Circuit Analysis using the Node and Mesh Methods We have seen that using Kirchhoff s laws and Ohm s law we can analyze any circuit to determine the operating conditions the currents and voltages The challenge of formal circuit analysis is to derive the smallest set of simultaneous equations that completely define the operating characteristics of a circuit In this lecture we will develop two very powerful methods for analyzing any circuit The node method and the mesh method These methods are based on the systematic application of Kirchhoff s laws We will explain the steps required to obtain the solution by considering the circuit example shown on Figure 1 R1 R3 Vs R2 R4 Figure 1 A typical resistive circuit The Node Method A voltage is always defined as the potential difference between two points When we talk about the voltage at a certain point of a circuit we imply that the measurement is performed between that point and some other point in the circuit In most cases that other point is referred to as ground The node method or the node voltage method is a very powerful approach for circuit analysis and it is based on the application of KCL KVL and Ohm s law The procedure for analyzing a circuit with the node method is based on the following steps 1 Clearly label all circuit parameters and distinguish the unknown parameters from the known 2 Identify all nodes of the circuit 3 Select a node as the reference node also called the ground and assign to it a potential of 0 Volts All other voltages in the circuit are measured with respect to the reference node 4 Label the voltages at all other nodes 5 Assign and label polarities 6 Apply KCL at each node and express the branch currents in terms of the node voltages 7 Solve the resulting simultaneous equations for the node voltages 6 071 Spring 2006 Chaniotakis and Cory 1 8 Now that the node voltages are known the branch currents may be obtained from Ohm s law We will use the circuit of Figure 1 for a step by step demonstration of the node method Figure 2 shows the implementation of steps 1 and 2 We have labeled all elements and identified all relevant nodes in the circuit n1 R1 n2 R3 Vs R2 n3 R4 n4 Figure 2 Circuit with labeled nodes The third step is to select one of the identified nodes as the reference node We have four different choices for the assignment In principle any of these nodes may be selected as the reference node However some nodes are more useful than others Useful nodes are the ones which make the problem easier to understand and solve There are a few general guidelines that we need to remember as we make the selection of the reference node 1 A useful reference node is one which has the largest number of elements connected to it 2 A useful reference node is one which is connected to the maximum number of voltage sources For our example circuit the selection of node n4 as the reference node is the best choice equivalently we could have selected node n1 as our reference node The next step is to label the voltages at the selected nodes Figure 3 shows the circuit with the labeled nodal voltages The reference node is assigned voltage 0 Volts indicated by the ground symbol The remaining node voltages are labeled v1 v2 v3 v1 R1 v2 R3 Vs R2 v3 R4 6 071 Spring 2006 Chaniotakis and Cory 2 Figure 3 Circuit with assigned nodal voltages For the next step we assign current flow and polarities see Figure 4 R1 v1 v2 i1 i3 Vs R3 i2 R2 v3 R4 i1 Figure 4 Example circuit with assigned node voltages and polarities Before proceeding let s look at the circuit shown on Figure 4 bit closer Note that the problem is completely defined Once we determine the values for the node voltages v1 v2 v3 we will be able to completely characterize this circuit So let s go on to calculate the node voltages by applying KCL at the designated nodes For node n1 since the voltage of the voltage source is known we may directly label the voltage v1 as v 1 Vs 4 1 and as a result we have reduced the number of unknowns from 3 to 2 KCL at node n2 associated with voltage v2 gives i1 i 2 i 3 4 2 The currents i1 i2 i3 are expressed in terms of the voltages v1 v2 v3 as follows Vs v 2 R1 v2 i2 R2 v2 v3 i3 R3 i1 6 071 Spring 2006 Chaniotakis and Cory 4 3 4 4 4 5 3 By combining Eqs 4 2 4 5 we obtain Vs v 2 v 2 v 2 v 3 0 R1 R2 R3 4 6 Rewrite the above expression as a linear function of the unknown voltages v2 and v3 gives 1 1 1 1 1 4 7 v 2 Vs v 3 R3 R1 R1 R 2 R 3 KCL at node n3 associated with voltage v3 gives v2 v 3 v 3 0 R3 R4 4 8 or v 2 1 1 1 v 3 0 R3 R3 R4 4 9 The next step is to solve the simultaneous equations 4 7 and 4 9 for the node voltages v2 and v3 Although it is easy to solve Eqs 4 8 and 4 9 directly it is useful to rewrite them in matrix form as follows 1 1 1 v2 R1 R 2 R 3 1 v 2 R3 v 3 1 R3 Vs 1 1 v 3 R3 R4 1 R1 4 10 0 Or R1 R1 1 R 2 R 3 R1 R3 v 2 i R1 R1 v3 R3 R4 R1 R3 or equivalently 6 071 Spring 2006 Chaniotakis and Cory 4 Vs 0 4 11 Aiv V 4 12 In defining the set of simultaneous equations we want to end up with a simple and consistent form The simple rules to follow and check are Place all sources current and voltage on the right hand side of the equation as inhomogeneous drive terms The terms comprising each element on the diagonal of matrix A must have the same sign For example there is no combination RR12 RR13 If an element on the diagonal is comprised of both positive and negative terms there must be a sign error somewhere If you arrange so that all diagonal elements are positive then the off diagonal elements are negative and the matrix is symmetric Aij A ji If the matrix does not have this property there is a mistake somewhere Putting the circuit equations in the above form guarantees that there is a solution consisting of a real set of currents Once we put the equations in matrix form and perform the checks detailed above the solutions then there is a solution if the det A 0 The unknown voltage v k are given by vk det Ak 4 13 det A Where Ak is the matrix A …