ECE 2100
Circuit Analysis
Lesson 29
Chapter 10: Sinusoidal Steady State Analysis (Circuit
Theorems in the Phasor Domain) FINISH
Chapter 11: AC Power Analysis
(Instant & Ave Power; Max Ave Power Transfer;
Effective or RMS Value)
Daniel M. Litynski, Ph.D.

ECE 2100
Circuit Analysis
Lesson 28
Chapter 10: Sinusoidal Steady State Analysis
(Circuit Theorems in the Phasor Domain)
Chapter 11: AC Power Analysis
(Instantaneous & Average Power)
Daniel M. Litynski, Ph.D.
http://homepages.wmich.edu/~dlitynsk/

Chapter 10
Sinusoidal Steady-State
Analysis
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
3

Sinusoidal Steady-State Analysis
Chapter 10
10.1 Basic Approach
10.2 Nodal Analysis
10.3 Mesh Analysis
10.4 Superposition Theorem
10.5 Source Transformation
10.6 Thevenin and Norton Equivalent Circuits
4

10.1 Basic Approach (1)
Steps to Analyze AC Circuits:
1.
Transform the circuit to the phasor or frequency domain .
2.
Solve the problem using circuit techniques
(nodal analysis, mesh analysis, superposition, etc.).
3.
Transform the resulting phasor to the time domain.
Time to Freq
Solve variables in Freq
Freq to Time
5

10.2 Nodal Analysis (1)
Example 1
Using nodal analysis, find v of figure below.
1 and v
2 in the circuit
Answer: v
1
(t) = 11.32 sin(2t + 60.01
 ) V v
2
(t) = 33.02 sin(2t + 57.12
 ) V
6

10.3 Mesh Analysis (1)
Example 2
Find I o in the following figure using mesh analysis.
Answer:
I o
= 1.194
 65.44
 A
7

10.4 Superposition Theorem (1)
When a circuit has sources operating at different frequencies,
• The separate phasor circuit for each frequency must be solved independently, and
• The total response is the sum of timedomain responses of all the individual phasor circuits.
8

10.4 Superposition Theorem (2)
Example 3
Calculate v o in the circuit of figure shown below using the superposition theorem.
V o
= 4.631 sin(5t – 81.12
 ) + 1.051 cos(10t – 86.24
 ) V
9

10.5 Source Transformation (1)
10

10.5 Source Transformation (2)
Example 4
Find Io in the circuit of figure below using the concept of source transformation.
I o
= 3.288
 99.46
 A
11

10.6 Thevenin and Norton
Equivalent Circuits (1)
Thevenin transform
Norton transform
12

10.6 Thevenin and Norton
Equivalent Circuits (2)
Example 5
Find the Thevenin equivalent at terminals a–b of the circuit below.
Z th
= 12.4 – j3.2  V
TH
= 18.97
 -51.57
 V
13

ECE 2100
Circuit Analysis
Lesson 29
Chapter 11: AC Power Analysis
(Instant & Ave Power; Max Ave Power Transfer;
Effective or RMS Value)
Daniel M. Litynski, Ph.D.

Chapter 11
AC Power Analysis
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
15

AC Power Analysis
Chapter 11
11.1 Instantaneous and Average Power
11.2 Maximum Average Power Transfer
11.3 Effective or RMS Value
11.4 Apparent Power and Power Factor
11.5 Complex Power
11.6 Conservation of AC Power
11.7 Power Factor Correction
11.8 Power Measurement
16

11.1 Instantaneous and
Average Power (1)
• The instantaneous power, p(t) p ( t )
 v ( t ) i ( t )

V m
I m cos ( w t
  v
) cos ( w t
  i
)

1
2
V m
I m cos (
 v
  i
)

1
2
V m
I m cos ( 2 w t
  v
  i
)
Constant power
Sinusoidal power at 2 w t p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.
17