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1 AC Nodal and Mesh Analysis Discussion D11.1 Chapter 4 4/10/2006.

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Presentation on theme: "1 AC Nodal and Mesh Analysis Discussion D11.1 Chapter 4 4/10/2006."— Presentation transcript:

1 1 AC Nodal and Mesh Analysis Discussion D11.1 Chapter 4 4/10/2006

2 2 AC Nodal Analysis

3 3 How did you write nodal equations by inspection?

4 4 Writing the Nodal Equations by Inspection The matrix G is symmetric, g kj = g jk and all of the off-diagonal terms are negative or zero. The i k (the k th component of the vector i) = the algebraic sum of the independent currents connected to node k, with currents entering the node taken as positive. The g kj terms are the negative sum of the conductances connected to BOTH node k and node j. The g kk terms are the sum of all conductances connected to node k.

5 5 v1v1 v2v2 v3v3 Example with resistors

6 6 For steady-state AC circuits we can use the same method of writing nodal equations by inspection if we replace resistances with impedances and conductances with admittances. Let's look at an example.

7 7 Problem 4.31 in text Change impedances to admittances

8 8

9 9 Matlab Solution

10 10 Nodal Analysis for Circuits Containing Voltage Sources That Can’t be Transformed to Current Sources If a voltage source is connected between two nodes, assume temporarily that the current through the voltage source is known and write the equations by inspection.

11 11 Problem 4.33 in text Note: V 2 = 10 assume I 2

12 12

13 13 Problem 4.33 in text Note: V 2 = 10 assume I 2

14 14 Matlab Solution

15 15 AC Mesh Analysis

16 16 How did you write mesh equations by inspection?

17 17 The matrix R is symmetric, r kj = r jk and all of the off-diagonal terms are negative or zero. Writing the Mesh Equations by Inspection The v k (the k th component of the vector v) = the algebraic sum of the independent voltages in mesh k, with voltage rises taken as positive. The r kj terms are the negative sum of the resistances common to BOTH mesh k and mesh j. The r kk terms are the sum of all resistances in mesh k.

18 18 Example with resistors

19 19 For steady-state AC circuits we can use the same method of writing mesh equations by inspection if we replace resistances with impedances and conductances with admittances. Let's look at an example.

20 20 Problem 4.38 in text: Find I 1 and I 2

21 21 Matlab Solution

22 22 What happens if we have independent current sources in the circuit? 1.Assume temporarily that the voltage across each current source is known and write the mesh equations in the same way we did for circuits with only independent voltage sources. 2.Express the current of each independent current source in terms of the mesh currents and replace one of the mesh currents in the equations. 3.Rewrite the equations with all unknown mesh currents and voltages on the left hand side of the equality and all known voltages on the r.h.s of the equality.

23 23 Problem 4.40 in text: Find I 0 Assume you know V 2 Note I 2 = -2

24 24

25 25 Matlab Solution

26 26 Problem 4.40 in text: Find I 0 Assume you know V 2 Note I 2 = -2


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