1 / 25

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).

shino
Download Presentation

Chapter 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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=

More Related