This document discusses techniques for solving problems in basic electric circuits using mesh and nodal analysis. It provides two example problems applying these techniques, presenting the circuit diagrams and step-by-step working to find current and voltage values. Additional references for circuit analysis techniques are also listed.

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Session 9C: Focus
 Solving Problems
◦ Mesh Analysis
◦ Nodal Analysis

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S9C_Problem 1: Using Mesh Analysis
Note: This is problem 27 on page 114, Figure 4.53 in the Ref 1 book (by Hayt)
 Find v1 acrosss 3Ω resistor: v1 =480 mV
i1 = 2 A
Voltage controlled dependent voltage source:
3i3 = v1 - i2 = v1 v1 = 3i3 = -i2
i3 = -i2 / 3
KVL for the mesh with i3
2(i3 - i1) + 3i3 + 5(i3 - i2) = 0
- 2i1 - 5i2 + (3 + 5 + 2) i3 = 0
- 2 (i1) - 5i2 + 10 (-i2/3) = 0
- 25i2 = 6i1 = 6 (2) = 12
i2 = -12/25 = -0.48 v1 = -i2 = 0.48V

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S9C_Problem 1: Using Nodal Analysis
Note: This is problem 27 on page 114, Figure 4.53 in the Ref 1 book (by Hayt)
 Find v1 acrosss 3Ω resistor:
vA
vB vC
v1 = vA - vC
vB = 4 v1
KCL for Node vA
2 = (vA - vB) /2 + (vA - vC) /3
12 = 3vA - 3vB + 2vA - 2vC
12 = 5vA - 3vB - 2vC
KCL for Node vC
v1 = (vC - vB) /5 + (vC - vA) /3
15 (vB /4) = 3vC - 3vB + 5vC - 5vA
v1 = vB /4
15 v1 = 3vC - 3vB + 5vC - 5vA
15 vB = 12vC - 12vB + 20vC – 20vA
20vA + 27vB - 32vC = 0
vB = 4 v1 = 4 (vA - vC)
4vA - vB - 4vC= 0
For Node vB

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S9C_Problem 1: Using Nodal Analysis
Note: This is problem 27 on page 114, Figure 4.53 in the Ref 1 book (by Hayt)
 Find v1 acrosss 3Ω resistor: v1 =480 mV
vA
vB vC
12 = 5vA - 3vB - 2vC
20vA + 27vB - 32vC = 0
4vA - vB - 4vC= 0
Solving these equations: HOME WORK
vA = 5.6 V vC = 5.12 V
Eliminate vB

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S9C_Problem 2: Mesh Analysis
Note: This is problem 42 on page 117, Figure 4.67 in the Ref 1 book (by Hayt)
 Find the mesh currents: i1 = -1.219 A, i2 = -562.5 mA, i3 = 781.3 mA
Is there a supermesh? Yes (i1 and i3)
When the current source 2A is removed,
For considering the supermesh,
Will there be current through 3Ω,
which is in series with 2A current source?
Express 2A in terms of mesh currents: 2 = i3 - i1
No
KVL for the supermesh
-1 + 7(i1 - i2) + 3 (i3 - i2) + 2i3 = 0
7i1 - 10 i2 + 5i3 = 1
KVL for the mesh with i2
7(i2 - i1) + 1 i2 + 3 (i2 - i3) = 0
– 7 i1 + 11 i2 - 3 i3 = 0
2 + i1 = i3
7i1 - 10 i2 + 5 (2 + i1) = 1
12i1 - 10 i2 = -9
– 7 i1 + 11 i2 - 3 (2 + i1) = 0
– 10 i1 + 11 i2 = 6

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 Solve to get the voltage labeled v = 1.731 V
 Use both Nodal and Mesh analysis to get the above answer
 Hint: Supernode and Supermesh techniques will be required
to solve it, while using Nodal and Mesh analyses
Ref
Note: This is Exercise problem 23 on page 113, Figure 4.49 in the Ref 1 book (by Hayt)
Which is a reference node?
Center
Node
Home Work

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Session 9C: Summary
 Solving Problems
◦ Mesh Analysis
◦ Nodal Analysis

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References
Ref 1 Ref 2