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Chapter 4

Chapter 4. Techniques of Circuit Analysis. Planar Circuit. It is a circuit that can be drawn on a plane with no crossing branches. Figure (a) A planar circuit. (b) The same circuit redrawn to verify that it is planar. Figure A nonplanar circuit (branches are crossing).

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Chapter 4

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  1. Chapter 4 Techniques of Circuit Analysis

  2. Planar Circuit It is a circuit that can be drawn on a plane with no crossing branches. Figure (a) A planar circuit. (b) The same circuit redrawn to verify that it is planar.

  3. Figure A nonplanar circuit (branches are crossing)

  4. Terms for Describing Circuits

  5. Table (Continued)

  6. Figure 3.3 A circuit illustrating nodes, branches, meshes, paths, and loops

  7. Example 3.1 For the circuit shown below, identify • All nodes • All essential nodes. • All branches • All essential branches. • All meshes • Two paths that are not loops or essential branches. • Two loops that are not meshes.

  8. The nodes are: a, b, c, d, e, f, g Essential nodes: b, c, e, g Branches: v1, v2, R1, R2, R3, R4, R5, R6, R7, I Essential Branches: v1-R1, R2-R3, v2-R4, R5, R6, R7, I All meshes: v1-R1-R5-R3-R2, v2-R2-R3-R6-R4, R5-R6-R7, R7-I Two paths that are not loops or essential branches: R1-R5-R6 and v2-R2 Two loops that are not meshes: v1-R1-R5-R6-R4-v2, I-R5-R6

  9. Node-Voltage Method Follow the following steps: • Redraw the circuit so that no branches cross over,

  10. Mark clearly the essential nodes on the circuit diagram. The circuit below has 3 essential nodes. Select the node connected to the largest number of branches as a reference node (shown with arrow below). 4. Define the node voltages on the diagram (v1 and v2 in the circuit diagram below.) This means we need 2 equations only. Always the number of equations is equal the number of essential nodes -1

  11. 6. Apply KCL at each of the essential nodes. Do not apply KCL at the reference node. Figure : Computation of the branch current i.

  12. Example 3.2 Find v1 and v2 using node-voltage method.

  13. Solution: Follow the same steps as explained before we need to form 2 equations. Applying KCL at node 1 gives: Applying KCL at node 2 gives: Solving the above two equations gives

  14. Solution of Previous Example using MATLAB: >a=[1.7 -0.5; -0.5 0.6]; >b=[10 ; 2]; > x=inv(a)*b; %This is the solution vector x that contains i1 and i2

  15. Example 3.3 • Use the node-voltage method to find the branch currents ia, ib and ic in the circuit shown. • Find the power associated with each source, and state whether the source is delivering or absorbing power.

  16. Solution: We have 2 essential nodes, the lower node is chosen as our reference. Therefore we only need one equation. Follow the steps!! Applying KCL at node 1 gives:

  17. Node- voltage method with dependent sources Example 3.4: Use the node-voltage method to find the power dissipated in the 5 ohm resistor shown in the circuit below.

  18. Solution: We have 3 essential nodes with the lower one chosen as reference. Therefore, we only need 2 equations. Follow the preparation steps!! Applying KCL at node 1: (1) Applying KCL at node 2: (2) Substituting in Eq. 2 gives (3) Solving Equations (1) and (3) gives

  19. Node-Voltage Method, Special case #1 • We will consider as a special case # 1 if a voltage source is connected between an essential node and the reference node. This simplifies the node-voltage method because the voltage of the essential node here is the same as the voltage of the source. • Example on this special case follows.

  20. Example 3.5 In the circuit shown, use node-voltage method the power absorbed by the 100 V source, and the power delivered by the 5A source.

  21. Solution: This is special case # 1 since the voltage source lies between an essential node and reference node. Prepare the circuit!! (label v1, v2 and the reference). Since this is a special case #1, v1 can be found directly. V1=100 V (we do not need to apply KCL at essential node 1) Applying KCL at essential node 2 gives: Power absorbed by the 100 V source= Power delivered by the 5A source=5V2=625W

  22. Node-Voltage Method, Special case #2 (Supernode) • We will consider as a special case # 2 if a voltage source is connected between two essential nodes (and none of these nodes can be a reference node). • A supernode is a combination of two essential nodes connected by a voltage source, as shown below. • Sum of all currents leaving a supernode is zero.

  23. Nodes 2 and 3 are connected by a voltage source. Hence, nodes 2 and 3 can be combined to form a supernode as shown below.

  24. Example 3.6 In the circuit shown, find the power delivered by the 50V source. Solution: Prepare the circuit!!! 4 essential nodes. We need 3 equations.

  25. This problem includes special case #1 and special case #2. V1=50 (special case #1) KCL at supernode (2,3): Power delivered by 50 V source=

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