Page 1
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
[email protected]
Engineering 43
Series/Parallel,Dividers,Nodes & Meshes
Page 2
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Series Parallel
Up To Now We Have Studied Circuits That Can Be Analyzed With One Application Of KVL Or KCL
We will see That In Some Situations It Is Advantageous To Combine Resistors To Simplify The Analysis Of A Circuit
Now We Examine Some More Complex Circuits Where We Can Simplify The Analysis Using Techniques:• Combining Resistors• Ohm’s Law
Page 3
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Resistor Equivalents Series
• Resistors Are In Series If TheyCarry Exactly The Same Current
Parallel• Resistors Are In Parallel If
They have Exactly the Same Potential Across Them
NS RRRRR 321
NP RRRRR
11111
321
Page 4
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Conductance Equivalents
ReCall: G = 1/R For SERIES
Connection
For PARALLELConnection
NS
NS
GGGGG
RRRRR
11111
321
321
NP
NP
GGGGG
RRRRR
321
321
11111
GS = 1.479 S
GP = 15 S
Page 5
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Combine ResistorsExample:Find RAB
6k||3k = 2k
(10K,2K)SERIES
SERIESk3
kkk 412||6
k12
k5
(4K,2K)SERIES
kkk 36||6 (3K,9K)SERIES
kkk 312||4
Page 6
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
More Examples Step-1: Series
Reduction Step-2: Parallel
Reduction
k9
9 kΩ
kkk 69||18
kkk 1066
k22
Page 7
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example w/o Redrawing
Step-2: 12k 12k = 6k Step-3: 3k 6k = 2k
Step-1: 4k↔8k = 12k
Step-4: 6k (4k↔2k) = 3k = RAB
kkk 612||12
kkk 26||3
)24(||6 kkk
12k
Page 8
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Series-Parallel Resistor Circuits
Combing Components Can Reduce The Complexity Of A Circuit And Render It Suitable For Analysis Using The Basic Tools Developed So Far• Combining Resistors In SERIES Eliminates
One NODE From The Circuit• Combining Resistors In PARALLEL
Eliminates One LOOP From The Circuit
Page 9
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
S-P Circuit Analysis Strategy
Reduce Complexity Until The Circuit Becomes Simple Enough To Analyze
Use Data From Simplified Circuit To Compute Desired Variables In Original Circuit • Hence Must Keep Track Of Any
Relationship Between Variables
Page 10
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example – Ladder Network
Find All I’s & V’s in Ladder Network• 1st: S-P Reduction
k12kk 12||4
k6
kk 6||63
2
36
IkVk
VI
b
a
3I
• 2nd: S-P Reduction– Also by Ohm’s Law
Page 11
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ladder Network cont.• Final Reduction; Find Calculation Starting
Points
VmAkV
mAkk
VI
a 30.13
0.139
121
VVIkV
mAIIIImAk
V
k
VI
bb
a
5.13
5.05.06
3
6
3
33212
• Now “Back Substitute” Using KVL, KCL, and Ohm’s Law– e.g.; From Before
Page 12
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Voltage Divider
tiRv
tiRv
R
R
2
1
2
1
tiRRtv
tiRtiRtv
21
21
21 RR
tvti
Ohm’s Law in KVL
Find i(t) by
Ohm’s Law
KVL ON THIS LOOP
Page 13
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Voltage Divider cont.
Now Sub i(t) Into Ohm’sLaw to Arrive at TheVoltage Divider Eqns
21
221
1 21and
RR
tvRv
RR
tvRv RR
tvR
tvRv
R
tvvR RR
2
22
1 0and0
000
21
0
00and
00
1112 21
R
tvvtv
R
tvRvR RR
Quick Chk → In Turn, Set R1, R2 to 0
KVL ON THIS LOOP
Page 14
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
V-Divider Summary
Governing Equations
)(21
11
tvRR
RvR
)(21
22
tvRR
RvR
• The Larger the R, The Larger the V-drop
Example• Gain/Volume Control
– R1 is a VariableResistor Called aPotentiometer, or “Pot” for Short
Page 15
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Volume Control Example• Case-I → R1 = 90 kΩ
2.25V93090
30
)(
2
21
22
Vkk
kV
tvRR
RV
V4.593020
30
)(
2
21
22
Vkk
kV
tvRR
RV
• Case-II → R1 = 20 kΩ 30kΩ
9 V
Page 16
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Practical Example Power Line
Also
8.25% of Pwr Generated is Lost to Line Resistance!* How to Reduce Losses?
Power Dissipated by the Line is a LOSS
Using Voltage Divider
kV 367
kV 400Ω 16.5183.5
Ω 183.5load
V MW 7345.183kA2 2
load
load2
load
P
RIP
MW 665.16kA2 2line
line2
loadsrcLOSS-line
P
RIPPP
Page 17
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Equivalent Circuit
The Equivalent Circuit Concept Can Simplify The Analysis Of Circuits• For Example, Consider A Simple
Voltage Divider
+-
1R
2R
Sv
i
21 RR
vi S
+-Sv 21 RR
i
SERIES Resistors → 1R 2Rº
21 RR
– As Far As TheCurrent IsConcerned BothCircuits AreEquivalent The One On The Right Has Only One Resistor
Page 18
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Schematic vs. Physical
Sometimes, For Practical Construction Reasons, Components That Are Electrically Connected May Be Physically Quite Apart• Each Resistor Pair Below Has the SAME
Node-to-Node Series-Equivalent Circuit
Page 19
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
COMPONENT SIDE
CONNECTOR SIDE
ILLUSTRATING THE DIFFERENCEBETWEEN PHYSICAL LAYOUT ANDELECTRICAL CONNECTIONS
PHYSICAL NODE
PHYSICAL NODE
SECTION OF 14.4 KB VOICE/DATA MODEM
CORRESPONDING POINTS
Page 20
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Generalization Multiple v-Sources
Voltage Sources In SeriesCan Be AlgebraicallyAdded To Form AnEquivalent Source• We Select The Reference
Direction To Move AlongThe Path
i(t)+-
+-
+ -
+-
+ -
1R
2R
1Rv
2Rv
1v
2v
3v
4v
5v
01542321 vvvvvvv RR
– Voltage Rises AreSubtracted From Drops
Apply KVL
Page 21
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Multiple v-Source Equivalent
Collect All SOURCES On One Side
The Equivalent Circuit:• V-source in Series
ADD directly
2154321 RR vvvvvvv
21 RReq vvv
+-eqv
1R
2R
Page 22
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Generalization Mult. Resistors
Apply KVL (rise = Σdrops)
tiRv kRik
Now by Ohm’s Law
And Define RS
Then Voltage Division For Multiple Resistors
KVL
tvR
Rv
S
kRk
• [Rk/RS] is the Divider RATIO
Page 23
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example
Find: I, Vbd, P30kΩ
Apply KVL & Ohm
VVIkV bdbd 10 KVLby So 0][2012
WWARIP 3001030)1030()10( 43242
Solving for I
Now Vbd
Finally, The 30 kΩ Resistor Power Dissipation
APPLY KVLTO THIS LOOP
bdV
Page 24
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Examples
Find: I, Vbd
• Use KVL and Ohm’s Law
APPLY KVLTO THIS LOOP
mAIkIkI 05.004012806
VVVkIV bdbd 1001240
V3
VVS 9320
201525
Find VS by V-Divider • The V20k Divider Eqn
201525
203
SVV
• Solving for VS
Page 25
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
When In Doubt → ReDraw
From The Last DiagramIt Was Not ImmediatelyObvious That This Wasa V-Divider Situation• UnTangle/Redraw at Right
VVVV
VVV
adS
Sad
9320
251520
20
251520
or
3251520
20
Page 26
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Single Node-Pair (SNP) Circuits
SNP Circuits Are Characterized By ALL the Elements Having The SAME VOLTAGE Across Them → They Are In PARALLEL
V
SNP Example
V
EXAMPLE OF SINGLE NODE-PAIR
This Element is INACTIVE
• The Inactive Element Has NO Potential Across it → SHORT Circuited
Page 27
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
UnTangling Reminder
Nodes Can Take STRANGE Shapes
LowDistortion
PowerAmplifier
NODE → A region of Constant Electrical Potential
e.g.; a group of connected WIRES is ONE Node
Page 28
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW
SOME PHYSICAL NODES
COMPONENT SIDE CONNECTION SIDE
Page 29
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Current Divider
Basic Circuit
The Current i(t) Enters The Top Node then Splits, or DIVIDES, into the the Currents i1(t) and i2(t)
Apply KCL at Top Node
Use Ohm’s Law to Replace Currents
APPLY KCL
Page 30
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Current Divider cont.
Basic Circuit
By KCL & Ohm The Current Division
Define PARALLEL Resistance
tvR
tip
1
tiRR
RRtv
21
21
pR
Page 31
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Current Divider Example
By I-Divider
For This Ckt Find: I1, I2, Vo
When in doubt… REDRAW the circuit to Better Visualize the Connections
2-Legged Divider is
more Evident
24V
kΩ802
IVo
Page 32
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Real World Example
Car Stereo and Circuit Model
Use I-Divider to Find Current thru the 4Ω Speakers
Thus the Speaker Power
Power Per Speaker by Joule
mA215 mA215
Page 33
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Current & Power Example
Find I2 by I-Divider OR KCL• Choose KCL
For This Ckt Find: • I1, I2,
• P40k Power ABSORBED by 40 kΩ Resistor
By I-Divider
KCL
mA
mAI
21
1640120
1201
mAI
ImAI
4
016
2
12
The 40k Power by RI2
WmWP
millikmilli
mAP
k
k
76.55760
k4012
40
2
240
Page 34
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx34
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Generalization: Multi i-Sources
KCL on Top Node:
Given Single Node-Pair Ckt w/ Multiple Srcs
The Equivalent Ckt Combine Src Terms To Form Equivalent Source
KCL
Page 35
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx35
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Generalization: Multi i-Sources
By Analysis and Electrical Physics of KCL
Thus CURRENT Sources in PARALLEL ADD directly • Compare to VOLTAGE Sources in SERIES
which also ADD Directly
=Oiiiii 6431
Page 36
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx36
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
i-Source Example
Combine Srcs to Yield Equivalent Ckt
For This Ckt Find Vo, and the Power Supplied by the I-Srcs
Vo by Ohm’s Law V10mA5k2 Srcpo IRV
mA10 mA15k3
k6
-
+
OV
kkk
kkRp 2
36
3*6
OVpR
mA5
mW150
5mA1V10
mW100mA10V10
15
10
m
m
P
P
Use PASSIVE SIGN Convention for Power
Page 37
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx37
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Generalization: Multi Resistors
KCL on Top Node:
Given Single Node-Pair Ckt w/ Multiple R’s
Ohm’s Law atEach Resistor
The EquivalentResistance & v(t)
po
N
K Kp
RtitvRR
1
11
tiR
Rti
Rtvti
tiRtv
oK
pK
KK
op
KCL
Page 38
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx38
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Multi-R Example
Find Rp
For This Ckt Find i1, and the Power Supplied by the I-Source
Recall the General Current Divider Eqn
mA8
k4 k20 k5
1i
kΩ2kΩ2
1
kΩ20
415
kΩ5
1
kΩ20
1
kΩ4
11
p
p
R
R
tiR
Rti
Rtvti
tiRtv
oK
pK
KK
op
Page 39
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx39
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Multi-R Example cont.
Find v for Single-Node-Pair by Ohm
Find i1, by Divider • Take Care with
Passive Sign Conv
mA8k4 k20 k5
1i
mA4mA8kΩ4
kΩ21 i
v
V16kΩ2mA8
psrc Riv
mW128mA8V16 srcsrc viP
• Note: this time For Passive Sign Convention CURRENT Direction assigned as POSITIVE Find Psrc by v•i
Page 40
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx40
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Multi-R: Alternative Approach
The Ckt After the R-Combination
Start by Combining R’s NOT associated with i1
Now Have 1:1 Current Divider so
mA8
k4 k20 k5
1i20k||5k
mA8
k4 k41i
mA42
1kΩ4kΩ4
kΩ4
1
1
src
src
ii
ii
Page 41
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx41
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Multi-R, Multi-Isrc SNP
Soln Game Plan: Convert The Problem Into A Basic CURRENT DIVIDER By Combining Sources And Resistors
Given Single Node-Pair Ckt: Find IL
1mA
mA2mA4mA1,
eqSI
Combine Sources• Assume DOWN =
POSITIVEeqSI ,
Page 42
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx42
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Multi-R, Multi-Isrc SNP cont.
Next Combine Parallel Resistors
Given Single Node-Pair Ckt: Find IL
IL by 3:1 I-Divider
Then the Equivalent Circuit →
Note MINUS Sign
Page 43
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx43
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The SAME Ckt Can Look Quite Different
k3
k3
k6
k6
A
B
C
mA9
A
B
C
k6
k6
k3
k3mA9
k6
k3
k6
k3
A
CB
9mA
I1 I2
3mA
3mAmA993
12
1
II
I
I1
I2
I1
I2
Page 44
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx44
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
UnTangling Utility Redrawing A Circuit May, Sometimes, Help To
Better Visualize The Electrical Connections
k3k3 k6k6
A
B
C
mA9I1
I2
• Be FAITHFUL to the Node-Connections
k6
k3
k6
k3
A
CBI1 I2
Page 45
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx45
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Another Example
Alternatives for P• By vi & passive sign:
For This Ckt Find the I-SrcPower, P20
Use ||-Resistance
mA2020 VIVP
k2 k4 k3
mA20
+
V_
• By Joule and Energy Balance
220 mA20 PR RPPj
kΩ13
12
kΩ12
436
kΩ3
1
kΩ4
1
kΩ2
11
p
p
R
R
suppliedmW13
4800
mA20k1312
20
220
P
P
Page 46
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx46
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Nodal Analysis (based on KCL)
A Systematic Technique To Determine Every Voltage and Current in a Circuit
The variables used to describe the circuit will be “Node Voltages”• The voltages of each node Will Be
Determined With Respect To a Pre-selected REFERENCE Node– The Reference Node is Often Referred to as
Ground (GND) Or
COMMON
Page 47
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx47
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Consider Resistor Ladder
Goal: Determine All Currents & Potentials in this “Ladder” Network
Plan• Use Series/Parallel Transformation to Find I1
• Back-Substitute Using KVL, KCL, Ohm to Find Rest
Page 48
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx48
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Series-Parallel Transformations Xform1
• Combine 3 Resistors at End of Network
k6
kk 6||6
3I
k12kk 12||4
][6
][3
2
3
kIV
kIV
a
b
Xform2• Combine 3 Resistors
at End of Network
Note By Ohm’s Law
Page 49
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx49
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Xform cont. Xform3
• To Single-Loop Ckt
VkIV
mAk
VI
a 33
112
12
1
1
mAk
VI a 5.0
62
mAIII 5.0213
Now Back Substitute• Recall
• By KCL
k
VI
12
121
Page 50
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx50
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Xform cont.
Recall Xform2
mAkVI
VkIV
b
b
375.04
Then
5.13
4
3
VkIV
mAIII
c 375.03
Law sOhm'by Then
125.0
5
435
In Summary• I1 = 1 mA
• I2 = I3 = 0.5 mA
• I4 = 0.375 mA
• I5 =0.125 mA
• Va = 3 V
• Vb = 1.5 V
• Vc = 0.375 V
Finally by KCL
Page 51
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx51
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Node Analysis Perspective
ba
ba
VVV
VVV
3
3 0
KVL KVL KVL
REFERENCE
as
as
VVV
VVV
1
1 0
cb
cb
VVV
VVV
5
5 0
In General: Vx5 = Vx−V5 = Vx−0 = Vx • Then the KVL Eqns
Take Node-5 As the Ref, →V5 = 0, Always
Page 52
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx52
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Node Analysis cont
If We Know Va, Vb, and Vc, Then • Can Calc V1, V2, V3 by KVL, Then
– Use Ohm’s Law to Find I1→I5
i.e., If we Know All Node Potentials, Then Can Calc All Branch Currents
Theorem: IF ALL NODE VOLTAGES WITH RESPECT TO A COMMON REFERENCE NODE ARE KNOWN, THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL VARIABLE FOR THE CIRCUIT
Page 53
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx53
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ALWAYS Define Reference Node
The Statement V1 = 4V is Meaningless• UNTIL The
Designation of a REFERENCE NODE
V4
V2
12V
VVVVVV 6242112
By Convention The Ground (GND) Symbol Indicates the Reference Point• ALL Node Voltages
are Measured Relative to GND
Page 54
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx54
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Strategy for Node Analysis1. Identify All Nodes And
Select A Ref. Node
2. Identify Known Node Voltages
3. at Each Node With Unknown Voltage Write A KCL Equation• e.g., (Sum Of
Current Leaving) =0
4. Replace Currents in Terms Of Node V’s
0:@ 321 IIIVa
0369
k
VV
k
V
k
VV baaaS
0:@ 543 IIIVb
0:@ 65 IIVc
0943
k
VV
k
V
k
VV cbbba
039
k
V
k
VV ccb
Yields Algebraic Eqns In The Node Voltages
Final Desired Eqn Set
REFERENCE
SV
aV
bV
cV
Page 55
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx55
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Node Equation Mechanics
When Writing Node Equations• At Each Node We
Can Choose Arbitrary Directions for Currents
• Then select any form of KCL
When The Currents Are Replaced In Terms Of The Node Voltages The Node Eqns That Result Are The Same Or Equiv.
a b c
d
aV
bV
cV
dV
1R 3R
2R2I3I1I
00
0LEAVING CURRENTS
321321
R
VV
R
VV
R
VVIII cbdbba
00
0 NODE INTO CURRENTS
321321
R
VV
R
VV
R
VVIII cbdbba
Page 56
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx56
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Node Eqn Mechanics cont. When Writing The Node Equations
• Use Ohm’s Law to Write The Equation Directly In Terms Of The Node Voltages
• BY Default Use KCL In The Form Sum-of-currents-leaving = 0– But The Reference Direction For The Currents
Does NOT Affect The Node Equation
a b c
d
aV
bV
cV
dV
1R 3R
2R '2I
'3I'
1I
00
0LEAVING CURRENTS
321
'3
'2
'1
R
VV
R
VV
R
VVIII bcdbab
00
0 NODE INTO CURRENTS
321
'3
'2
'1
R
VV
R
VV
R
VVIII bcdbab
Page 57
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx57
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ckts w/ Independent Sources
At Node-1• Using Resistances
00
2
21
1
1
R
vv
R
viA
Using Conductances Eliminates Tedious Division Operations
Replacing R’s w/ G’s• At Node-1
0)( 21211 vvGvGiA
• At Node-2
0OUTi
Page 58
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx58
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Node Analysis of Indep Src Ckts
ReOrder Terms in Eqns for iA & iB
The Manipulation Of Systems Of Algebraic Equations Can Be Efficiently Done Using Matrix Analysis• c.f., MTH-6 or ENGR-25 (MATLAB)
The Model For The Circuit is a System Of Algebraic Equations
Page 59
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx59
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example Write the KCL Eqns
• @ Node-1 We Visualize The Currents Leaving And Write the KCL Eqn
Similarly at Node-2
03
12
4
122
R
vv
R
vvi
• Could Use (i Entering Node) Just as well
Page 60
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx60
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
KCL Eqn Example Write KCL At
Each Node In Terms Of Node Voltages• 3 Nodes Implies 2
KCL Equations
mA15
A
B
C
k8 k8k2 k2AV
BV
Mark the nodes (to insure that None is missing)
Select C asReference
01582
@ mAk
V
k
VA AA
01528
@ mAk
V
k
VB BB
VV
VV
B
A
24
24
Solving by Algebra, Find:
Two simple eqns in Two Unknowns
Page 61
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx61
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Linear Algebra Analysis
kRRkR
mAimAi BA
6,12
4,1
Given
321
Recall R=1/G, Then Insert Numerical Values, and Change to Time Independent Notation (All CAPS)
The Math Model
The Node Eqns in Conductance Form
Page 62
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx62
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Linear Algebra Analysis cont. The Numerical Model
Multiply the 1st Eqn by 4kΩ to Find V1 in Terms of V2
Back Sub into 2nd eqn
Then V2
And V1
Alternatively, Multiply Both Sides of Math Model by LCD in kΩ
k12
k6
• R.H.S. of Eqn Now in Volts
• V1, V2 CoEffs are No.s
Page 63
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx63
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Linear Algebra Analysis cont. The “Clean” Eqns
2242
11223
21
21
VVV
VVV
Proceed with Gaussian Elimination• Add Eqns to Eliminate V2
VV
VV
6
122
1
1
• Back Substitute to Find V2
VV
VV
VVV
15
302
2426
2
2
2
V6
V15
Page 64
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx64
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Use Matrix Algebra Recall The Math Model
From MTH-6 the Form of Matrix Multiplication
IGV • In this Case
G V I
The Matrix Eqn Soln
IGV 1• In this Case
Calculating the Matrix Inverse, G-1, is NOT Trivial• Use Matrix Manipulation
– Adjoint Matrix– Determinant Calculation
• Or use MATLAB
Page 65
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx65
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
GV = I By MATLAB
Construct the Coefficient Matrix G >> G = [1/4e3 -1/6e3;
-1/6e3 1/3e3]
G =
1.0e-003 *
0.2500 -0.1667 -0.1667 0.3333
Construct the Constraint Vector, I
>> I = [1e-3; -4e-3]I = 0.0010 -0.0040
Matrix Inversion by “Left” Division for V
>> V = G\IV = -6 -15
Page 66
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx66
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example Ckt w/ V-controlled Isrc
Write Node Equations
Treat Dependent Source as a Normal Current-Source• Node Eqns
Express Controlling Variable In Terms Of Node Voltages
Page 67
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx67
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example Ckt w/ V-controlled Isrc
4 Eqns in 4 Unknowns• Solve Using Most
Convenient Method– Choose SUB &
GAUSSIAN ELIM
Sub for vxin •vx Isrc
Continue w/ Gaussian Elim OR UseMatrix Algebra
Page 68
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx68
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solve Using MATLAB
Define Components (m-file Node_Anal_0602.m)
]/[2
,4,2,4
,2,1
4
321
VA
mAimAikR
kRRkR
BA
R1 = 1000; R2 = 2000; R3 = 2000;R4 = 4000; %resistances in OhmsiA = 0.002; iB = 0.004; %sources in AmpsAlpha = 2; %gain of dependent source in Siemens
Page 69
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx69
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solve Using MATLAB cont
Define Coefficient Matrix
]/[2
,4,2,4
,2,1
4
321
VA
mAimAikR
kRRkR
BA
G=[(1/R1+1/R2), -1/R1, 0; % first Matrix row-1/R1,(1/R1+alpha+1/R2),-(alpha+1/R2); % 2nd row0, -1/R2,(1/R2+1/R4)] %third row.
G = 0.0015 -0.0010 0 -0.0010 2.0015 -2.0005 0 -0.0005 0.0008
Page 70
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx70
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solve Using MATLAB cont
Define Constraint Vector
]/[2
,4,2,4
,2,1
4
321
VA
mAimAikR
kRRkR
BA
I=[iA;-iA;iB]; Solve by Left/Back Division; V in voltsV=G\I % end with carriage return and get the ReadBackV = 11.9940 15.9910 15.9940
V 994.11 V 991.15V 994.15
Page 71
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx71
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loop Analysis (Based on KVL)
Loop Analysis is The 2nd Systematic Technique To Determine All Currents And Voltages In A Circuit• IT Is the DUAL To Node Analysis
– It First Determines All Currents In A Circuit And Then It Uses Ohm’s Law To Compute Voltages
• There Are Situations Where Node Analysis Is Not An Efficient Technique And Where The Number Of Equations Required By Loop Analysis Is Significantly Smaller
Page 72
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx72
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loop Analysis Illustration
Apply Node Analysis Observe
Need 3 Eqns to Find All Node Potentials; 24
Notice There is Only ONE Current Flowing Thru All Components• A Single Loop Ckt• Can Use Ohm’s Law to
Determine Voltages
Apply KVL for Clockwise Loop Starting at GND
+-
+-
1R 2R
3RV18
V12
2RV 1RV
3RV
I
3V2V1V
4V
GND
0][18][12 321 RRR VVVVV
• Have 4 Non-Ref Nodes• One SuperNode• One Node Connected to
GND Thru a V-src
Page 73
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx73
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loop Analysis Illustration cont
Now Use Ohm’s Law to Express V’s In Terms of the Loop Current
By KVL
Note:• Recalling that V=IR Allows
Writing the Ohm Eqn “by Inspection” for a Single Loop Ckt
The Loop Generates a SINGLE Eqn to Yield the Loop CURRENT
+-
+-
1R 2R
3RV18
V12
2RV 1RV
3RV
3V2V1V
4V
GND
KVL
0][18][12 321 IRVIRIRV
I
Page 74
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx74
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loops, Meshes, Loop-Currents
Each Component Is Characterized By The • VOLTAGE
ACROSS It• CURRENT THRU It
A Loop Is A Closed Path That Does Not Go Twice Over Any Node
This Circuit Has 3 Loops1. fabef
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
1I
2I
3I
2. ebcde
3. fabcdef
Page 75
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx75
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loops, Meshes cont.
A MESH is a LOOP That Does Not Enclose Any OTHER Loop• This Ckt Has
Meshes– fabef
A Loop Current is a Fictitious or Virtual Current That is Assumed to Flow Around a Loop• The Loop Currents
of This Ckt– I1, I2, I3
Mesh Current = Current Within a Mesh Loop• e.g.: I1, I2
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
1I
2I
3I
– ebcde
Page 76
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx76
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loops, Meshes cont.
Claim• In a Circuit, the
Current Through Any Component Can Be Expressed In Terms of the (perhaps multiple) Loop Currents
Ckt Examples
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
1I
2I
3I32
21
31
III
III
III
cb
eb
fa
• The DIRECTION Of The Loop Currents is SIGNIFICANT
• FACT– Not ALL Loop
Currents are Required To Compute All The Currents Through Components
Page 77
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx77
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Loops, Meshes cont.
For Every Circuit There is a MINIMUM Number of Loop Currents Needed to Find Every Current in the System
• Such A Collection is Called the “MINIMAL SET” (of Loop Currents)
For a Given Circuit Let• B Number of
BRANCHES• N Number of NODES
The Minimum Number of Loop Currents is
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
1I
3I
)1( NBL
Page 78
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx78
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Illustration For This Ckt
• B = 7• N = 6• L = 7-(6-1) = 2
Need Two Loop Currents• The Currents Shown
are MESH Currents– Hence They are
Independent and form a Minimal Set
Determination of Loop Currents• KVL on Left Mesh
• KVL on Right Mesh
• Using Ohm’s Law
Page 79
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx79
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Illustration cont. Substituting and
Rearranging
We Obtain in MATRIX FORM the Loop Equations for This Circuit
Page 80
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx80
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Mesh Practice Write The Mesh
Equations Some Bookkeeping
• 8 BRANCHES• 7 NODES• L = 8-(7-1) = 2
– Need TWO Mesh Currents
This is MESH Current Practice• Choose as the
Two Loops Meshes– i1– i2
Identify All Voltage Drops
KVL on Bottom Mesh
4Rv
1Rv
2Rv
5Rv
3Rv
02211 RSRS vvvv
033452 RSRRR vvvvv
KVL on Top Mesh
Page 81
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx81
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Mesh Practice cont. Now Use Ohm’s Law To
Find The Mesh Current Equation Set
4Rv
1Rv
2Rv
5Rv
3Rv
11Ri
52Ri
42Ri
221 )( Rii
32Ri
Page 82
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx82
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Develop a ShortCut Whenever An Element
Has More Than One Loop Current Flowing Through It We Calculate NET Current In The DIRECTION of TRAVEL
Draw The Mesh Currents• Orientation can be
arbitrary, But Conventionally Defined as CLOCKWISE
NOW • Write KVL For Each Mesh • Apply Ohm’s Law To Every
Resistor
+-
+ -
V 1
V 2R 1
R 2 R 3
R 4R 5
WRITE THE MESH EQUATIONS
1I
2I
Page 83
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx83
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Develop a ShortCut cont At Each Loop must Follow
The Passive Sign Convention Using Loop Current REFERENCE DIRECTION• This Defines the Polarity of
the Voltage Drops
Then KVL for Meshes 1 & 2
Note The NET Currents
+-
+ -
V 1
V 2R 1
R 2 R 3
R 4R 5
WRITE THE MESH EQUATIONS
1I
2I
0)( 51221111 RIRIIRIV
0)( 21242322 RIIRIRIV
)( 21 II
)( 12 II
Page 84
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx84
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example Use Loop Analysis
to Find Io
SHORTCUT: • POLARITIES ARE
NOT NEEDED.• Apply Ohm’s Law To Each
Element As KVL Is being written
KVL for Meshes 1 & 2
Collect Like Terms & Solve
mAIVkI
kIkI
kIkI
21612
-----------------------
Add and2396
12612
22
21
21
mAIkIkI4
561212 121
mAIIIo 4
321
Page 85
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx85
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example - Alternate Alternative Loop Current
Selection KVL for Mesh1 & Loop2
Collect Like Terms & Solve In This Case one mesh
and one loop• Io = I1
– This Selection is More Efficient than 2 small meshes; Only Need to Find l1
oImAIkI
kIkI
kIkI
4
31824
-----------------
Substract and2996
312612
11
21
21
Page 86
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx86
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Circuits w/ Indep. I-Sources Find Both Vo & V1
There is NO Relationship Between V1 and the 2 mA Source Current However ...• The Mesh-1 Current is
CONSTRAINED by the 2mA Source– Thus the Mesh-1 Eqn
mAI 21
In General• Current Sources That Are
NOT SHARED By Other Meshes (Or Loops) Serve To DEFINE a Mesh (Loop) Current And Reduce The Number Of Required Equations
Page 87
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx87
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Circuits w/ Indep. I-Sources cont The Mesh-2 KVL Eqn
Then
ReArranging Find Equivalent Eqn
VV
mAkVmAkV
5.10
75.06224
1
1
To Obtain V1 Apply KVL To Any Closed Path That Includes V1
VkIkI 282 21 Use I1 Constraint
to Calculate I2
][2
96
4
3
8
2)2(2
2
2
VIkV
mAk
VmAkI
O
211 6240 IkVIkV
KVL
Page 88
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx88
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example
Two Mesh Currents Are Defined By Current Sources
Only Need Eqn for Mesh-3
mAImAI 24 21
Page 89
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx89
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont
Collect Terms for eqn-3 Finally Use KVL to Calculate VoVkIkIkI 31242 321
mAk
mAkmAkVI
4
1
12
)2(4)4(233
Then I3
KVL FOR Vo
036 3 oVVkI
V 2
33mA
4
16
VkVo
Page 90
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx90
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
STOP Here if Short on Time
Page 91
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx91
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Eqns by “Inspection” If The Circuit Contains
Only INDEPENDENT VOLTAGE Sources Then The Mesh Equations Can Be Written “By Inspection”• MUST HAVE All Mesh
Currents With The Same Orientation
In loop “k”• The Coefficient Of Ik Is The
Sum Of Resistances Around The Loop
The Right Hand Side Is The Algebraic Sum Of Voltage Sources Around The Loop• VoltageRise = POSITIVE
on R.H.S. of eqn
Page 92
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx92
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Eqns by “Inspection” cont The Coefficient Of Ij Is
The Sum Of Resistances COMMON To Both k and j and With a NEGATIVE Sign
In This Example• Loop1: k = 1, j = 2
VkIIk 12612 21
VIkIk 396 21
• Loop2: k = 2, j = 1
Page 93
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx93
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Equation Practice Loop-1
• Coefficient of I1
Similarly The Coeffs for Loop-2
• Coefficient of I2
V
kI
kkI
I
6RHS
3coeff
39coeff
0coeff
3
2
1
kkI 641
02 I
kI 63
][6 V
• Coefficient of I3
• Right Hand Side (RHS)
In Summary for Loop1
Page 94
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx94
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Equation Practice cont.
In Summary for Loop-2
Applying the Method to Loop-3 Yields
Solve 3-Eqns in 3-Unknowns Using Normal Linear Algebra, or MATLAB, Techniques
Page 95
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx95
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example Use Mesh Eqns to
Determine Vo
1. Draw the Mesh Currents
2. Write the Mesh Eqns for Mesh-1 & Mesh-2
1I 2I
][32)242( 21 VkIIkkk
)36()62(2 21 VVIkkkI
Divide Both Sides of Both Eqns by 1kΩ• Units on RHS become
V/kΩ, or mA
Solve System of Eqns
VkIV
mAImAI
mAII
mAII
o 5
336then
10
113330
-----------------
Add and4982
328
2
22
21
21
Page 96
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx96
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example 1. Draw the Mesh
Currents
2. Write Mesh Eqns by KVL
Or Eqns by Inspection
12k
6k
4k 4k
2k
12V
9V
WRITE THE MESH EQUATIONS
1I
2I
3I
4I
0)(61212 :1 MESH 311 IIkVkI
0)(4)(412 :2 MESH 3242 IIkIIkV
0)(4)(69 :3 MESH 2313 IIkIIkV
02)(49 :4 MESH 424 kIIIkV
VkIkI 12618 31
VkIkIkI 12448 432
VkIkIkI 91046 321
VkIkI 964 42
Calculate Currents Using Multi-Eqn Solver Tools• 4 Eqns in 4 unknowns
– Solve using Standard Linear Algebra Techniques
– Perfect for MATLABVI
R
Page 97
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx97
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
All Done for Today
A DifferentType ofNODE
BRANCHES Connect to NODES
Page 98
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx98
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard Work
Let’s Use KCL to Derive the Req for N Parallel Resistors
Done Previously
Page 99
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx99
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
[email protected]
Engineering 43
AppendixΔ↔WYE
& others
Page 100
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx100
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Recall Passive Sign Convention
i
Rv 'Rv
'i
R
vvi Nm LAW SOHM'
R
vvi mN 'LAW SOHM'
iivv RR ' Thus '
If V’ Drops R←L• i’ by Passive Sign
Convention
If V Drops L→R• i by Passive Sign
Convention
Page 101
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx101
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Single Loop Ckts - Background
Using KVL And KCL We Can Write Enough Equations To Analyze ANY Linear Circuit
Begin The Study Of Systematic And Efficient Ways Of Using The Fundamental KCL & KVL Circuit Laws• This Time →
Single LOOP Circuits
Page 102
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx102
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Power of Loop Analysis Consider this
Circuit
ab c
def
1
2 3
4
56
6 branches6 nodes1 loop
ALL ELEMEN TS I N SER I ESON LY ON E CUR R EN T
Alternative Analyses• Write N-1 (5) KCL Equations• Determine Only the SINGLE Loop Current
– An Easy Choice
The Plan for Loop Analysis• Begin With The Simplest One-Loop Circuit• Extend Results To Multiple-Source
and Multiple-Resistor Circuits
Page 103
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx103
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
“Inverse” Voltage Divider
The Std V-Divider1R
+- 2RSV
OV
SO VRR
RV
21
2
Inverse
Divider OS VR
RRV
2
21
kV 500 458.3kV220
220202
21
S
OS
V
VR
RRV
• Use Inverse Divider
Example Find VS
Page 104
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx104
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example
Find• Vx, Vab
• P3Vx
– The Power Absorbedor Supplied By theDependent Source
Apply KVL VVVV XXX 203412
I
abV
+
-
xV3-+ V4
k4
-
+
XV
ab
SV
VVS 12=+ -
VVVV abXab 10034
VVVVV
VVV
abab
XSab
100212
0
1
2
3
1
2
3
IVP XVX3)3(
mAk
VI 1
4
4
mWmAVPXV 6][1][23)3(
Find DS Power• Passive Sign Conven.
• Ohm’s Law
• Then Pwr ABSORBED
Page 105
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx105
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example
Find: VDA, VCD, I• Apply KVL & Ohm
k30+-
+ -
V9k20
k10
A B C
DE
I
mAk
VI
k*Ik*Ik*I-
05.060
3
0103092012
VVIkV DADA 5.110*1012
12
1
2
VmAkIkVCD 5.105.030*30
12V
• Ohm’s Law
Page 106
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx106
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard Work
Let’s Work This Problem
1 2 0 m A 4 k 4 k
8 k
I o
Page 107
[email protected] • ENGR-43_Lec-02a_SP_VI-Divide_NodeMesh.pptx107
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inverse Series Parallel Combo
Find R: Simple Case• Constraints
– VR = 600 mV
– I = 3A– Only 0.1Ω R’s Available
Recall R = V/I 2.0
3
6.0
A
VR
1.01.02 availRR
Since R>Ravail, Then Need to Run in Series
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
More Complex Case Find R for Constraints
• VR = 600 mV
• I = 9A• Only 0.1Ω Resistors Available
m 67.660667.09
6.0
A
VR
R 0.1║(0.1↔0.1)
Either of These 0.1Ω R-Networks Will Work
33.33 mΩ
Page 109
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Effect Of Resistor Tolerance The R Spec: 2.7k, ±10% What Are the Ranges for
Current & Power?• I = V/R
mAk
VImA
k
VImA
k
VInom 115.4
7.29.0
10367.3
7.21.1
10704.3
7.2
10maxmin
mW
k
VPmW
k
VPmW
k
VPnom 15.41
7.29.0
1067.33
7.21.1
1004.37
7.2
10 2
max
2
min
2
• P = V2/R
• For Both I & P the Tolerance.: -9.1%, +11.1% – Asymmetry Due to Inverse Dependence on R
Page 110
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Final Example
Find VS
• Straight-Forward
VB
IBIS
VmAkVIkVV
mAmAmAImAI
mAk
V
k
VI
VmAkV
SBS
BS
BB
B
915.020620
15.005.01.01.0
05.0120
6
120
61.060
• Or, Recognize As Inverse V-Divider
Page 111
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Final Example cont
Inverse DividerCalculation
VB
IBIS
Vk
kkV
R
kRVV
kkkR
VmAkV
BS
B
940
20406
20
40120||60
Before As61.060
||
||
||
Page 112
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Wye↔ Transformations
This Circuit Has NoSeries or ParallelResistors
If We Could MakeThe Change Below Would Have Series-Parallel Case
Page 113
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Y↔ Xforms cont Then the Circuit Would Appear as Below and
We Could Apply the Previous Techniques
Page 114
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Wye↔ Transform Eqns
→Y (pg. 58)
→Y
←YR ac =
R 1||(
R 2↔
R 3) R
ab = Ra + R
b
321
13
321
32
321
21
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
a
accbba
c
accbba
b
accbba
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
3
2
1
←Y (pg. 58)
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
↔Y Application Example
Find IS
a
c
1R
2R
3R
Connection
Calc IS
Req
mAk
V
R
VI
kkkkkkR
eq
SS
eq
2.110
12
10)62(||936
• Use the →Y Eqns toArrive at The ReducedDiagram Below
a
c
b
b
Page 116
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Another ↔Y Example
For this Ckt Find Vo
Convert this Y to Delta
Keep This Node-Pair
Page 117
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Another ↔Y Example cont Notice for Y→Δ in
this Case Ra = Rb = Rc = 12 kΩ• Only need to Calc
ONE Conversion
k
k
kk
R
RRRRRRRRR
b
accbba 3612
12123321
The Xformed Ckt
4mA 36k
36k
36k
12k
12kOV
• ||-R’s Form a Current Divider
Page 118
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Another ↔Y Example cont
The Ckt After ||-Reductions
4mA 36k
36 ||12 9k k k
OV
9kOI
Can Easily Calc the Current That Produces Vo
36 84
36 18 3O
kI mA mA
k k
Then Finally Vo by Ohm’s Law
89 9 24
3O OV k I k mA V
Page 119
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
“BackSubbing” Example
Given I4 = 0.5 mA, Find VO
mA5.0
Page 120
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
BackSubbing Strategy
Always ask: What More Can I Calculate?• In the Previous Example Using Ohm’s Law,
KVL, KCL, S-P Combinations - Calc:
2
432
3
4
2
3
6
IkV
IIIk
VI
IkV
a
b
b
11
521
5
46
4
IkVIkV
IIIk
VI
VVV
xzO
xz
baxz
mA5.0