1 / 25

# Chapter 4 - PowerPoint PPT Presentation

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).

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Chapter 4' - shino

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Chapter 4

Techniques of Circuit Analysis

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)

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

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.

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

• Redraw the circuit so that no branches cross over,

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

6. Apply KCL at each of the essential nodes. Do not apply KCL at the reference node.

Figure : Computation of the branch current i.

Example 3.2 KCL at the reference node.

Find v1 and v2 using node-voltage method.

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

Solution of Previous Example using MATLAB: to form 2 equations.

>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

Example 3.3 to form 2 equations.

• 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.

Solution: to form 2 equations.

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:

Node- voltage method with dependent sources to form 2 equations.

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

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

Node-Voltage Method, Special case #1 chosen as reference. Therefore, we only need 2 equations. Follow the preparation steps!!

• 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.

Example 3.5 chosen as reference. Therefore, we only need 2 equations. Follow the preparation steps!!

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

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

Node-Voltage Method, Special case #2 ( lies between an essential node and reference node.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.

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.

Example 3.6 nodes 2 and 3 can be combined to form a

In the circuit shown, find the power delivered by the 50V source.

Solution: Prepare the circuit!!! 4 essential nodes. We need 3 equations.

This problem includes special case #1 and special case #2. nodes 2 and 3 can be combined to form a

V1=50 (special case #1)

KCL at supernode (2,3):

Power delivered by 50 V source=