![By applying KCL at node 2,
I2 =
V2
R3
+
V2 − V1
R2
I2 = −V1
1
R2
+ V2(
1
R2
+
1
R3
) (2)
Solve (1) and (2) for node voltages by any standard method.
By placing (1) and (2) in matrix form, we have
(
1
R1
+
1
R2
) −
1
R2
−
1
R2
(
1
R2
+
1
R3
)
V1
V2
=
I1
I2
[G][V ] = [I]
This can be solved for V1 and V2 by matrix inversion.
[V ] = [G]−1
[I]](https://image.slidesharecdn.com/circuitanalysis-230820043723-d5dcd38d/85/Circuit_Analysis-pdf-3-320.jpg)
![Solve (5) and (6) for I1 and I2 by any standard method. They can
also be solved by matrix inversion.
(R1 + R3) −R3
−R3 R2 + R3
I1
I2
=
V1
−V2
[R][I] = [V ]
This can be solved for I1 and I2 by matrix inversion.
[I] = [R]−1
[V ]
Notice that the matrix R is symmetric in this case.](https://image.slidesharecdn.com/circuitanalysis-230820043723-d5dcd38d/85/Circuit_Analysis-pdf-11-320.jpg)
Circuit_Analysis.pdf
- 1. Nodal Analysis I In nodal analysis, node voltages are found by solving a set of equations. I Choosing node voltages rather than element voltages as variables reduces the number of equations. It uses the following steps. 1. Select a node as the reference node. Assign voltages to other nodes with respect to the reference node. 2. Apply KCL at every node except reference node. 3. Solve the set of equations for node voltages.
- 2. Circuit with Current Sources R1 1 R2 2 R3 I1 I2 0 Let 0 be the reference node and the voltages of 1 and 2 be V1 and V2 respectively. By applying KCL at node 1, I1 = V1 R1 + V1 − V2 R2 I1 = V1( 1 R1 + 1 R2 ) − V2 1 R2 (1)
- 3. By applying KCL at node 2, I2 = V2 R3 + V2 − V1 R2 I2 = −V1 1 R2 + V2( 1 R2 + 1 R3 ) (2) Solve (1) and (2) for node voltages by any standard method. By placing (1) and (2) in matrix form, we have ( 1 R1 + 1 R2 ) − 1 R2 − 1 R2 ( 1 R2 + 1 R3 ) V1 V2 = I1 I2 [G][V ] = [I] This can be solved for V1 and V2 by matrix inversion. [V ] = [G]−1 [I]
- 4. Circuit with Voltage Sources + − VA 1 R1 2 R2 0 R3 3 + − VB There are three nodes. Let V1, V2 and V3 be the node voltages. Since a voltage source VA is connected between 1 and 0, V1 = VA To find V2 and V3, we need to apply KCL at node 2 and 3. As we do not know the current through the voltage source VB, combine 2 and 3 and form a super node.
- 5. Applying KCL for super node, V2 − VA R1 + V2 R2 + V3 R3 = 0 V2( 1 R1 + 1 R2 ) + V3 1 R3 = VA R1 (3) We also know that V2 − V3 = VB (4) Solve (3) and (4) for V2 and V3 by any standard method. Whenever there is a voltage source (dependent or independent) between two non reference nodes, form a super node by combining them and apply KCL for the super node.
- 6. Let us consider the same circuit with R4 between 2 and 3. + − VA 1 R1 2 R2 0 R3 3 + − VB R4 Applying KCL for super node, V2 − VA R1 + V2 R2 + V3 R3 + V2 − V3 R4 + V3 − V2 R4 = 0 The last two terms cancel each other.
- 7. Since R4 does not contribute anything, super node is formed by including it. + − VA 1 R1 2 R2 0 R3 3 + − VB R4 Super node method requires both KCL and KVL to solve for node voltages.
- 8. Test yourself Find vx using nodal analysis. 2 Ω 9 Ω − + vx + − 22 V 11 A KCL at super node, v1 2 + v2 9 = 11 and v1 − v2 = 22 Solving this, v2 = 0 V. Since vx = v2 = 0 V. 2 Ω 1 9 Ω − + vx + − 22 V 11 A 2
- 9. Mesh Analysis I It is another way of analyzing circuits I It uses KVL to find mesh currents. I It is applicable only to planar circuit. Mesh analysis uses the following steps 1. Assign mesh currents to all the meshes. 2. Apply KVL to each of the meshes. 3. Solve the resulting simultaneous equations for mesh currents.
- 10. Circuit with Voltage Sources + − V1 R1 R3 + − V2 R2 I1 I2 Applying KVL for mesh 1 −V1 + I1R1 + (I1 − I2)R3) = 0 I1(R1 + R3) − I2R3 = V1 (5) Applying KVL for mesh 2 (I2 − I1)R3) + I2R2 + V2 = 0 −I1R3 + I2(R2 + R3) = −V2 (6)
- 11. Solve (5) and (6) for I1 and I2 by any standard method. They can also be solved by matrix inversion. (R1 + R3) −R3 −R3 R2 + R3 I1 I2 = V1 −V2 [R][I] = [V ] This can be solved for I1 and I2 by matrix inversion. [I] = [R]−1 [V ] Notice that the matrix R is symmetric in this case.
- 12. Circuit with Current Source + − V1 R1 Ix R3 + − V2 R2 As we do not know the voltage across the current source, a super mesh is formed by excluding the current source and any elements connected in series.
- 13. + − V1 R1 Ix R3 + − V2 R2 + − V1 R1 + − V2 R2 I1 I2
- 14. By applying KVL for super mesh, −V1 + I1R1 + I2R2 + V2 = 0 I1R1 + I2R2 = V1 − V2 (7) We also know that I2 − I1 = Ix (8) Solve (7) and (8) for mesh currents I1 and I2. 1. Super mesh requires the application of both KVL and KCL. 2. Super mesh has no current of its own. Whenever there is a current source (dependent or independent) between two meshes, form a super mesh by excluding the current source and any elements connected in series.
- 15. Test yourself Find vx using mesh analysis. 2 Ω 9 Ω − + vx + − 22 V 11 A By writing KVL for super mesh, 2I1 + 22 + 9I2 = 0 and I1 − I2 = −11 Solving this I2 =0 A and vx = 0 V. 2 Ω 9 Ω − + vx + − 22 V I1 I2