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Basic Circuit Analysis Methods
(KVL and KCL method, Node method)
6.002xCIRCUITS ANDELECTRONICS
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Remember, our EECS playgroun
Review
Observe the lumpedmatter discipline LMD
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Lumped circuit element+
-
power consumed by element =
Review
i
v
vi
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LMD allows us to create thelumped circuit abstraction
Review
+!! R1
R3R2
V
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KVL:
For all loops
KCL:
For all nodes
0=j j
0=j ji
ReviewReview
Maxwells equations simplify toalgebraic KVL and KCL under LMD!
dlE =
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DE
+!!
Review
R1 R4R3
R2 R5
d
c
b
a
V0
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Lets Begin by Building aToolchest ofAnalysis Techniques
+!!
R1 R4
R3
R2 R5
Analyzingcircuit me
Find allelemenand is
V0
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Method 1: Basic KVL, KCL method ofCircuit analysis
Goal: Find all element vs and is
1. write element v-i relationships(from lumped circuit abstraction)
2. write KCL for all nodes
3. write KVL for all loopslots of unknownslots of equationslots of fun
solve
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!
i+
-
Element e
Current is taken to bepositive going into thepositive voltage terminal
Method 1: Basic KVL, KCL method of Circuit analy
Then powerconsumedby element e
is = vi
This
convecalledAssocvaria
discip
Goal: Find all element vs and isLabeling element vs and is
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Method 1: Basic KVL, KCL method ofCircuit analysis
For R
For voltage source
For current source
You will need this for step 1: Element Relations
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Lets Apply KVL, KCL Method to this
The Demo Circuit
+!!
R1 R4
R3
R2 R5
V0
Goal: Find all element vs and is
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KVL, KCL Example
a
d
c
+!!
Note the use of associated var
R4
R5R2
R1
R3bV0
Label all vs and is
0
12 u
a
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Step 1 of KVL, KCL Method12 unknowns
1. Element relationshipsiv,
c
1
+
2
+
0i
+!!V00+
a
5050, ii
L1
a
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Step 2 of KVL, KCL Method
12 unknowns
2. KCL at the nodes
(use conventiosum currents the node)
5050, ii
c
1
+
2
+
0i
+!!V00+
a
L1
a
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Step 3 of KVL, KCL Method12 unknowns
3. KVL for loops
(use conventyou go arounassign first sign to each
5050, ii
c
1
+
2
+
0i
+!!V00+
a
L1
KVL KCL M th d
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KVL, KCL Method1. Element v, i relationships
v0 = V0v1 = i1R1
v2 = i2R2
v3 = i3R3v4 = i4R4
v5 = i5R5
2. KCL at the nodes
redundant
0410=++ iii
0132=+ iii
0435=
iii
0520= iii
a:
b:
d:
c:
3. KVL for loops
431=+ vvv
210=++ vvv
253
=+ vvv
540=++ vvv
L1:
L2:
L3:
L4:
Method 3 the node method will be much better!
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Other Analysis MethodsMethod 2 Apply element combination rules
A R1 R2 R3 RN
B G2G1 GN
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Method 2 Apply element combination ru
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Method 2 Apply element combination ru
+!!
Example
R1
R3R2
V
Method 3 Node analysis
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1.
2.
3.
4.
5.
Select reference node ( ground) fromwhich voltages are measured.
Label voltages of remaining nodes withrespect to ground. These are theprimary unknowns.
Write KCL for all but the ground node,substituting device laws and KVL.
Solve for node voltages.
Back solve for branch voltages andcurrents (i.e., the secondary unknowns).
ParticularapplicatioKVL, KCLmethod
Method 3 Node analysis
6.002xworkhorse
Method 3 Node analysis
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Example: Old Faithful, plus current source
+!!
R1 R4R3
R2R5
V0
Method 3 Node analysis
1. Select rground n
2. Label nowith resground.
g
I1
St 3 f N d M th dV0
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Step 3 of Node Method
For convenience, writei
iR
G1
=
To avoid mistakes, use convention
E.g., always sum the currents leaving a node
+!! e1
R
V0
V0
3. Writnodes
devic
Step 4 of Node Method
V0
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Step 4 of Node Method
Move constant terms to RHS & collect unknowns
2 equations, 2 unknowns Solve for es
(compare units)
0)()()( 21321101 =++ GeGeeGVe
0)()()( 152402312 =++ IGeGVeGee
KCL at e1
KCL at e2
4. Solvvolt
+!! e1
R
V0
0
Step 5 of Node Method
V0
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Step 5 of Node Method
5. Bacbran
curr
+!! e1
R
V0
0
e1 e2Once you have solved for and ,easy to find branch vs and is
For example:
i+
v1
Revisit Step 4 of Node Method for Cultural Intere
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In matrix form:
=
++
++
04
1
2
1
5433
3321
VG
VG
e
e
GGGG
GGGG
conductivitymatrix
unknownnode
voltages
Revisit Step 4 of Node Method for Cultural Intere
)()()( 10323211 GVGeGGGe =+++
140543231 )()()( IGVGGGeGe +=+++
4. Solve for no
Step 4 of Node Method 4 S l f
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+
=
++
++
104
01
2
1
5433
3321
IVG
VG
e
e
GGGG
GGGG
( )( ) 23543321
104
01
3213
3543
2
1
GGGGGGG
IVG
VG
GGGG
GGGG
e
e
++++
+
++
++
=
Solve
5G
3G
4G
3G
2
3G
5G
2G
4G
2G
3G
2G
5G1
G4
G1
G3
G1
G
1I0V4G3G0V1G5G4G3G
1e
++++++++
++++
=
( )( ) ( )( )
5343
2
3524232514131
104321013
2
GGGGGGGGGGGGGGGGG
IVGGGGVGGe
++++++++
++++
=
(same denominator)
Nin
nein th
Step 4 of Node Method 4. Solve for no
Step 4 of Node Method V0
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E.g., solve for , given
K2.8
1
G
G
5
1=
K9.3
1
G
G
4
2=
K5.1
1G
3= 0
1=I
026.0 Ve =
If , thenVV 30= 02
8.1 Ve =
Step 4 of Node Method
+!! e1
R
V0
e2