Basic Laws of Electric Circuits Nodal Analysis Lesson
Basic Laws of Electric Circuits Nodal Analysis Lesson 6
Basic Circuits Nodal Analysis: The Concept. • Every circuit has n nodes with one of the nodes being designated as a reference node. • We designate the remaining n – 1 nodes as voltage nodes and give each node a unique name, vi. • At each node we write Kirchhoff’s current law in terms of the node voltages. 1
Basic Circuits Nodal Analysis: The Concept. • We form n-1 linear equations at the n-1 nodes in terms of the node voltages. • We solve the n-1 equations for the n-1 node voltages. • From the node voltages we can calculate any branch current or any voltage across any element. 2
Basic Circuits Nodal Analysis: Concept Illustration: Figure 6. 1: Partial circuit used to illustrate nodal analysis. Eq 6. 1 3
Basic Circuits Nodal Analysis: Concept Illustration: Clearing the previous equation gives, Eq 6. 2 We would need two additional equations, from the remaining circuit, in order to solve for V 1, V 2, and V 3 4
Basic Circuits Nodal Analysis: Example 6. 1 Given the following circuit. Set-up the equations to solve for V 1 and V 2. Also solve for the voltage V 6. Figure 6. 2: Circuit for Example 6. 1. 5
Basic Circuits Nodal Analysis: Example 6. 1, the nodal equations. Eq 6. 3 6 Eq 6. 4
Basic Circuits Nodal Analysis: Example 6. 1: Set up for solution. Eq 6. 3 Eq 6. 4 Eq 6. 5 Eq 6. 6 7
Basic Circuits Nodal Analysis: Example 6. 2, using circuit values. Figure 6. 3: Circuit for Example 6. 2. Find V 1 and V 2. At v 1: Eq 6. 7 At v 2: 8 Eq 6. 8
Basic Circuits Nodal Analysis: Example 6. 2: Clearing Equations; From Eq 6. 7: V 1 + 2 V 1 – 2 V 2 = 20 or Eq 6. 9 3 V 1 – 2 V 2 = 20 From Eq 6. 8: 4 V 2 – 4 V 1 + V 2 = -120 or -4 V 1 + 5 V 2 = -120 Solution: V 1 = -20 V, 9 Eq 6. 10 V 2 = -40 V
Basic Circuits Nodal Analysis: Example 6. 3: With voltage source. Figure 6. 4: Circuit for Example 6. 3. At V 1: Eq 6. 11 At V 2: Eq 6. 12 10
Basic Circuits Nodal Analysis: Example 6. 3: Continued. Collecting terms in Equations (6. 11) and (6. 12) gives Eq 6. 13 Eq 6. 14 11
Basic Circuits Nodal Analysis: Example 6. 4: Numerical example with voltage source. Figure 6. 5: Circuit for Example 6. 4. What do we do first? 12
Basic Circuits Nodal Analysis: Example 6. 4: Continued At v 1: Eq 6. 15 At v 2: Eq 6. 16 13
Basic Circuits Nodal Analysis: Example 6. 4: Continued Clearing Eq 6. 15 4 V 1 + 100 – 10 V 2 = -200 or 14 V 1 – 10 V 2 = -300 Eq 6. 17 Clearing Eq 6. 16 4 V 2 + 6 V 2 – 60 – 6 V 1 = 0 or -6 V 1 + 10 V 2 = 60 14 V 1 = -30 V, V 2 = -12 V, I 1 = -2 A Eq 6. 18
Basic Circuits Nodal Analysis: Example 6. 5: Voltage super node. Given the following circuit. Solve for the indicated nodal voltages. super node Figure 6. 6: Circuit for Example 6. 5. 15 When a voltage source appears between two nodes, an easy way to handle this is to form a super node. The super node encircles the voltage source and the tips of the branches connected to the nodes.
Basic Circuits Nodal Analysis: Example 6. 5: Continued. Constraint Equation V 2 – V 3 = -10 At V 1 At super node 16 Eq 6. 20 Eq 6. 21 Eq 6. 19
Basic Circuits Nodal Analysis: Example 6. 5: Continued. Clearing Eq 6. 19, 6. 20, and 6. 21: 7 V 1 – 2 V 2 – 5 V 3 = 60 Eq 6. 22 -14 V 1 + 9 V 2 + 12 V 3 = 0 Eq 6. 23 V 2 – V 3 = -10 Eq 6. 24 Solving gives: V 1 = 30 V, 17 V 2 = 14. 29 V, V 3 = 24. 29 V
Basic Circuits Nodal Analysis: Example 6. 6: With Dependent Sources. Consider the circuit below. We desire to solve for the node voltages V 1 and V 2. Figure 6. 7: Circuit for Example 6. 6. In this case we have a dependent source, 5 Vx, that must be reckoned with. Actually, there is a constraint equation of Eq 6. 25 18
Basic Circuits Nodal Analysis: Example 6. 6: With Dependent Sources. At node V 1 19 At node V 2 The constraint equation:
Basic Circuits Nodal Analysis: Example 6. 6: With Dependent Sources. Clearing the previous equations and substituting the constraint VX = V 1 - V 2 gives, Eq 6. 26 Eq 6. 27 which yields, 20
circuits End of Lesson 6 Nodal Analysis
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