is rms value of the voltage
The peak is
Time domain representation

73. The line-to-line voltage between conductors of the transmission line are the
phasor sums of the phase voltages
Thus the line-to-line voltages are larger than the phase voltages:

74. 6.9.1 Wye-Connected Loads
Transmission of power from generator to the load
The three impedances making up the load are identical (Balanced load)
Hence the current returning through the neutral wire is zero,
and the neutral may be removed

75. 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
Power in terms of line-to-line voltages and load current gives

76. Example 6.24 Consider a balanced, wye-connected load where each load impedance is and 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

77. 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