Nodal Analysis and Mesh Analysis in Electrical Circuits, Study notes of Basic Electronics

Download Nodal Analysis and Mesh Analysis in Electrical Circuits and more Study notes Basic Electronics in PDF only on Docsity! 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. Circuit with Current Sources R1 1 R2 2 R3I1 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) 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. 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. 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. 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) 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. 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.