1. AC CIRCUIT ANALYSIS USING PHASORS AND EQUIVALENT IMPEDANCE CONCEPT
Week 9
AC CIRCUIT ANALYSIS USING PHASORS AND EQUIVALENT IMPEDANCE CONCEPT

2. Learning Outcomes After completing this topic you will be able to:
Apply the phasor concept in ac-driven circuit analysis.
Apply the phasor i-v relationships of the resistor, capacitor, and inductor in steady-state ac circuit analysis.
Apply the equivalent impedance concept rule to find the magnitude and phase relationships of currents and voltages in ac circuits.
Draw phasor diagrams for the circuit voltages and currents.
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3. Phasors and Circuit Analysis
To use phasors for circuit analysis, we first need to know how KVL, KCL, and Ohm’s law in apply in the phasor domain.

4. Kirchhoff’s Voltage Law in the Phasor Domain
Consider the following time-domain circuit where a sinusoidally varying voltage source v1(t) is driving a current i(t) through the circuit and establishing voltage drops v2(t), v3(t).... Vn(t) across the rest of the circuit elements.
+ v2(t) -
+ v3(t) -
+ vn(t) -
+ v1(t) -

5. Recall that KVL states that the sum of voltages around a closed path is equal to zero.
0 = v1(t) + v2(t) + …. + vn(t)
For a sinusoidally driven circuit, we can write

6. In general, ejt ≠ 0, so
This result tells us that KVL applies in the phasor domain as well as the time domain.
+ v2(t) -
+ v3(t) -
+ vn(t) -
Time-domain circuit
+ v1(t) -
Phasor transform
+ V2 -
+ V3 -
+ Vn -
+ V1 -
Phasor-domain circuit
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7. Kirchhoff’s Current Law in the Phasor Domain
Recall that KCL states that the sum of currents leaving (or entering) a node is equal to zero.
0 = i1(t) + i2(t) + …. + in(t)
i(t)
i1(t)
i2(t)
in(t)

8. Using Euler’s identity ej = cos + jsin, we can rewrite the previous equation as

9. In general, ejt ≠ 0, so
This result shows that KVL applies in the phasor domain as well as the time domain.
Phasor-domain KCL equation
Time-domain KCL equation
i(t)
i1(t)
i2(t)
iN(t)

10. Driving Point Impedance of a Two-Terminal Network
The driving point impedance of a two-terminal circuit (box) is the impedance seen by a source V connected across its input terminals. The value of the driving point impedance is given by the ratio of the voltage and current phasor: Z
Note
When Z applies to a 2-terminal circuit (rather than simple component) it is known as the equivalent impedance, or alternatively, as the driving point impedance of the box.

11. Driving Point Impedance of Series-Connected Impedances
Consider the following circuit where voltage source V is supplying current I to a set N impedances connected together in series and lumped in an imaginary box. Suppose we wish to find the equivalent impedance seen by the source V. To do this we need to find the ratio V/I of the box. Z
Application of KVL to the circuit leads us to the equation
Therefore,
Equivalent impedance of box seen by source V is given by the ratio of V/I. Z

12. Driving Point Impedance of Parallel-Connected Impedances
For the circuit shown below, application of KCL leads to the node current equation
Now, application of Ohm’s law to the impedances gives us the following relationships Z1 Z2 ZN

13. Thus, upon substitution of the above expressions into the KCL equation, we obtain
We can rewrite the above equation as

14. Hence, we obtain the result
This means we can replace the N parallel-connected impedances with a single impedance, Z. Z ZN Z1 Z2

15. For the simple case of two impedance connected in series, we obtain
This can be rewritten in the more convenient form as Z2 Z1 Z

16. Worked Example 1
Determine the driving-point impedance of the circuit at a frequency of 20 Hz.
= 100μF

17. Solution

18. Worked Example 2
Determine the driving-point admittance of the circuit at a frequency of 400 Hz.

19. Worked Example 3
Determine the input impedance of the circuit at  = 10 rad/sec. Z1 Z3

20. Solution
Let
Z1 = impedance of the 2-mF capacitor in series with the 20-Ω resistor
Z2 = impedance of the 4-mF capacitor
Z3 = impedance of the 2-H inductor in series with the 50-Ω resistor Z1 Z3

21. Z1 Z3

22. Exercise
Determine the driving-point impedance of the circuit at a frequency of 20 Hz. Z I + V -

23. Worked Example 4
Determine the driving point impedance of the circuit at a frequency of 40 kHz.

24. Solution Convert the resistance and capacitance into impedances.
ZR = 25 Ω
ZC = -j19.89 Ω Z
Therefore, driving point impedance of box is

25. Phasor Analysis Steps
Phasor analysis (a.k.a sinusoidal steady-state analysis) consists basically of four steps.
Transform all independent sources to phasors
Calculate the impedance of the passive circuit elements
Apply analysis methods learned in Electric Circuit Analysis I
Apply inverse phasor transform to obtain time-domain expression for currents and voltages of interest.
Note
All of the sources must have the SAME frequency! 25

26. Worked Example 5 For the circuit shown below, find:
Angular frequency (in radians per second)
Impedance of R in Ω
Impedance of L in Ω
Impedance of C in Ω
Driving point impedance in Ω
Phasor voltage and current
Find particular response i(t) 26

27. Solution Angular frequency  = 1000 rad/s Impedance of R in Ω:
ZR = R Ω
Impedance of L in Ω:
ZL = jL = j x 1000 x 25 x 10-3 = j25 Ω
Impedance of C in Ω:
ZC = 1/(jC) = 1/(j x 1000 x 10 x 10-6) = -j100 Ω

28. Z Driving point impedance,
Zin = ZR + ZC + ZL = 50 – j100 + j25 = 50 - j75 Ω
= 90.14-56.14° Z

29. Find phasor voltage and current.
VS = P{vS(t)} = P{35cos(1000t + 30°)} = 3530° V
I = P{i(t)} I
Zin VS
Application of KVL to the circuit leads us to the equation
- VS + IZin = 0
Hence,
I = VS
Zin
3530°
90.14-56.14° =
0.38886.14° A

30. Find particular response i(t).
i(t) = 0.388cos(1000t °) A
Phasor diagram: VL VS VR VC I
75°
30°
15°

31. Worked Example 6
Determine v(t) and i(t).

32. Solution
Draw the phasor equivalent circuit. ;
Therefore, VS I
j2 Ω

33. Solution 2. Find equivalent impedance seen by the source.
3. Find circuit current.
4. Find phasor voltage drop V across the inductor.
5. Find time-domain current and voltage.

34. Worked Example 7 Determine Current i(t) voltage drop vL(t)
voltage drop vR(t)

35. Solution Draw the phasor equivalent circuit. ; ; Therefore, I 250 Ω +VL -
+ VR - VS
j250 Ω

36. Solution 2. Find equivalent impedance seen by the source.
3. Find circuit current.
4. Find phasor voltage drop VL across the inductor.
5. Find phasor voltage drop VR across the resistor.

37. Solution
6. Find time-domain current and voltages.

38. Worked Example 8
Find the steady-state expression for io(t) if ig(t) = 125cos(500t) mA
R1 = 50 Ω
R2 = 250 Ω
L = 1 H
C = 20 F R1 R2 C L

39. Solution Angular frequency  = 500 rad/s Impedance of R1 in Ω:
ZR1 = 50 Ω
Impedance of R2 in Ω:
ZR2 = 250 Ω
Impedance of L in Ω:
ZL = jL = j x 500 x 1 = j500 Ω
Impedance of C in Ω:
ZC = 1/(jC) = 1/(j x 500 x 20 x 10-6) = -j100 Ω

40. Driving point impedance,
Zin = (ZR1 + ZC) // (ZR2 + ZL)
= (50 – j100)//(250 + j500)
= 75 - j100 Ω
= 125-36.87
Zin

41. Find phasor voltage and current.
Io = P{io(t)} = P{125cos(500t )} = 1250° mA
Vo = P{vo(t)}
Application of Ohm’s law to the circuit leads us to the equation
Zin Io Vo
Vo = IoZin
Hence,
Vo = IoZin
= 125 x 10-3 0 x 125-36.87 = -53.13 V

42. Now, VO
15.625-53.13
IG = = = 0.028  A
250 + j500
ZR2 + ZL
Find particular response iG(t).
iG(t) = 0.028cos(500t °) A

43. Exercise
Find the steady-state expression for vo(t) if vg(t) = 64cos(8000t)

44. Solution Angular frequency  = 8000 rad/s Impedance of R in Ω:
ZR1 = 2kΩ
Impedance of L in Ω:
ZL = jL = j x 8000 x 500 x 10-3 = j4500 Ω
Impedance of C in Ω:
ZC = 1/(jC) = 1/(j x 8000 x x 10-9) = -j4000 Ω

45. Driving point impedance,
Zin = ZC + ZR // ZL
= -j (2000//j4500)
= 1670 – j4742 Ω
= 502870.6 ZC ZL ZR
ZIN

46. Find phasor voltage and current.
Vg = P{vg(t)} = P{64cos(8000t )} = 640° V
I = P{i(t)} I
Application of KVL to the circuit leads us to the equation
Zin VS
- Vg + IZin = 0
Hence,
I = Vg
Zin
640°
502870.6° =
0.013-70.6° A

47. Find phasor voltage drop VL across the inductor.
Find particular response vO(t).
vo(t) = cos(8000t °) A

48. Concept of Reference Phasor
When no reference angle is given either for the supply voltage or the supply current, we can define our own reference angle.
The following examples show the use of the reference phasor concept in ac circuit analysis.

49. Worked Example 8
An a.c. circuit consists of a 25  resistor in series with a 40 F capacitor connected across a 240 V 50 Hz supply. Sketch the circuit and calculate the current in the circuit, and its phase angle relative to the supply voltage.

50. Solution
The time-domain circuit referred to by the question is shown in the following figure. Its phasor equivalent is also show. I vS R C vR vC i VR R VS VC ZC
(b)
(a)

51. Notes
The supply voltage is given as an rms value. Because our definition of phasor uses peak value for the magnitude of the phasor, we need to multiply the rms value with 2 to convert it into peak value.
The question does not give any information on the phase angle of the supply voltage. To represent the supply voltage with a phasor, we need to assign any arbitrary angle to it. For simplicity, we choose 0 as the phase angle for the supply voltage phasor. That is, we write
By assigning an angle to the supply voltage phasor, it means that we have chosen it to be the reference phasor. The angles of other phasors in the circuit are measured with respect to this reference angle.

52. The impedance seen by the source is
The supply current is
Therefore,

53. The time domain current is
The phasor diagram for circuit is shown in Figure 8. Since the supply voltage phasor has a phase angle of 0o and the current phasor has a phase angle of o, the current phasor lags behind the supply voltage phasor by 72.56o. VS Re Im I

54. Worked Example 9
A 10 Ω resistor is connected in series with a 120 μF capacitor across a 20 V, 100 Hz supply. Calculate
impedance of the circuit,
current, and
voltage drop across each component.
Solution
The phasor domain circuit is shown in the figure below. VS R C VR VC I

55. The total circuit impedance is given by
Substituting in the given values for R and C into the above equation, we obtain
Therefore,

56. Let VS be the reference phasor so that we can write
The circuit current is given by
Therefore,

57. The phasor voltage drop VR is given by
The phasor voltage drop VC is given by

58. Worked Example 10
A resistor and a capacitor are connected in series across a 240 V, 50 Hz supply. If the current flow in the circuit is 5 A, leading the applied voltage by 45.6o, calculate
impedance of the circuit
phase angle,
the resistance and the capacitance
the voltage drop across each component.

59. The phasor domain circuit is shown in the figure.
Solution
The phasor domain circuit is shown in the figure.
Let VS be the reference phasor so that we can write
Given that the phasor current I has a magnitude of 5 A and leads the applied voltage by 45.6o, so we can write VS R C VR VC I
Therefore,

60. Referring to the impedance triangle shown below, we obtain
Again referring to the impedance triangle in Figure 10, we obtain
giving Z Re Im R
XC= 1/ωC φ

61. Worked Example 11
A circuit having a resistance of 25 Ω and an inductance of 12.5 mH is connected to a 100 V, 400 Hz a.c. supply. Calculate the current in the circuit, and its phase angle relative to the supply voltage.
Solution
The circuit referred to by the question is shown in the following figure.
We note that in this problem we are not given the circuit but have to draw it ourselves and provide our own labels and reference directions for the circuit’s current and voltages.
100 V rms
400 Hz
25 Ω
12.5 mH

62. Solution
To use phasor analysis technique to solve the problem, we first need to label the circuit current and the various voltage drops in the circuit and assign a reference direction for each. The labelled time-domain circuit is shown in the figure.
25 Ω
12.5 mH i vR vL
Note that since we have not been given the phase angle of the supply voltage, we have conveniently assumed it to be zero. In other words, we have assumed the supply voltage as the reference phasor.

63. The phasor domain circuit is shown in the figure below.
jXL = j2πx400x12.5 x10-3 = j Ω
25 Ω
jXL = j Ω I VR VL
Note
To convert rms value of a voltage or a current to peak value, multiply the given rms value with

64. Ohm’s law gives
Substituting the constraint equations obtained above into the KVL equation we get
or,

65. Therefore,
The time domain current is

66. The phasor diagram for circuit is shown in the following figure
The phasor diagram for circuit is shown in the following figure. Since the supply voltage phasor has a phase angle of 0o and the current phasor has a phase angle of o, the current phasor lags behind the supply voltage phasor by 51.49o. VS Re Im I

67. Exercise Given the following circuit, determine
impedance of the circuit
the source current
the voltage drop across each component.
Sketch the impedance and phasor diagram.
8 V rms
20 kHz
3.3 kΩ
15 mH

68. Given the following circuit, determine impedance of the circuit
Exercise
Given the following circuit, determine
impedance of the circuit
the source current
the voltage drop across each component.
Sketch the impedance and phasor diagram.
4 kΩ
XL = 6 kΩ

69. Exercise
A series circuit consisting of a resistance of 50 Ω, an inductance of 0.4 H, and a capacitance of 100 F is connected to a 240 V, 50 Hz single-phase supply.
Calculate
(a) the impedance of the circuit
(b) The current flowing, and
(c) The p.d. across each component
(d) The phase angle

70. END