ECE 2100
Circuit Analysis
Lesson 27
Chapter 9
Impedance & Admittance,
Circuit Theorems in the Phasor Domain
Daniel M. Litynski, Ph.D.
http://homepages.wmich.edu/~dlitynsk/
ECE 2100
Circuit Analysis
Lesson 26
Chapter 14
Sec 14.1, 14.5, 14.7
Series RLC Circuit
RC Low Pass Filter
Daniel M. Litynski, Ph.D.
http://homepages.wmich.edu/~dlitynsk/
9.5 Impedance and Admittance (1)
• The impedance Z of a circuit is the ratio of the phasor
voltage V to the phasor current I, measured in ohms Ω.
V
I
Z
R
jX
where R = Re, Z is the resistance and X = Im, Z is the
reactance. Positive X is for L and negative X is for C.
• The admittance Y is the reciprocal of impedance,
measured in siemens (S).
Y
1
Z
I
V
3
9.5 Impedance and Admittance (2)
Impedances and admittances of passive elements
Element
R
L
C
Impedance
Z
Admittance
Y
R
Z
j L
Z
1
j C
Y
Y
1
R
1
j L
j C
4
9.5 Impedance and Admittance (3)
0; Z
Z
j L
0
;Z
0; Z
Z
1
j C
;Z
0
5
9.5 Impedance and Admittance (4)
After we know how to convert RLC components
from time to phasor domain, we can transform
a time domain circuit into a phasor/frequency
domain circuit.
Hence, we can apply the KCL laws and other
theorems to directly set up phasor equations
involving our target variable(s) for solving.
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9.5 Impedance and Admittance (5)
Example 8
Refer to Figure below, determine v(t) and i(t).
vs
5 cos(10t )
Answers: i(t) = 1.118cos(10t – 26.56o) A; v(t) = 2.236cos(10t + 63.43o) V
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9.6 Kirchhoff’s Laws
in the Frequency Domain (1)
• Both KVL and KCL hold in the phasor domain
or more commonly called frequency domain.
• Moreover, the variables to be handled are
phasors, which are complex numbers.
• All the mathematical operations involved are
now in the complex domain.
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9.7 Impedance Combinations (1)
• The following principles used for DC circuit
analysis all apply to AC circuit.
• For example:
a. voltage division
b. current division
c. circuit reduction
d. impedance equivalence
e. Y-Δ transformation
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9.7 Impedance Combinations (2)
Example 9
Determine the input impedance of the circuit in figure below
at ω =10 rad/s.
Answer: Zin = 32.38 – j73.76
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ECE 2100
Circuit Analysis
Chapter 10
Sinusoidal Steady-State
Analysis
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
11
Sinusoidal Steady-State Analysis
Chapter 10
10.1
10.2
10.3
10.4
10.5
10.6
Basic Approach
Nodal Analysis
Mesh Analysis
Superposition Theorem
Source Transformation
Thevenin and Norton Equivalent Circuits
12
10.1 Basic Approach (1)
Steps to Analyze AC Circuits:
1.
2.
3.
Transform the circuit to the phasor or frequency
domain.
Solve the problem using circuit techniques
(nodal analysis, mesh analysis,
superposition, etc.).
Transform the resulting phasor to the time
domain.
Time to Freq
Solve
variables in Freq
Freq to Time
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10.2 Nodal Analysis (1)
Example 1
Using nodal analysis, find v1 and v2 in the circuit
of figure below.
Answer:
v1(t) = 11.32 sin(2t + 60.01) V
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v2(t) = 33.02 sin(2t + 57.12) V
10.3 Mesh Analysis (1)
Example 2
Find Io in the following figure
using mesh analysis.
Answer: Io = 1.19465.44 A
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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.
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10.4 Superposition Theorem (2)
Example 3
Calculate vo in the circuit of figure shown below
using the superposition theorem.
Vo = 4.631 sin(5t – 81.12) + 1.051 cos(10t – 86.24) V
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10.5 Source Transformation (1)
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10.5 Source Transformation (2)
Example 4
Find Io in the circuit of figure below using the
concept of source transformation.
Io = 3.28899.46 A
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10.6 Thevenin and Norton
Equivalent Circuits (1)
Thevenin transform
Norton transform
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10.6 Thevenin and Norton
Equivalent Circuits (2)
Example 5
Find the Thevenin equivalent at terminals a–b of
the circuit below.
Zth =12.4 – j3.2 
VTH = 18.97-51.57 V
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