Engineering 43
Impedance
KCL & KVL
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Review → V-I in Phasor Space
 Resistors
No Phase Shift
V
I
R
 Inductors
i(t) LAGS
V
I
  90
L
 Capacitors
i(t) LEADS
I  CV90
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Impedance
 For each of the passive
components, the
relationship between
the voltage phasor and
the current phasor is
algebraic (previous sld)
 Consider now the
general case for an
arbitrary 2-terminal
element
 The Frequency Domain
Analog to Resistance is
IMPEDANCE, Z
Engineering-43: Engineering Circuit Analysis
3
V
Z
I
 Since the Phasors V & I
Have units of Volts and
Amps, Z has units of
OHMS
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Impedance cont.
 Since V & I are
COMPLEX, Then Z is
also Complex
Z
V VM  v VM


( v   i ) | Z |  z
I
I M  i I M
 Impedance is NOT a
Phasor
• It’s Magnitude and Phase
Do Not Change
regardless of the
Location within The
Circuit
Engineering-43: Engineering Circuit Analysis
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 However, Z IS a
COMPLEX NUMBER
that can be written in
polar or Cartesian form.
• In general, its value
DOES depend on the
Sinusoidal frequency
Z ( )  R  jX ( )
 R  RESISTive component
 X ( )  REACTive component
• Note that the
REACTANCE, X, is
a function of ω
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Impedance cont.2
 Thus
Z z  R  jX
 The Magnitude and
Phase
Z  R2  X 2
X
 z  tan 1
R
 Where
R  Z cos  z
X  Z sin  z
Engineering-43: Engineering Circuit Analysis
5
 Summary Of PassiveElement Impedance
Element
R
L
C
Phasor Eq. Impedance
V  RI
ZR
V  jLI
Z  jL
1
1
V
I
Z
jC
jC
 Examine ZC
1
j
j
ZC 


jC jjC  1C
1
1
 ZC  j
 XC 
C
C
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
KVL & KCL Hold In Phasor Spc
 v2 (t ) 


v1 ( t )
v3 ( t )


KVL: v1(t )  v2 (t )  v3 (t )  0
vi (t )  VMie
j ( t  i )
, i  1,2,3
i0 (t )
Similarly for the Sinusoidal
Currents ...
V1  V2  V3  0 Phasors!

V1
V3


Engineering-43: Engineering Circuit Analysis
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i3 (t )
ik (t )  I Mk e j ( t k ) , k  0,1,2,3
VM 11  VM 2 2  VM 33  0

i2 (t )
KCL :  i0 (t )  i1 (t )  i2 (t )  i3 (t )  0
KVL : (VM 1e j1  VM 2 e j 2  VM 3e j3 )e jt  0
 V2 
i1 (t )
 I 0  I1  I 2  I 3  0
I0
I1
I2
I3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Series & Parallel Impedances
 Impedances (which have units of Ω)
Combine as do RESISTANCES
• The SERIES Case
I
 V1 
 V2 
Z1
Z2
I
Zs  Z1  Z2
Z s  k Z k
• The Parallel Case
I

Z1
I

Z2 V
V


Engineering-43: Engineering Circuit Analysis
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Zp 
Z1Z 2
Z1  Z 2
1
1
 k
Zp
Zk
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Admittance
 The Frequency Domain
Analog of
CONDUCTANCE is
ADMITTANCE
• Admittance is Thus
Inverse Impedance
1
Y   G  jB (Siemens)
Z
• G  CONDUCTance
• B  SUSCEPTance
 Find G & B In terms of
Resistance, R, and
Reactance, X
Engineering-43: Engineering Circuit Analysis
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1
1
Y 
Z R  jX
 Multiply Denominator by
the Complex Conjugate
1 R  jX
R  jX
Y
 2
R  jX R  jX R  X 2
R
 G 2
R X2
X
 B 2
R X2
 Note that G & R
and X & B are
NOT Reciprocals
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Series & Parallel Admittance
 Admittance Element Phasor Eq. Impedance
Summarized
R
L
C
V  RI
V  j L I
1
V
I
j C
ZR
Z  jL
Z
Admittance
1
Y  G
R
1
Y
j L
1
jC
Y  j C
 Admittances (which have units of Siemens)
Combine as do CONDUCTANCES
 The SERIES Case
1
1

Ys
k Yk
Engineering-43: Engineering Circuit Analysis
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 The PARALLEL Case
Yp   Yk
k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Complex Numbers in MATLAB
 MATLAB recognizes
complex numbers
these in these forms
• Rectangular
• Exponential
 Can Use “i” or “j” for
√(-1)
 MATLAB Always
returns “i” for √(-1)
 Sometimes need “*”
Engineering-43: Engineering Circuit Analysis
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>> phiR = 23*pi/180 % 23deg in Rads
phiR =
0.4014
>> Z1 = 7 + i*23 % if i or j
BEFORE, then need *
Z1 =
7.0000 +23.0000i
>> Z2 = 11 - 13j
Z2 =
11.0000 -13.0000i
>> Z3 = 43*exp(j*phiR) % Need *
Z3 =
39.5817 +16.8014i
>> Z4 = 37*exp(0.61j)
Z4 =
30.3270 +21.1961i
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Phasors in MATLAB
 MATLAB Does NOT
Recognize Phasor
NOTATION
• But it DOES handle
Complex
Exponentials
 e.g.:
>> phi8 = 17*pi/180
phi8 =
0.2967
>> Z7 = 29*exp(j*phi7)
Z7 = 21.2093 -19.7780i
>> Z8 = -53*exp(j*phi8)
Z8 = -50.6842 -15.4957i
>> Zsum = Z7 + Z8
Zsum = -29.4749 -35.2737i
Z 7  29  43
>> Zdif = Z7 - Z8
Zdif = 71.8934 - 4.2823i
Z8  5317
>> Zprod = Z7*Z8
Zprod = -1.3814e+003 +6.7378e+002i

Engineering-43: Engineering Circuit Analysis
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>> phi7 = -43*pi/180
phi7 =
-0.7505

>> Zquo = Z7/Z8
Zquo = -0.2736 + 0.4739i
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Phasors in MATLAB
 MATLAB Always
returns Complex
No.s in
RECTANGULAR
Form
 Can Recover
Magnitude & Phase
Using Commands
• abs(Z)
• angle(Z)
Engineering-43: Engineering Circuit Analysis
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>> Zquo = Z7/Z8
Zquo = -0.2736 + 0.4739i
>> Asum = abs(Zsum)
Asum =
45.9674
>> phi_sum = angle(Zsum)
phi_sum =
-2.2669
>> phi_sumd = phi_sum*180/pi
phi_sumd = -129.8824
>> Aquo = abs(Zquo)
Aquo =
0.5472
>> phi_quo = angle(Zquo)
phi_quo =
2.0944
>> phi_quod = phi_quo*180/pi
phi_quod = 120.0000
>> Zquo_test = Aquo*exp(j*phi_quo)
Zquo_test =
-0.2736 + 0.4739i
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
MATLAB: a+jb ↔ A∟φ
 BMayer MATLAB
Functions
function Zrectd = Rectab(Mag, phi_deg)
% B. Mayer 22Apr09 * ENGR43
% finds for POLAR COMPLEX number Z the
Rectangular Equivalet
%% note that phi is in DEGREES
%
a = Mag*cosd(phi_deg);
b = Mag*sind(phi_deg);
Zrectd = a + j*b
function Phasor = MagPh(Zr)
% B. Mayer 22Apr09 * ENGR43
% finds for RECTANGULAR COMPLEX number Z
%% Magnitude
%% Phase Angle in DEGREES
Magnitude = abs(Zr);
Phase_deg = angle(Zr)*180/pi;
Phasor = [Magnitude, Phase_deg];
Engineering-43: Engineering Circuit Analysis
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 Example
>> Z1r = 13 - 19j
Z1r =
13.0000 -19.0000i
>> Phasor1 = MagPh(Z1r)
Phasor1 =
23.0217
-55.6197
>> Phasor2 = [43 -127]
Phasor2 =
43 -127
>> Zr2 = Rectab(Phasor2(1),
Phasor2(2))
Zr2 =
-25.8780 -34.3413i
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
MATLAB Equivalent Functions
 Rectangular to Polar  Polar to Rectangular
 Both use RADIANS only
Engineering-43: Engineering Circuit Analysis
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Phasor Diagrams
 As Noted Earlier
Phasors can be
Considered as
VECTORS in the
Complex Plane
• See Diagram at Right
 Phasors Obey the
Rules of Vector
Arithmetic
• Which were orignially
Developed for Force
Mechanics
Engineering-43: Engineering Circuit Analysis
15
Imaginary
b
A

a
Real
 See Next Slide for
Review of Vector
Addition
• Text Diagrams follow the
PARALLELOGRAM
Method
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Vector Addition
 Parallelogram Rule
For Vector Addition
 Examine Top & Bottom of
The Parallelogram
• Triangle Rule For
Vector Addition
• Vector Addition is
Commutative
PQ  QP
C
B
C
B
• Vector Subtraction →
Reverse Direction of
The Subtrahend
P  Q  P   Q
Engineering-43: Engineering Circuit Analysis
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Example  Phasor Diagram
 For The Ckt at Right,
Draw the Phasor
Diagrams as a function
of Frequency
 First Write KCL
V
V
IS  
 jCV
R jL
1

1
I S  V 
 jC 
 R jL

I S  V k Yk  Admittance s
 Now we can Select ANY
Phasor Quantity, I or V,
as the BaseLine
Engineering-43: Engineering Circuit Analysis
17
 That is, we Can Select
ONE Phasor to have a
ZERO Phase Angle
• In this Case Choose V
 Next Examine
Frequency Sensitivity of
the Admittances
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Example  Phasor Diagram cont
 The KCL
V
V
IS  
 jCV
R jL
 This Eqn Shows That
as ω increases
• YL DEcreases
• YC INcreases
 Now Rewrite KCL using
Phasor Notation
 Examining the Phase
Angles Shows that in
the Complex Plane
• IR Points RIGHT
• IL Points DOWN
• IC Points UP
VM 0 VM   90
IS 

 CVM 90
R
L
 As ω Increases, IC
as 1 j  1  90; j  190
begins to dominate IL
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Example  Phasor Diagram cont.2
| I L || I C |
 Case-I: ω=Med so That
•
YL  YC
I C  jCV
IC  I L  0
V
IL 
jL
 IS  IR
 Case-II: ω=Low so That
• YL  2YC
 The Circuit is Basically
INDUCTIVE
Engineering-43: Engineering Circuit Analysis
19
 Case-III: ω=Hi so That
• YC  2YL
 The Circuit is Basically
CAPACITIVE
| I L || I C |
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
KCL & KVL for AC Analysis
 Simple-Circuit Analysis
• AC Version of Ohm’s Law → V = IZ
• Rules for Combining Z and/or Y
• KCL & KVL
• Current and/or Voltage Dividers
 More Complex Circuits
• Nodal Analysis
• Loop or Mesh Analysis
• SuperPosition
Engineering-43: Engineering Circuit Analysis
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Methods of AC Analysis cont.
 More Complex Circuits
• Thevenin’s Theorem
• Norton’s Theorem
• Numerical Techniques
– MATLAB
– SPICE
Engineering-43: Engineering Circuit Analysis
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Example
 For The Ckt At Right,
Find VS if
Vo  8V45
 Solution Plan: GND at
Bot, then Find in Order
• I3 → V1 → I2 → I1 → VS
 I3 First by Ohm
I3 
VO V 
 445 A
2
 Then V1 by Ohm
V1  (2  j 2)I 3 
 8  45 445
 Then I2 by Ohm
I2 
V1
11.314V0

 5.657  90( A)
j 2
290
 Then I1 by KCL
I1  I 2  I 3  5.657  90  445
I1   j5.657  (2.828  j 2.828)( A)
I1  2.828  j 2.829( A)
V1  11.3140(V )
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Example cont.
Z eq
 Then VS by Ohm & KVL
VS  2I1  V1  2(2.828  j 2.829)  11.3140
VS  16.97  j5.658(V )
VS  17.888V  18.439
 Note That we have
I1 and VS
 Thus can find the
Circuit’s Equivalent
Impedance
VS
Z eq 
I1
Engineering-43: Engineering Circuit Analysis
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 Then Zeq
17.888V  18.439
Z eq 
2.828  j 2.829A
Z eq  4.00  j 2.00
Z eq  4.47226.56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Nodal Analysis for AC Circuits
 For The Ckt at Right
Find IO
 Use Node Analysis
• Specifically a SuperNode
that Encompasses The
V-Src  KCL at SN
V1
V2
V2
 20 

0
1  j1
1 1  j1
 And the SuperNode
Constraint
V2  V1  60
or
V1  V2  60
Engineering-43: Engineering Circuit Analysis
24
 The Relation For IO
V2
IO 
( A)
1
 In SuperNode KCL
Sub for V1
V2  60
V
 20  V2  2  0
1  j1
1  j1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Nodal Analysis cont.
 Solving for For V2
 1
1 
6
V2 
1

2


1

j
1
1

j
1
1  j1


 The Complex Arithmetic
V2
(1  j1)  (1  j1)(1  j1)  (1  j1) 2(1  j1)  6

(1  j1)(1  j1)
1  j1
4
V2
 8  j2
1 j
V2 
 Or
8  j 2 1  j   5  j 3
1
4
2
Engineering-43: Engineering Circuit Analysis
25
 Recall
2
V2 V 
IO 
1
3
5
I O    j ( A)
2
2
I O  2.92 A  30.96
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Loop Analysis for AC Circuits
 Same Ckt, But Different
Approach to Find IO
 Note: IO = –I3
 Constraint: I1 = –2A0
 The Loop Eqns
LOOP 2 :
(1  j )( I1  I 2 )  60  (1  j )( I 2  I 3 )  0
LOOP 3 : (1  j )( I 2  I 3 )  1I 3  0
 Solution is I3 = –IO
 Recall I1 = –2A0
Engineering-43: Engineering Circuit Analysis
26
 Simplify Loop2 & Loop3
L2 : 2 I 2  (1  j ) I 3  6  (1  j ) I1
2 I 2  (1  j ) I 3  6  (1  j )( 2)
L3 : (1  j ) I 2  (2  j ) I 3  0
 Two Eqns In Two
Unknowns
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Loop Analysis cont
 Isolating I3
(1  j )
2

 2(2  j ) I 3  (1  j )(8  2 j )
I2
 Then The Solution
I3 
10  6 j
4
 I0 
5 3
 j ( A)
2 2
 Could also use a
SuperMesh to Avoid
the Current Source
CONSTRAINT : I 2  I1  20
SUPERMESH : (1  j )I1  60  1(I 2  I 3 )  0
MESH 3 : 1(I 3  I 2 )  (1  j )I 3  0
Engineering-43: Engineering Circuit Analysis
27
 The Next Step is to
Solve the 3 Eqns for
I2 and I3
 So Then Note
IO  I 2  I3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Recall Source SuperPosition
I
I L2
1
L
=
V
1
L
 Circuit With Current
Source Set To Zero
 By Linearity
IL  I  I
2
L
Engineering-43: Engineering Circuit Analysis
28
VL2
 Circuit with Voltage
Source set to Zero
• OPEN Ckt
1
L
+
• SHORT Ckt
VL  V  V
1
L
2
L
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
AC Ckt Source SuperPosition
 Same Ckt, But Use
Source SuperPosition
to Find IO
 Deactivate V-Source
 The Reduced Ckt
 Combine The Parallel
(1  j )(1  j )
Impedances
Z '  (1  j ) || (1  j ) 
1
(1  j )  (1  j )
Engineering-43: Engineering Circuit Analysis
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
AC Source SuperPosition cont.
 Find I-Src Contribution
to IO by I-Divider
Z '  1
I '0  20
 The V-Src Contribution
by V-Divider
1
 10( A)
11
 Now Deactivate the
I-Source (open it)
1 1  j 
Z "  1 || (1  j ) 
1  1  j 
Engineering-43: Engineering Circuit Analysis
30
Z "  1 || (1  j )
"
Z
V1"  "
60(V )
Z 1 j
Z"
I  V 1  "
60( A)
Z 1 j
"
O
"
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
AC Source SuperPosition cont.2
 Sub for Z”
1 j
2 j
"
I0 
6 ( A)
1 j
1 j
2 j
1 j
I 
6
(1  j )  3  j
"
0
6 6
I   j ( A)
4 4
"
0
 Finally SuperPose the
Response Components
Engineering-43: Engineering Circuit Analysis
31
 The Total Response
3 3
I 0  I  I  1   
2 2
'
0
"
0
5 3
 I 0   j ( A)
2 2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt

j

Multiple Frequencies
 When Sources of Differing FREQUENCIES
excite a ckt then we MUST use SuperPosition
for every set of sources with NON-EQUAL
FREQUENCIES
 An Example
V2
V1
 We Can Denote the Sources as Phasors
V1  100V0
& V2  50V  10
 But canNOT COMBINE them due to
DIFFERENT frequencies
Engineering-43: Engineering Circuit Analysis
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Multiple Frequencies cont.1
 Must Use
SuperPosition for
EACH Different ω
V1
 V1 first (ω = 10 r/s)
Z L ,10  j 10 1
 V2 next (ω = 20 r/s)
Z L , 20  j 20 1
 The Frequency-1
Domain
Phasor-Diagram
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Multiple Frequencies cont.2
 The Frequency-2
Domain
Phasor-Diagram
V2
 Recover the Time
Domain Currents

 Finally SuperPose


it   i' t   i" t   7.07 A cos 10t  45  2.24 A cos 20t  73.43
– Note the MINUS sign from CW-current assumed-Positive
Engineering-43: Engineering Circuit Analysis
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt

Source Transformation
 Source transformation is a good tool to reduce
complexity in a circuit
• WHEN IT CAN BE APPLIED
 “ideal sources” are not good models for
real behavior of sources
• A real battery does not produce infinite current
when short-circuited
 Resistance → Impedance Analogy
ZV
+
-
RV
VS
a
ZI
a
RI
b
IS
b
THE MODELS ARE EQUIVALENT S WHEN
ZV  Z I  Z
RV  RI  R
VS  RI S
VS  ZI S
Improved model
Improved model
for voltage source for current source
Engineering-43: Engineering Circuit Analysis
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Source Transformation
 Same Ckt, But Use
Source Transformation
to Find IO
 Start With I-Src
 Then the Reduced Circuit
V ' 8  2 j
 Next Combine the Voltage
Sources And Xform
Engineering-43: Engineering Circuit Analysis
36
VS'
8 2 j
IS 

Z Series 1  j
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Source Transformation cont
 The Reduced Ckt
 Now Combine the
Series-Parallel
Impedances
1 j 2
Z p  (1  j ) || (1  j ) 
 1
11 j  j
 The Reduced Ckt
 IO by I-Divider
IO  I S 
8 2 j
1 j
Zp
1 4  j 4  j 1  j 


2 1  j 1  j 1  j 
 IO 

5  j3
2
Engineering-43: Engineering Circuit Analysis
37
IS 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
5  j3
2
WhiteBoard Work
 Let’s Work This Nice
Problem to Find VO
Engineering-43: Engineering Circuit Analysis
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
All Done for Today
Charles
Proteus
Steinmetz
Delveloper of Phasor Analysis
Engineering-43: Engineering Circuit Analysis
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Engineering-43: Engineering Circuit Analysis
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
Engineering-43: Engineering Circuit Analysis
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
WhiteBoard Work
 Let’s Work this Nice
Problem
i1 (t)
i2 (t)
3.33 F
20
is(t)
6mH
10
iS t   100mA cos5000t  8.13
 I  100mA8.13
 See Next Slide for
Phasor Diagrams
Engineering-43: Engineering Circuit Analysis
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
P8.29 Phasor Diagrams
 Tip-To-Tail Phasor
(Vector) Addition
I 2  84.98mA109.44
I S  99.8mA8.11
I S  100mA8.13
I 2  84.98mA109.44
I1  143.3mA  27.41
I1  143.3mA  27.41
Engineering-43: Engineering Circuit Analysis
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt
WhiteBoard Work
 Let’s Work Some
Phasor Problems
2
z
4
10mH
Engineering-43: Engineering Circuit Analysis
44
500F
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-44_Lec-08-2_Impedance.ppt