This document provides an overview and summary of key concepts related to sinusoidal steady-state analysis including phasors, impedance, admittance, and their applications. The key topics covered are: 1) Definition and graphical representation of phasors 2) Voltage-current relationships for resistors, inductors, and capacitors in the frequency domain 3) Definition of impedance as the ratio of voltage to current and how to combine impedances in series and parallel 4) Definition of admittance as the reciprocal of impedance and its components of conductance and susceptance 5) Examples of calculating impedance and admittance for circuits containing resistors, inductors, and capacitors.

1. LECTURE 04
Chapter 10: Sinusoidal Steady-State Analysis
1

2. Lecture 04: Agenda
1. The Phasor
1. The Resistor
2. The Inductor
3. The Capacitor
4. Summary
2. RL Circuit
3. Impedance
4. Admittance
5. Example
2

3. 1. The Phasor
3

4. 1. The Phasor (Graphical Representation)
4
M
j
(Imaginary
Axis)
Msinθ
θ
MCosθ σ (RealAxis)

5. 1. The Phasor (Graphical Representation)
5
j
(Imaginary
Axis)
M

6. Phasor: Real & Imaginary Components
6
M
cos(ωt)
M Sin(ωt)
M ω
θ
jω
Imaginary
Axis
Sine Component
σ Real (Axis)
Cosine Component

7. 1. The Phasor
7
• i(t) as a time-domain
representation and terming the
phasor I a frequency-domain
representation
• It should be noted that the
frequency-domain expression of a
current or voltage does not
explicitly include the frequency

8. 1.1 The Resistor
8

9. 1.2. The Inductor
9
V Leads I by 90o in an Inductor

10. 1.3. The Capacitor
10
• The equivalent expression in the
frequency domain is obtained once
more by letting v(t) and i(t) be the
complex quantities, taking the
indicated derivative, suppressing ejωt,
and recognizing the phasors V and I
• Doing this, we find
• I leads V by 90° in a capacitor
• This, of course, does not mean that a current response is present one-
quarter of a period earlier than the voltage that caused it!
• We are studying steady-state response, and we find that the current
maximum is caused by the increasing voltage that occurs 90° earlier
than the voltage maximum

11. 1.4. Summary - Comparison of Time-
Domain and Frequency-Domain Voltage-
Current Expressions
11

12. 2. RL Circuit
12

13. 13
2/28o 2.06/13.96o A.
½/-62o
𝟐. 𝟎𝟔 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟑. 𝟗𝟔𝒐) 𝑨.

14. Practice Problem
14

15. Practice Problem
15
• Solution: For ω = 1 rad/sec,
• Zc = - j Ω ;
• Vc = 2 28o x - j = 2
ZL = jωL = j x1x 2 = j2 Ω
- 62o = 0.939 – j1.766 V
• IR2 = Vc / R = 2 - 62 o / 2 = 1 - 62o A
• Is = 1 - 62 + 2 28o = 2.235 + j0.056 = 2.236 1.435o A
31.25o A
• IR1 = 3 53o + 2.236 1.435o = 4.041 + j2.452 = 4.73
iR1 (t) = 4.73 cos ( t + 31.25o)A
• Vs = IR1 x R1 + ILx ZL
= 4.73 31.25o + 3 53o x 2 90o
= - 0.75 + j6.06 = 6.11 97.06o Volts
vs(t) = 6.11 cos ( t + 97.06o) V

16. 3.1 Impedance
• The current-voltage relationships for the three passive elements
in the frequency domain are (assuming that the passive sign
convention is satisfied)
• If these equations are written as phasor voltage/phasor current
ratios
• We find that these ratios are simple quantities that depend on
element values (and frequency also, in the case of inductance
and capacitance)
• We treat these ratios in the same manner that we treat
resistances, provided we remember that they are complex
quantities
16

17. 3.3 Impedance
17
• Let us define the ratio of the phasor voltage to the phasor current as
impedance, symbolized by the letter Z
• The impedance is a complex quantity having the dimensions of ohms
• Impedance is not a phasor and cannot be transformed to the time
domain by multiplying by e jωt and taking the real part
• Instead, we think of an inductor as being represented in the time
domain by its inductance L and in the frequency domain by its
impedance jωL
• A capacitor in the time domain has a capacitance
C; in the frequency domain, it has an impedance
1/jωC
• Impedance is a part of the frequency domain and
not a concept that is a part of the time domain

18. 3.4 Series Impedance Combinations
• The validity of Kirchhoff’s two laws in the frequency domain
leads to the fact that impedances may be combined in series and
parallel by the same rules we established for resistances
• For example, at ω = 10 × 103 rad/s, a 5 mH inductor in series
with a 100 μF capacitor may be replaced by the sum of the
individual impedances
•The impedance of the inductor is; ZL = jωL = j50 Ω
and the impedance of the capacitor is
𝐶
𝒁 =
1
𝑗 𝜔𝐶 𝜔𝐶
= −𝑗
= −𝑗1 Ω
The impedance of the series combination is therefore
Zeq = ZL + ZC = j50 –j1=j49 Ω
18

19. 3.5 Parallel Impedance Combinations
• The parallel combination of the 5 mH inductor and the 100 μF
capacitor at ω = 10,000 rad/s is calculated in exactly the same fashion
in which we calculated parallel resistances:
𝒁𝑒
𝑞
= 𝑍𝐿×𝑍𝐶
= (𝑗50)(−𝑗)
= 50
𝑍𝐿+𝑍𝐶 𝑗50−𝑗 𝑗49
= −𝑗1.020 Ω
• The impedance of the parallel combination is therefore; Zeq = -j 1.02 Ω
• At ω = 5000 rad/s, the parallel equivalent is −j2.17 Ω
• We may choose to express impedance in either rectangular form
(Z = R + jX) or polar form (Z = |Z| /θ )
• In rectangular form, we can see clearly the real part which arises only
from real resistances, and an imaginary component, termed the
reactance, which arises from the energy storage elements
19

20. 3.6 Reactance
• Both resistance and reactance have units of ohms, but reactance
will always depend upon frequency
• An ideal resistor has zero reactance; an ideal inductor or
capacitor is purely reactive (i.e., characterized by zero
resistance)
• Can a series or parallel combination include both a capacitor
and an inductor, and yet have zero reactance?
• Sure! Consider the series connection of a 1 Ω resistor, a 1 F
capacitor, and a 1 H inductor driven at ω = 1 rad/s
Zeq = 1 − j (1)(1) + j (1)(1) = 1
• At that particular frequency, the equivalent is a simple 1 Ω
resistor
• However, even small deviations from ω = 1 rad/s lead to
nonzero reactance
20

21. 4. Admittance
• Although the concept of impedance is very useful, and familiar
in a way based on our experience with resistors, the reciprocal
is often just as valuable
• We define this quantity as the admittance Y of a circuit element
or passive network, and it is simply the ratio of current to
voltage: The real part of the admittance is the conductance G,
and the imaginary part is the susceptance B
• All three quantities (Y, G, and B) are measured in Siemens
• The real part of the admittance is the conductance G, and the
imaginary part of the admittance is the susceptance B
• Thus,
21

22. 7.2 Admittance
• By Rationalization,
• Equating the real and imaginary parts gives
• Showing that G ≠ 1/R as it is in resistive circuits
22

23. Impedance & Admittance (2/3) Ex 10.6
23
Solution:
The impedance of the inductor is; ZL = jωL = jx5x2 = j10 Ω
and the impedance of the 200 mF capacitor is
𝐶
𝒁 =
1 −
𝑗
𝑗 𝜔𝐶 5X0.2
= = −𝑗1 Ω
The impedance of the 500 mF capacitor is
𝐶
𝒁 =
1 −
𝑗
𝑗 𝜔𝐶 5X0.5
= = −𝑗0.4 Ω

24. Impedance & Admittance (2/3) Ex 10.6
24

25. Impedance & Admittance (3/3)
25

26. Impedance & Admittance (3/3)
26

27. Impedance & Admittance (3/3)
27

28. Impedance Example
• Find the input impedance of the circuit in Fig.Assume that the circuit
operates at ⍵ = 50 rad/s.
Solution:
Z1 = Impedance of the 2-mF capacitor
Z2 = Impedance of the 3- resistor in series with the10-mF capacitor
Z3 Impedance of the 0.2-H inductor in series with the 8 Ω resistor
Then
The input impedance is
Thus,
28

29. Admittance Example
• Determine the admittance Y for the circuit
Solution:
• Since it is a parallel circuit, we
will find the admittance of all
components and find Y by adding
the individual admittances
• The admittance of 2 Ω resistor is 𝑌𝑅 = 1 1
= = 0.5 S
𝑅 2
𝐿
• The admittance of j4 Ω inductor is 𝑌 =
1 1
Z𝐿 𝑗4
𝐶
• The admittance of -j5 Ω capacitor is 𝑌 =
= = −𝑗0.25 S
1 1
Z𝐶 −𝑗5
= = 𝑗0.2 S
• The total admittance Y = 𝑌𝑅 + 𝑌𝐿 + 𝑌𝐶 = 0.5 −𝑗0.25 + 𝑗0.2
Y = 0.5 −𝑗0.05 S
The impedance 𝒁 = 1
=
1
𝑌 0.5 −𝑗0.05
= 1.98 + 𝑗0.198 Ω