Post on 15-Mar-2020
EENG223 Mesh Analysıs2
Mesh Analysis
Nodal analysis was developed by applying KCL at each non-reference node.Mesh analysis is developed by applying KVL around meshes/loops in the circuit.Mesh analysis results in a system of linear equations which must be solved for unknown currents.
EENG223 Mesh Analysıs3
Mesh Analysis
quantity of interest is currenta mesh is a loop that does not contain another loop within itwork for planar circuit onlyplanar circuit -> no branch passes over or under other branchM-meshes -> assign clockwise current for each meshapply KVL around each mesh
EENG223 Mesh Analysıs5
Steps of Mesh Analysis
1. Identify meshes.2. Assign a current to each mesh.3. Apply KVL around each loop to get an
equation in terms of the loop currents.4. Solve the resulting system of linear
equations.
EENG223 Mesh Analysıs7
Steps of Mesh Analysis
1. Identify mesh (loops).2. Assign a current to each mesh.3. Apply KVL around each loop to get an
equation in terms of the loop currents.4. Solve the resulting system of linear
equations.
EENG223 Mesh Analysıs9
Steps of Mesh Analysis
1. Identify mesh (loops).2. Assign a current to each mesh.3. Apply KVL around each loop to get an
equation in terms of the loop currents.4. Solve the resulting system of linear
equations.
EENG223 Mesh Analysıs10
Voltages from Mesh Currents
R
I1
+ –VR
VR = I1 R
R
I1
+ –VRI2
VR = (I1 - I2 ) R
EENG223 Mesh Analysıs11
KVL Around Mesh 1
1kΩ
1kΩ
1kΩ
V1 V2I1 I2+–
+–
+ - +
-
-V1 + I1 1kΩ + (I1 - I2) 1kΩ = 0I1 1kΩ + (I1 - I2) 1kΩ = V1
EENG223 Mesh Analysıs12
KVL Around Mesh 2
1kΩ
1kΩ
1kΩ
V1 V2I1 I2+–
+–
--
+
+
(I2 - I1) 1kΩ + I2 1kΩ + V2 = 0(I2 - I1) 1kΩ + I2 1kΩ = -V2
EENG223 Mesh Analysıs13
Steps of Mesh Analysis
1. Identify mesh (loops).2. Assign a current to each mesh.3. Apply KVL around each loop to get an
equation in terms of the loop currents.4. Solve the resulting system of linear
equations.
EENG223 Mesh Analysıs14
Matrix Notation
The two equations can be combined into a single matrix/vector equation.
−
=
Ω+ΩΩ−
Ω−Ω+Ω
2
1
2
1
k1k1k1k1k1k1
VV
II
EENG223 Mesh Analysıs15
Solving the Equations
Let: V1 = 7V and V2 = 4VResults:
I1 = 3.33 mAI2 = -0.33 mA
FinallyVout = (I1 - I2) 1kΩ = 3.66V
EENG223 Mesh Analysıs19
Current Sources
The current sources in this circuit will have whatever voltage is necessary to make the current correct.We can’t use KVL around the loop because we don’t know the voltage.What to do?
EEE 223 Mesh Analysıs20
Current Sources
The 4mA current source sets I2:I2 = -4 mA
The 2mA current source sets a constraint on I1 and I3:
I1 - I3 = 2 mAWe have two equations and three unknowns. Where is the third equation?
EENG223 Mesh Analysıs21
1kΩ
2kΩ
2kΩ
12V 4mA
2mA
I0
I1 I2
I3The Supermesh surrounds this source!
The Supermesh
does not include this
source!
+–
EENG223 Mesh Analysıs22
KVL Around the Supermesh
-12V + I3 2kΩ + (I3 - I2)1kΩ + (I1 - I2)2kΩ = 0
I3 2kΩ + (I3 - I2)1kΩ + (I1 - I2)2kΩ = 12V
EENG223 Mesh Analysı s23
Matrix Notation
The three equations can be combined into a single matrix/vector equation.
−=
Ω+ΩΩ−Ω−Ω−
V12mA2mA4
1k2k2k1k2k101
010
3
2
1
III
EENG223 Mesh Analysıs24
Solve Using MATLAB
>> A = [0 1 0; 1 0 -1;2e3 -1e3-2e3 2e3+1e3];
>> v = [-4e-3; 2e-3; 12];>> i = inv(A)*vi = 0.0012
-0.0040-0.0008