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Mesh Analysis
Examples
Example 1
2 6.364A
40V6.364A
20V1.818A
4
6I1 I2
Using Mathcad
mesh 1: 20 4I1 6 I1 I2 0
mesh 2: 2I2 6 I2 I1 40 0
Find I1 I2 1.818
6.364
Using PSpice
2 6.364A
40V6.364A
20V
1.818A
41.818A
0
6
4.546A
I1 = -1818 A I2 = -6.364 A
Example 2: By Inspection
8.863A
12V6.549A
46.235A 1
1
24V
18V
30 1
2
6V 2
I2I1
I3
Using Mathcad: By InspectionR I V
R
4 3( )
0
3
0
1 1 2( )
1
3
1
2 3 1 1( )
V
18 6
24 18
12 24
I lsolve R V( ) I
6.235
8.863
6.549
Using PSpice
0
12V6.549A
4
6.235A1
8.863A
1
24V
18V
3 1
2
6V 2
Example 3
4k2mA
3V
4k
6k
2k
4mA
-2mA
4mA
I1
Using Mathcad
2 k 4 m 2k 4k 6k( ) I1 4k( ) 2 m( ) 3 0
Find I1 250
Mesh 1:
Using PSpice
4k
6.000mA
2mA
2.000mA
3V
250.0uA
0
4k
2.250mA
6k
250.0uA2k
3.750mA4mA
4.000mA
I1 = 250 A
Example 4: Supermesh
3
20V 3
6A
46
I2I1
Mesh 1 and mesh 2 form supermesh
I1 = -1.6 A I2 = 4.4 AAns:
Using Mathcad
Supermesh 1& 2: 20 4I1 6I2 0
Auxilliary : I2 I1 6
Find I1 I2 1.6
4.4
Using PSpice
34.400A
20V
1.600A
3
4.400A
6A
6.000A
41.600A
0
6
6.000A
I1 = -1.6 A I2 = 4.4 AAns:
Example 5: Supermesh
2.5Ib20V
2Ia+-
8
80
40
5A
Ia
Ib
Example 5: Supermesh
2.5Ib20V
2Ia+-
8
80
40
5A
Ia
Ib
I1
5A I2
I3
I1 = 5.431A I2 = -3.621A I3 = -1.81AAns:
Using Mathcad: equations
Controlling variable 1: Ib I2Controlling variable 2: I1 I3 Ia
Supermesh 1&2: 80 I1 5 8 I1 I3 40 I2 I3 20 0
Auxilliary: I2 I1 2.5 Ib
mesh 3: 8 I3 I1 40 I3 I2 2Ia 0
Using Mathcad: Results
Find I1 I2 I3 Ia Ib
5.431
3.621
1.81
7.241
3.621
Using PSpice
G1
2.5*I(Vb)
9.052A
20V
3.621A
Vb0Vdc
3.621A
2*I(Va)1.810A
+-
0
8
7.241A
80
431.0mA
40
1.810A
5A
5.000A
Va
0Vdc
I1 = 5.431A I2 = -3.621A I3 = -1.81AAns:
Find Va and Ix
Problem 1
i1
3A
6V
-
2 2
4
81
i2
i3
+
Va
Ix
+
-
Using Mathcad
Supermesh 1&2: 6 2 I1 I3 4 I2 I3 8I2 0
Auxilliary: I1 I2 3
mesh 3: 2 I3 I1 2I3 4 I3 I2 0
Find I1 I2 I3 3.474
0.474
1.105
Ix = -I1 = -3.474 A
To Find Ix
i1
3A
6V
-
2 2
4
81
i2
i3
+
Va
Ix
+
-
Mesh 1: -6+(I1-I3)2+(1 3)+Va =0
To Find Va
i1
3A
6V
-
2 2
4
81
i2
i3
+
Va
Ix
+
-
+
-
+ -
+
-
V8
V4
V1
V2
-
+
Mesh 2: -(1 3)+ (I2-I3)4+8I2-Va =0 orVa = -1.737V
Problem 2
Find Vxy, Vc, and the power absorbed by all elements.
x
2.5Ib20V
2Ia +-
8
80
40
5A
Ia
Ib
I1
5A I2
I3
10
y
+
-
Vc
Using Mathcad: Equations
Controlling variable 1: Ib I2Controlling variable 2: I1 I3 Ia
Supermesh 1&2: 80 I1 5 8 I1 I3 40 I2 I3 20 0
Auxilliary: I2 I1 2.5 Ib
mesh 3: 8 I3 I1 40 I3 I2 2Ia 10I3 0
Using Mathcad: Results
Find I1 I2 I3 Ia Ib
5.625
3.75
1.563
7.188
3.75
To Find Vxy
Vxy, Vc
x
2.5Ib20V
2Ia +-
805A
Ib
I1
5A I2
I3
10
y
+
-
Vc
+ -
-Vxy-2Ia+10I3= 0
Vxy
Vxy = -2Ia+10I3
Vxy 2 Ia 10I3 Vxy 30
To Find Vc
x
2.5Ib20V
2Ia +-
805A
Ib
I1
5A I2
I3
10
y
+
-
Vc
+ -
+Vc+(I2-I3)40-20= 0
V40
Vc = -(I2-I3)40+20
40
Vc I2 I3 40 20 Vc 107.5
Power absorbed by active elements
x
2.5Ib20V
2Ia +-
8
80
40
5A
Ia
Ib
I1
5A I2
I3
10
y
+
-
Vc+
-
V5A
555 AA VPbcIb IVP 5.25.2
)(2020 bV IP
)(2 32 IIP aIa
For the active elements, we have to follow PSC to calculate the power absorbed.
Power absorbed by passive elements
x
2.5Ib20V
2Ia +-
8
80
40
5A
Ia
Ib
I1
5A I2
I3 10
+
-
Vc+
-
V5A
102310 IP
40)( 23240 IIP
80)5( 2180 IP
828 aIP
The power absorbed by the resistor is always positive; we can always use P = I2R
Power absorbed by all elements
P5A P2.5Ib P20V P2Ia P80 P8 P10 P40 0.000
P40 191.406P40 I2 I3 2 40
P10 24.414P10 I32 10
P8 413.281P8 Ia2 8
P80 31.25P80 5 I1 2 80
P2Ia 22.461P2Ia 2 Ia I3
P20V 75P20V 20 I2
P2.5Ib 1.008 103P2.5Ib Vc 2.5 Ib
P5A 250P5A V5A 5
V5A 50V5A I1 5 80