ELT221
1
Nodal & Loop Analysis - Nodal Analysis - Loop Analysis - Mesh Analysis
Enzo Paterno
EP Revision 12/9/14
ELT221
2
NODAL ANALYSIS
Enzo Paterno
In a nodal analysis, the variables in the circuit are selected to be the node voltages. The node voltages are defined with respect to a common point in the circuit (i.e. ground) and assumed to be positive voltages. For example the circuit below has four node voltages, (VS , Va , Vb , Vc ) with respect to the ground node.
Once the node voltages in a circuit are defined , we can calculate the current through any resistive element using Ohm’s law:
RVV
I NN −+ −=
VS Va Vb Vc + - I R
+
ELT221
3
NODAL ANALYSIS
Enzo Paterno
In a nodal analysis, KCL equations are used so that the variables are the unknown node voltages. For a network consisting of N nodes, N 1 Linearly Independent, (LI), KCL equations are required to determine the N 1 node voltages (i.e. Nth node is ground). Example: 3 nodes 2 LI equations required
B
B
BB
iRR
VRV
iRV
RVV
iiiiiiv
=
+−
=−−
=−⇒+=
322
2
1
3
2
2
21
32322
11
:@
A
A
A
iR
VRR
V
iR
VVRV
iiiv
=
−
+
=−
+
+=
22
211
2
21
1
1
211
111
:@
ELT221
4
NODAL ANALYSIS
Enzo Paterno
Let R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ, IA = 1 MA, and IB = 4 mA.
321
321
22
211
1
1016
141
10161
61
121
111
:@
−
−
=−
+
=
−
+
=
−
+
xVk
Vk
xk
Vkk
V
iR
VRR
V
v
A
321
321
322
21
2
1043
161
10461
61
61
111
:@
−
−
=−
+
=
+−
=
+−
xVk
Vk
xkk
Vk
V
iRR
VR
V
v
B
ELT221
5
NODAL ANALYSIS
Enzo Paterno
321 101
61
41 −=
−+ xV
kV
k3
21 1043
161 −=
−+ xV
kV
k
We have a 2x2 linear system. Solve for V1 and V2. How do we solve an n x n linear system?
TI-83 Plus
Method of substitution Method of elimination Gaussian elimination Cramer’s rule Matrix SW calculator MATLAB software tool JAVA Applet
ELT221
Enzo Paterno 6
GAUSSIAN ELIMINATION
Two simultaneous linear equations form a 2x2 linear system:
This 2x2 linear system can be represented in matrix form:
[ ][ ] [ ]KxMKK
DCBA
x
x=→
=
2
1
2
1
Using matrix theory, the solution to this linear system can be found by:
[M]-1 is the inverse of matrix [M]
M is a square matrix, x and K are column matrices (vectors)
A x1 + B x2 = K1 C x1 + D x2 = K2
[ ] [ ] [ ]KMx1−
=
ELT221
Enzo Paterno 7
Below is a general system of m equations with n unknowns:
This system can be expressed as a matrix equation: Ax = b
The system solution is given by: x = A-1 b
GAUSSIAN ELIMINATION
ELT221
8
MATRIX ALGEBRA REVIEW
Enzo Paterno
( ) ( ) ( )
[ ]TCAAdj
AAAdjA
MijjiCij
caAA
abBTAaA
ba
m
kkjk
nxmjiijmxnij
m
i
r
j
n
kkjik
=
=
+−=
==
== →=
−
=
= = =
∑
∑∑∑
)(
)()1(
)det(
1
11
1 1 1
(Amn Bnr = Cmr)
)()()det( 12212211
2221
1211
AAAAAA
AAAA
A
−==
=
Matrix Multiplication
Transpose
Determinant
Cofactor
Inverse
Adjoint
−
−=
=
−
1121
12221
2221
1211
1AAAA
AA
AAAA
A
ELT221
9
NODAL ANALYSIS – GAUSSIAN ELIMINATION
Enzo Paterno
321 101
61
41 −=
−+ xV
kV
k3
21 1043
161 −=
−+ xV
kV
k
=
−
−
−
−
3
3
104101
21
31
61
61
41
xx
VV
kk
kk
−
−
=
−
−
−
3
3
1
104101
31
61
61
41
21
xx
kk
kkVV
−−
=
−−
=
−
−
156
104101
5.4336
21
3
3
xx
VV V1 = - 6 V
V2 = - 15 V
ELT221
10
NODAL ANALYSIS – CRAMER’S RULE
Enzo Paterno
V
kk
kk
kx
kx
V 6
31
61
61
41
31104
61101
3
3
1 −=
−
−
−
−
=
−
−
321 101
61
41 −=
−+ xV
kV
k3
21 1043
161 −=
−+ xV
kV
k
V
kk
kk
xk
xk
V 15
31
61
61
41
10461
10141
3
3
2 −=
−
−=
−
−
ELT221
11
NODAL ANALYSIS
Enzo Paterno
SV aV bV cV
0:@
321
321
=++−+=
IIIIIIVa
0:@
543
543
=++−+=
IIIIIIVb
kV
kkkV
kV
kVV
kV
kVV
cba
cbbab
991
41
31
31
0943
=
+++−
=−
++−
kV
kV
kkkV
kVV
kV
kVV
sba
baasa
9331
61
91
0369
=−
++
=−
++−
Vs is a constant Vs = 12 v
031
91
91
39:@
=
+−
=−
kkV
kV
kV
kVVV
cb
ccbC
ELT221
12
NODAL ANALYSIS
Enzo Paterno
VVVVVVVVV
kVV
kV
kVV
5.16)6(124
02
012612
2
2
32212
32212
==−+=
=−++−
=−
++−
VVVV
612
3
1
−==
V1 =12 V V3 = -6 V Only one KCL needed
ELT221
13
NODAL ANALYSIS
Enzo Paterno
V1
V2 V3
ELT221
14
NODAL ANALYSIS
Enzo Paterno
V1
I1
I3 I2
723413:14433672
68412
:@
321
21311
21311
3211
=−−−+−=−
−+
−=
−+=
VVVEqVVVVV
kVV
kVV
kV
IIIV
V2 V3
ELT221
15
NODAL ANALYSIS
Enzo Paterno
V2
I3
I4 I5
V1
V3 06134:2
46644846
:@
321
23221
23221
5432
=+−+−=−
+−
=−
+=
VVVEqVVVVV
kV
kVV
kVV
IIIV
ELT221
16
NODAL ANALYSIS
Enzo Paterno
V1
V2 V3
I2
I4 I6
042:322
884
:@
321
33132
33132
6243
=−+=−+−
=−
+−
=+
VVVEqVVVVVk
VkVV
kVV
IIIV
ELT221 NODAL ANALYSIS
17 Enzo Paterno
=
−−
−−
0072
42161343413
3
2
1
VVV
−−
−−=
−
0072
42161343413 1
3
2
1
VVV
723413:1 321 =−− VVVEq
06134:2 321 =+− VVVEq
042:3 321 =−+ VVVEq
32 VVVo −=
ELT221 NODAL ANALYSIS
18 Enzo Paterno
vVvVvVvV
22.010.432.478.7
0
3
2
1
====
ELT221 NODAL ANALYSIS – Circuits with independent current sources
19 Enzo Paterno
Suppose that the network below has the following parameters: IA = 1 mA, IB = 4 mA, R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ. Determine all node voltages and branch currents.
22
121
2
21
1
1
211
111
:@
vR
vRR
I
Rvv
RvI
IIIV
A
A
A
−
+=
−+=
+=
232
12
232
21
32322
111
1:@
vRR
vR
I
vRR
vvI
IIIIIIV
B
B
BB
+−−=
−−
=
−=⇒+=
ELT221
20 Enzo Paterno
22
121
2
21
1
1
211
111
:@
vR
vRR
I
Rvv
RvI
IIIV
A
A
A
−
+=
−+=
+=
232
12
232
21
32322
111
1:@
vRR
vR
I
vRR
vvI
IIIIIIV
A
B
BB
+−−=
−−
=
−=⇒+=
321 101
61
61
121 −=−
+ xv
kv
kk3
21 10461
61
61 −−=
+−− xv
kkv
k
R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ
NODAL ANALYSIS – Circuits with independent current sources
ELT221
21 Enzo Paterno
−
−−
=
−
−−
−−
−−
3
31
33
33
2
1
104101
1033.1016.1016.1025.
xx
xxxx
VV
NODAL ANALYSIS – Circuits with independent current sources
ELT221
22 Enzo Paterno
Inverse
-6 -15
NODAL ANALYSIS – Circuits with independent current sources
ELT221
23 Enzo Paterno
Figure 3.5
NODAL ANALYSIS – Circuits with independent current sources
ELT221
24 Enzo Paterno
NODAL ANALYSIS – SUPERNODE
Consider the network below. To eliminate the problem of dealing with a current through a voltage source, we generate a dashed surface which here includes the 6v voltage source and is called a supernode, The KCL for the supernode is: 6mA = I1 + I2 + 4 mA
Eq1: 𝑉16𝑘
+ 𝑉212𝑘
= 2mA Eq2: V1 – V2 = 6v V1 = 6 + V2 Sub Eq3 into Eq1: V2 = 4v V1 = 10v
ELT221
25
MESH LOOP ANALYSIS
Enzo Paterno
In a loop analysis, the variables in the circuit are selected to be the branch currents and KVL equations are used so that the variables are the unknown currents. The number of LI KVL equations necessary to determine all the currents in a network with B branches, N nodes is: B – N + 1. For example the circuit below requires 7 – 6 + 1 = 2 LI KVL. We identify two independent mesh loops ABEF and BCDE. A mesh is a loop that does not contain any other loop within it.
ELT221
26 Enzo Paterno
( ) 1323211
132312111
213231111 0:@
S
S
S
vRiRRRivRiRiRiRi
RiRiRiRivABEF
−=+++−−=+−−−=−+−−
MESH LOOP ANALYSIS
( ) 2543231
232524231
313252422 0:@
S
S
S
vRRRiRivRiRiRiRiRiRiRiRiv
BCDE
=++−+=−−−+
=+−−−−
- +
( )( )
( )( )
−=
++−
++−
=++−+−=+++−
2
1
2
1
5433
3321
2543231
1323211
S
S
S
S
vv
ii
RRRRRRRR
vRRRiRivRiRRRi
+ -
ELT221
27 Enzo Paterno
MESH LOOP ANALYSIS
12612066612
21
211
=−=+−−
ikikikikik
+
+
39603663
21
212
=−=−+−−
ikikikikik
39612612
21
21
=−=−
ikikikik
6181212612
21
21
−=+−=−
ikikikik
612 2 =ikmA
ki 5.0
126
2 == mAk
i 25.11215
1 ==
Use Mesh analysis to solve for I1 and I2.
ELT221
28 Enzo Paterno
MESH LOOP ANALYSIS
+
+
39612612
21
21
=−=−
ikikikik
Use Mesh analysis to solve for I1 and I2.
=
−−
312
96612
2
1
ii
kkkk
mAk
ik
i
=
=
−− −
50.025.1
312
96612
2
11
ELT221
29 Enzo Paterno
MESH LOOP ANALYSIS
+
+
Use Mesh analysis to solve for I1 and I2.
mAk
ki 25.1
7209.0
96612
9003.06012.0
1 =−−
=
−−−−
= mAk
ki 5.0
96612
003.06012.012
2 =
−−
=
39612612
21
21
=−=−
ikikikik
ELT221
30 Enzo Paterno
MESH LOOP ANALYSIS
PSPICE Simulation
ELT221
31 Enzo Paterno
MESH LOOP ANALYSIS
i1 i2
Use Mesh analysis to solve for I1 and I2.
+
+
+
+
+
Mesh1 Mesh2
V + -
10055.120555.7100
21
211
=−+=+−−
ikikikikik
010505523
21
1222
=−+=+−−−
ikikikikikik
Mesh1: Mesh2:
ELT221
32 Enzo Paterno
MESH LOOP ANALYSIS
010510055.12
21
21
=−+=−+
ikikikik
mA
kkkkkk
i 10
10555.12
1005100
1 =
−−
−−
=
We use Cramer’s rule to find the currents i1 and i2:
Remark: This step could have been solved using Elimination by addition or Elimination by substitution techniques
bcaddcba
−=
2x2 determinant
mA
kkkk
kk
i 5
10555.1205
1005.12
2 =
−−
=
ELT221
33 Enzo Paterno
MESH LOOP ANALYSIS
i1
i2
Use Ohm’s law to solve for I1 and I2.
+
+
+
+
+ V + -
i2
RT = 7.5k + [ 5k // (3k + 2k) ] = 10k I1 = 100v / 10k = 10 mA I2 = 10 mA / 2 = 5 mA
ELT221
34 Enzo Paterno
MESH LOOP ANALYSIS
i1 i2 VOUT +
-
Mesh1 Mesh2
+
+
+
+
Use Mesh analysis to solve for VOUT.
V1
121
2111
20)1()1()1(
ViiiiiV
=−+=+−−
V2
Mesh1:
221
1222
30)1()1()2(
ViiiiiV
=+−=+−−
Mesh2:
ELT221
35 Enzo Paterno
MESH LOOP ANALYSIS
212
1
252
51
3112
12
VVVV
i +=
−−
−=
VOUT = 2 i2. We use Cramer’s rule to find the current i2:
2123254
522 VViRiVOUT +===∴
221
121
32
ViiVii
=+−=−+
If V1 = V2 = 10 v then VOUT = 4 + 8 = 12 V
ELT221
36 Enzo Paterno
MESH LOOP ANALYSIS
Use Mesh analysis to solve for I1, I2, and I3.
661006646
31
311
=+−=+−−−
ikikikikik
+
+
+ +
631209336
32
232
=−+=−+−+
ikikikikik
021360331266
321
23313
=−++=+−−+−
ikikikikikikikik
=
−−
−
066
21363120
6010
3
2
1
iii
kkkkk
kk
−−
−=
−
0
6
6
21363120
6010 1
3
2
1
k
k
iii
1:
2:
3:
ELT221
37 Enzo Paterno
MESH LOOP ANALYSIS
−−
−=
−
0
6
6
21363120
6010 1
3
2
1
k
k
iii
ELT221
38 Enzo Paterno
MESH LOOP ANALYSIS
−−
−=
−
0
6
6
21363120
6010 1
3
2
1
k
k
iii i1 = -0.678 mA
i2 = 0.468 mA i3 = -0.126 mA
ELT221
39 Enzo Paterno
MESH LOOP ANALYSIS
Use Mesh analysis to solve Vo.
There are two mesh currents. I1 is known & only one mesh equation is needed for I2.
mAk
i
ik
ikikikikik
75.086
284
28202262
2
2
21
122
==
=+−
=+−=+−−+
+
+
I1 = 2 mA
2:
Vo = 0.75 mA (6kΩ) = 4.5 v
ELT221
40 Enzo Paterno
MESH LOOP ANALYSIS
There are three mesh currents. I1 and I2 are known. Only one mesh equation is needed for I3.
I1 = 4 mA I2 = -2 mA
mAk
i
ik
ikikikikikikikik
25.012
331288
312420622443
3
3
321
31323
==
=++−
=+−−=−+−+−+
+
+
+
+
Use Mesh analysis to solve Vo.
Vo = - [0.25 mA (6kΩ) ] + 3 = 1.5 v
ELT221
41 Enzo Paterno
MESH LOOP ANALYSIS
Use Mesh analysis to solve for I1, I2, and I3.
12
122
13
133
240222
12601116
ikikVikikikV
ikikVikikVik
x
x
x
x
−==+−−
+−+==+−−−+
+ +
kikiki
ikikikikikikVikik x
2634
6124224126
123
1123
1213
−−−
=
−−−=−−==+−+
+
+
@3:
@2:
I2 = (4 - 2/3) = 3.33 mA
One could have used SuperMesh (take out the 4 mA )
Eq1:
Eq2:
Eq1 Eq2
I1 = 2 mA We see that 4mA = I2 – I3
mAik
iki
kimAki
32
244
266)4(4
33
3
33
−=⇒
−+
=
−−−+
=