Lesson 6Basic Laws of Electric CircuitsNodal Analysis

Basic CircuitsNodal Analysis: The Concept. Every circuit has n nodes with one of the nodes being designated as a reference node. We designate the remaining n 1 nodes as voltage nodes and give each node a unique name, vi. At each node we write Kirchhoffs current law in terms of the node voltages.1

Basic CircuitsNodal Analysis: The Concept. We form n-1 linear equations at the n-1 nodes in terms of the node voltages. We solve the n-1 equations for the n-1 node voltages. From the node voltages we can calculate any branch current or any voltage across any element.2

Basic CircuitsNodal Analysis: Concept Illustration:Figure 6.1: Partial circuit used to illustrate nodal analysis.Eq 6.13

Basic CircuitsNodal Analysis: Concept Illustration:Clearing the previous equation gives,Eq 6.2We would need two additional equations, from theremaining circuit, in order to solve for V1, V2, and V34

Basic CircuitsNodal Analysis: Example 6.1Given the following circuit. Set-up the equations to solve for V1 and V2. Also solve for the voltage V6.Figure 6.2: Circuit for Example 6.1.5

Basic CircuitsNodal Analysis: Example 6.1, the nodal equations.Eq 6.3Eq 6.46

Basic CircuitsNodal Analysis: Example 6.1: Set up for solution.Eq 6.3Eq 6.4Eq 6.5Eq 6.67

Nodal Analysis: Example 6.2, using circuit values.Figure 6.3: Circuit for Example 6.2.Find V1 and V2.At v1:At v2:Eq 6.7Eq 6.8Basic Circuits8

Nodal Analysis: Example 6.2: Clearing Equations; From Eq 6.7:V1 + 2V1 2V2 = 20or3V1 2V2 = 20From Eq 6.8:4V2 4V1 + V2 = -120or-4V1 + 5V2 = -120Eq 6.9Eq 6.10Solution: V1 = -20 V, V2 = -40 VBasic Circuits9

Basic CircuitsNodal Analysis: Example 6.3: With voltage source.Figure 6.4: Circuit for Example 6.3.At V1:At V2:Eq 6.11Eq 6.1210

Basic CircuitsNodal Analysis: Example 6.3: Continued.Collecting terms in Equations (6.11) and (6.12) givesEq 6.13Eq 6.1411

Basic CircuitsNodal Analysis: Example 6.4: Numerical example with voltagesource. Figure 6.5: Circuit for Example 6.4.What do we do first?12

Basic CircuitsNodal Analysis: Example 6.4: ContinuedAt v1:At v2:Eq 6.15Eq 6.1613

Basic CircuitsNodal Analysis: Example 6.4: ContinuedClearing Eq 6.154V1 + 10V1 + 100 10V2 = -200or14V1 10V2 = -300Clearing Eq 6.164V2 + 6V2 60 6V1 = 0or-6V1 + 10V2 = 60Eq 6.17Eq 6.18V1 = -30 V, V2 = -12 V, I1 = -2 A14

Basic CircuitsNodal Analysis: Example 6.5: Voltage super node. Given the following circuit. Solve for the indicated nodal voltages.Figure 6.6: Circuit for Example 6.5.When a voltage source appears between two nodes, an easy way to handle this is to form a super node. The super node encircles thevoltage source and the tips of the branches connected to the nodes.super node15

Basic CircuitsNodal Analysis: Example 6.5: Continued.At V1At supernodeConstraint EquationV2 V3 = -10Eq 6.19Eq 6.20Eq 6.2116

Basic CircuitsNodal Analysis: Example 6.5: Continued.Clearing Eq 6.19, 6.20, and 6.21:7V1 2V2 5V3 = 60-14V1 + 9V2 + 12V3 = 0V2 V3 = -10 Eq 6.22Eq 6.23Eq 6.24Solving gives:V1 = 30 V, V2 = 14.29 V, V3 = 24.29 V17

Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.Consider the circuit below. We desire to solve for the node voltagesV1 and V2.Figure 6.7: Circuit for Example 6.6.In this case we have a dependent source, 5Vx, that must be reckonedwith. Actually, there is a constraint equation of Eq 6.2518

Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.At node V1At node V2The constraint equation:19

Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.Clearing the previous equations and substituting the constraint VX = V1 - V2 gives,Eq 6.26Eq 6.27which yields,20

circuitsEnd of Lesson 6Nodal Analysis