1 Mesh Analysis Discussion D2.4a September 2006 Chapter 2 Section 2-8
1 Mesh Analysis Discussion D2.4a September 2006 Chapter 2 Section 2-8.
Dec 20, 2015
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1
Mesh Analysis
Discussion D2.4aSeptember 2006
Chapter 2Section 2-8
2
Mesh Analysis• Mesh analysis applies KVL to find unknown
currents. • It is only applicable to planar circuits (a circuit that
can be drawn on a plane with no branches crossing each other).
• A mesh is a loop that does not contain any other loops.
• The current through a mesh is known as the mesh current.
• Assume for simplicity that the circuit contains only voltage sources.
3
Mesh Analysis Steps
1. Assign mesh currents i1, i2, i3, … il, to the l meshes,
2. Apply KVL to each of the l meshes and use Ohm’s law to express the voltages in terms of the mesh currents,
3. Solve the l resulting simultaneous equations to find the mesh currents.
4
Example
Number of nodes, n =
Number of branches, b =
Number of loops, l =
1l b n
7
10
4
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
5
Example
The n-1 node voltages arevn = [vn1 vn2 vn3 vn4 vn5 vn6]
The b branch currents areib = [ib1 ib2 ib3 ib4 ib5 ib6 ib7 ib8 ib9 ib10]
The l loop currents arei = [i1 i2 i3 i4]
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
6
Example
Then we can calculate the b branch currents from
We will solve mesh equationsfor the l loop currentsi = [i1 i2 i3 i4]
1 3ib i
2 3 1ib ib i
4 2ib i
5 6 4ib ib i 7 3 1ib i i
8 2 4ib i i
9 1 2ib i i
10 3 4ib i i
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
7
ExampleWe can also solve for the n-1 node voltages from the loop currents (or branch currents)
1 3 3vn i r
22 3 3svn V i r
4 4 4 8vn i r r
13 1 2 7svn V i i r DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
5 4 8vn i r
16 svn V
8
Example
Apply KVL to each mesh
2 1 7 5 0sV v v v
2 6 7 0v v v
15 3 0sv V v
Mesh 1:
Mesh 2:
Mesh 3:
14 8 6 0sv v V v Mesh 4:
Solving mesh equationsfor the l loop currentsi = [i1 i2 i3 i4]
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
9
2 1 7 5 0sV v v v
2 6 7 0v v v
15 3 0sv V v
Mesh 1:
Mesh 2:
Mesh 3:
14 8 6 0sv v V v Mesh 4:
2 1 1 1 2 7 1 3 5( ) ( ) 0sV i r i i r i i r
2 2 2 4 6 2 1 7( ) ( ) 0i r i i r i i r
13 1 5 3 3( ) 0si i r V i r
Mesh 1:
Mesh 2:
Mesh 3:
14 4 4 8 4 2 6( ) 0si r i r V i i r Mesh 4:
Express the voltage in terms of the mesh currents:
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
10
Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
21 5 7 1 7 2 5 3( ) sr r r i r i r i V
7 1 2 6 7 2 6 4( ) 0r i r r r i r i
15 1 3 5 3( ) sr i r r i V
16 2 4 6 8 4( ) sr i r r r i V
2 1 1 1 2 7 1 3 5( ) ( ) 0sV i r i i r i i r
2 2 2 4 6 2 1 7( ) ( ) 0i r i i r i i r
13 1 5 3 3( ) 0si i r V i r
14 4 4 8 4 2 6( ) 0si r i r V i i r
11
Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
Vr r r r r i
r r r r r i
Vr r r i
r r r r i V
21 5 7 1 7 2 5 3( ) sr r r i r i r i V
7 1 2 6 7 2 6 4( ) 0r i r r r i r i
15 1 3 5 3( ) sr i r r i V
16 2 4 6 8 4( ) sr i r r r i V
12
Ri = k
R
k
iis an l x l symmetric resistance matrix
is a l x 1 vector of mesh currents
is a l x 1 vector of voltages representing “known” voltages
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
Vr r r r r i
r r r r r i
Vr r r i
r r r r i V
13
•The matrix R is symmetric, Rkj = Rjk and all of the off-diagonal terms are negative or zero.
Writing the Mesh Equations by Inspection
The ki (the ith component of the vector k) = the algebraic sum of the independent voltages in mesh i, with voltage rises taken as positive.
The Rkj terms are the negative sum of the resistances common to BOTH mesh k and mesh j.
The Rkk terms are the sum of all resistances in mesh k.
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
Vr r r r r i
r r r r r i
Vr r r i
r r r r i V
14
MATLAB Solution of Mesh Equations
1i R k
Ri = k
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
Vr r r r r i
r r r r r i
Vr r r i
r r r r i V
15
Test with numbers
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1
4
2
4
1
3
32
2V
4V 1i2i
3i4i
16
Test with numbers
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
1
2
3
4
7 4 1 0 4
4 9 0 2 0
1 0 4 0 2
0 2 0 7 2
i
i
i
i
Ri = k
17
MATLAB Run
1
2
3
4
2 4 1 4 1 0 4
4 3 2 4 0 2 0
1 0 3 1 0 2
0 2 0 2 4 1 2
i
i
i
i
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1
4
2
4
1
3
32
2V
4V 1i2i
3i4i
18
PSpice Simulation
MATLAB:
19
Let's write a general MATLAB program to solve this problem
1 2 3 4 5 6 7 8[ ]r r r r r r r r r
Inputs:
Find all voltages and currentsDC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
2
1
1
1 5 7 7 5 1
7 2 6 7 6 2
5 3 5 3
6 4 6 8 4
0
00
0 0
0 0
s
s
s
Vr r r r r i
r r r r r i
Vr r r i
r r r r i V
Vs1, Vs2
20
function mesh1(r,Vs1,Vs2)% PowerPoint mesh-1 example% Discussion D2.4% r is a 1 x 8 vector of resistances% Vs1 and Vs2 = the known dc voltage sources% mesh1(r,Vs1,Vs2)%R = [r(1)+r(5)+r(7) -r(7) -r(5) 0; -r(7) r(2)+r(6)+r(7) 0 -r(6); -r(5) 0 r(3)+r(5) 0; 0 -r(6) 0 r(4)+r(6)+r(8)]k = [Vs2; 0; -Vs1; Vs1]i = inv(R)*kvn = zeros(1,6);vn(1) = -i(3)*r(3);vn(2) = Vs2-i(3)*r(3);vn(3) = Vs1+(i(1)-i(2))*r(7);vn(4) = i(4)*(r(4)+r(8));vn(5) = i(4)*r(8);vn(6) = Vs1;vnib = zeros(1,10);ib(1) = -i(3);ib(2) = i(1);ib(3) = i(1);ib(4) = i(2);ib(5) = i(4);ib(6) = i(4);ib(7) = i(3)-i(1);ib(8) = i(2)-i(4);ib(9) = i(1)-i(2);ib(10) = i(3)-i(4);ib
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
21
MATLAB Run
DC
DC
1r
3r5r
7r
2r
6r
8r
4r
1v 2v
3v 4v
5v6v
7v
8v
+ +
+ +
++
+
+
-
-- -
-
-
-
-
1sV
2sV 1i 2i
3i 4i
2vn 3vn
4vn
5vn
6vn1vn
1ib
9ib
4ib3ib
2ib
8ib7ib10ib
5ib
6ib
22
What happens if we have independent current sources in the circuit?
1. Assume temporarily that the voltage across each current source is known and write the mesh equations in the same way we did for circuits with only independent voltage sources.
2. Express the current of each independent current source in terms of the mesh currents and replace one of the mesh currents in the equations.
3. Rewrite the equations with all unknown mesh currents and voltages on the left hand side of the equality and all known voltages on the r.h.s of the equality.
23
Example
Write mesh equations by inspection.
1
2
3
1 3 3 1 10
3 3 2 4 2 0
1 2 2 1 a
i
i
i v
DC 10V
1
3A
+ v -a
1i 2i
3i3 3i
24
1
2
4 3 0 7
3 9 0 6
1 2 1 9a
i
i
v
1
2
4 3 1 10
3 9 2 0
1 2 3 3 a
i
i
v
25
MATLAB Run
AAV
i1i2va
1
2
4 3 0 7
3 9 0 6
1 2 1 9a
i
i
v
DC 10V
1
3A
+ v -a
1i 2i
3i
26
PSpice Simulation
MATLAB:i1
va
i2
i1
va
i2
+ -
27
DC
+ v -a
1i 2i
3i1r
1sI
1vn2vn
3vn
5ib
1ib
2ib
4ib
6ib
3ib
1sV
2r
4r3r
Let's write a general MATLAB program to solve this problem
1 3 3 1 1 1
3 2 3 4 2 2
1 2 1 2 3
0s
a
r r r r i V
r r r r r i
r r r r i v
1 2 3 4[ ]r r r r r
Inputs:
Find all voltages and currents
Note that
1 1, s sV I
13 si I
28
Example
The n-1 node voltages arevn = [vn1 vn2 vn3]
The b branch currents areib = [ib1 ib2 ib3 ib4 ib5 ib6]
The l loop currents arei = [i1 i2 i3]
DC
+ v -a
1i 2i
3i1r
1sI
1vn2vn
3vn
5ib
1ib
2ib
4ib
6ib
3ib
1sV
2r
4r3r
29
Example
Then we can calculate the n-1 node voltages from
We will solve mesh equationsfor the loop currents and unknown voltage va
iiv = [i1 i2 va]
DC
+ v -a
1i 2i
3i1r
1sI
1vn2vn
3vn
5ib
1ib
2ib
4ib
6ib
3ib
1sV
2r
4r3r
11 svn V
2 2 4vn i r
3 1 2 3vn i i r
30
ExampleWe can also calculate the b branch currents from
1 1ib i
12 sib I
3 2ib i
6 1 2ib i i
14 1 sib i I
DC
+ v -a
1i 2i
3i1r
1sI
1vn2vn
3vn
5ib
1ib
2ib
4ib
6ib
3ib
1sV
2r
4r3r
15 2 sib i I
31
1 11 3 1 3 2 1( ) s sr r i r i r I V
13 1 2 3 4 2 2( ) 0sr i r r r i r I
11 1 2 2 1 2( ) s ar i r i r r I v
1 3 3 1 1 1
3 2 3 4 2 2
1 2 1 2 3
0s
a
r r r r i V
r r r r r i
r r r r i v
Expand matrix with13 si I
32
1 11 3 1 3 2 1( ) s sr r i r i r I V
13 1 2 3 4 2 2( ) 0sr i r r r i r I
11 1 2 2 1 2( ) s ar i r i r r I v
These can be written in matrix form as
1
1
1
1 11 3 3 1
3 2 3 4 2 2
1 2 1 2
0
0
1
s s
s
a s
V r Ir r r i
r r r r i r I
r r v r r I
33
function mesh2(r,Vs1,Is1)% PowerPoint mesh-2 example% Discussion D2.4% r is a 1 x 4 vector of resistances% Vs1 is a known dc voltage source % Is1 is a known dc current source% mesh2(r,Vs1,Is1)
R = [r(1)+r(3) -r(3) 0; -r(3) r(2)+r(3)+r(4) 0; -r(1) -r(2) 1]k = [Vs1-r(1)*Is1; -r(2)*Is1; (r(1)+r(2))*Is1]iiv = inv(R)*ki(1) = iiv(1);i(2) = iiv(2);ivn = zeros(1,3);vn(1) = Vs1;vn(2) = i(2)*r(4);vn(3) = (i(1)-i(2))*r(3);vnib = zeros(1,6);ib(1) = i(1);ib(2) = -Is1;ib(3) = i(2);ib(4) = i(1)+Is1;ib(5) = -i(2)-Is1;ib(6) = i(1)-i(2);ib
DC
+ v -a
1i 2i
3i1r
1sI
1vn2vn
3vn
5ib
1ib
2ib
4ib
6ib
3ib
1sV
2r
4r3r
34
Do same problem as before
[1 2 3 4]r
1 10Vs
mesh2(r,Vs1,Is1)
1 3Is
DC 10V
1
3A
+ v -a
1i 2i
3i
35
MATLAB Run
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