chapter iii circuit analysis techniques - guceee.guc.edu.eg/courses/electronics/elct301...
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Chapter III
Circuit Analysis Techniques
• Circuit Reduction & Source Transformation
• Node-Voltage method
• Mesh- Current method.
• Superposition methods.
• Thevenin’s and Norton’s circuits.
• Maximum Power Transfer theorem
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The Mesh- Current method
“LOOP Analysis”
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Mesh-Current Method (Loop Analysis)
• Nodal analysis was developed by applying KCL at each non-
reference node.
• Mesh-Current method is developed by applying KVL around meshes
in the circuit.
• Loop (mesh) analysis results in a system of linear equations which
must be solved for unknown currents.
• Reduces the number of required equations to the number of meshes
• Can be done systematically with little thinking
• As usual, be careful writing mesh equations – follow sign convention.
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Definitions
+
-
A
B C
Mesh: Loop that does not enclose other loops
Branch: Path between 2 essential nodes
How many mesh-currents?
No. of essential nodes Ne = 4
No. of branches
Be = 6
No. of Mesh-currents
M = Be –(Ne-1)
•Enough equations to get unknowns
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Steps of Mesh Analysis
1. Identify the number of basic 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.
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Steps of Mesh Analysis
2. Assigning Mesh Currents
Mesh 2
1kW
1kW
1kW
V1 V2Mesh 1
+
–
+
–
1kW
1kW
1kW
V1 V2I1 I2
+
–
+
–
1. Identifying the Meshes
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R
I1
+ –VR
VR = I1 R
R
I1
+ –VR
I2
VR = (I1 - I2 ) R
3. Formulation of voltage drop in terms of Mesh Currents
Steps of Mesh Analysis
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R2 W
R3 W
R1 W
V1 V2I1 I2
+
–
+
–
-V1 + I1 R1 + (I1 - I2) R3 = 0
I2 R2 + V2 + (I2-I1) R3 = 0
4. Writing Mesh-Current Equations
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Example
Find the power supplied by the sources using Mesh current method
M= 3-(2-1) = 2
• Assign mesh currents
• Write mesh equations
i1(12) + (i1-i2)2+ i1(4) -12= 0
i2(9) +8+ i2(3) + (i2-i1) 2= 0
• Solve mesh equations:
• I1 = 0.6129 A & I2 = - 0.4838 A
• P12V = -12 I1 & P9V = 8 I2
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Mesh current method Cases
Mesh 1
-10 + 4i1 + 6(i1-i2) = 0
Mesh 2
i2 = - 5A
No need to write a loop equation
Case I: When a current source exists only in one mesh
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Case II: Super Mesh
Case II: When a current source exists between two meshes (super Mesh)
4i1 + 2 (i1 – i3) + 2(i2 – i3) + 8i2 = 0
i2 – i1 = 5 A
-10 + 2(i3 – i2 ) + 2(i3 –i1) = 0
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𝑖2 − 𝑖1 = 4A
2 𝑖1 − 𝑖3 + 4𝑖1 + 8𝑖2 + 6 𝑖2 − 𝑖4 = 0
6 𝑖4 − 𝑖2 + 𝑖4 + 3 𝑖4 − 𝑖3 = 0
−30 + 2 𝑖3 − 𝑖1 + 3 𝑖3 − 𝑖4 = 0
Example
Source current
Super Mesh equation
Find the power supplied by the voltage source using the Mesh current method
Solving i3 = 8.561 A and the P30V = 30x8.561 = 256.83 W
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Case III: Mesh with Dependent Sources
−20 + 4 𝑖1
− 𝑖3
+ 2 𝑖1
− 𝑖2
= 0
4 𝑖3
− 𝑖1
+ 6𝑖3
+ 8 𝑖3
− 𝑖2
= 0
−10𝑖0
+ 2 𝑖2
− 𝑖1
+ 8 𝑖2
− 𝑖3
= 0
In addition to the 3 Mesh equations, an extra equation defining the control parameter
of the dependent source has to be formulated 𝒊𝟎
= 𝒊𝟑
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Example
Use Mesh current method to find io
𝑖1 = 4A
3 𝑖2 − 𝑖1 + 2𝑖2 − 5𝑖𝑜 + 𝑖2 − 𝑖3 = 0
5𝑖𝑜 + 𝑖3 − 𝑖2 + 4𝑖3 − 20 + 5 𝑖3 − 𝑖1 = 0
𝑖𝑜 = 𝑖1 − 𝑖3
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Example
Use nodal analysis and Mesh current method to find VD and if
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Solution (Nodal Analysis)
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Solution (Loop Analysis)
35 i1 -10i2 – 5i3 – 20i4 -175 if = 0
-10i1 + 110 i2 + 60 = 0
-5i1 + 205i3 – 200i4 – 60 = 0
-200i3 – 20i1 + 620i4 – 400i5 = 0
I5 = -0.625 VD
VD = 5 (i1 – i 3)
If = (i4 – i 3)