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).
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
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
Follow the following steps:
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
6. Apply KCL at each of the essential nodes. Do not apply KCL at the reference node.
Figure : Computation of the branch current i.
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:
>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
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:
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
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
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.
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.
V1=50 (special case #1)
KCL at supernode (2,3):
Power delivered by 50 V source=