Lecture_3 - ECE103 Spring 2008 Lecture 3 Methods of...

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ECE103 Spring 2008 Lecture 3 Methods of analysis Circuit analysis is based on the use of the Kirchhoff’s voltage for mesh analysis and Kirchhoff’s current law for nodal analysis. These two methods will allow us to write a set of equations that once solved will give the values of the currents and voltages in any linear circuit. There are many methods to solve simultaneous equations. One of them is the Cramer’s rule. We will use MATLAB to solve the equations and also PSPICE to simulate the circuits Nodal Analysis (without voltage sources) To simplify, firstly we will assume that circuits do not contain voltage sources. The method of nodal analysis consists in determining the voltages of the nodes in a circuit. Given a circuit with n nodes, without voltage sources, the nodal analysis involves taking the following three steps: 1- Select in the circuit a node of reference. Assign voltage values to the remaining (n-1) nodes (v 1 , v 2 , ….v n-1 ) relative to this reference node. 2- Apply the Kirchhoff’s current law to each of the (n-1) nodes and use the Ohm’s law to determine the branch currents in terms of the node voltages 3- Solve the resulting simultaneous equations to obtain the node voltages To understand these three steps we will describe the method solving an example. First we have to choose a reference node. All voltages for the rest of the nodes will be referred to the voltage of this reference node. The reference node is the ground node, this is defined as the zero potential. In the figure this is indicated as the node “0”. We now apply the KCL to each node. For node “1” we will have 1 2 1 2 0 I I i i - - - = We used in this equation the convention that currents flowing into the node are positive and currents flowing out of the node are negative. With the same convention, for node 2 we have: 0 1 2 I1 I2 i1 i2 i3
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2 2 3 0 I i i + - = To obtain the values of the currents we now apply the Ohm’s law. Considering that the current flows from the higher potential to the lower potential in a resistor, we can write ( ) 1 1 1 1 1 1 0 0 v i or i G v R - = = - ( ) 1 2 2 1 1 1 2 2 v v i or i G v v R - = = - ( ) 2 3 3 3 2 3 0 0 v i or i G v R - = = - Now we can substitute these expressions in the equations obtained with the KCL 1 1 2 1 2 1 2 1 2 1 2 0 v v v I I i i I I R R - - - - = - - - = 1 2 2 2 2 3 2 2 3 0 v v v I i i I R R - + - = + - = Written in terms of conductances we have ( ) 1 1 2 1 2 1 2 G v G v v I I + - = - ( ) 3 2 2 1 2 2 G v G v v I - - = these are two simultaneous equations that we must solve in order to obtain the values of v 1 and v 2 . Expressed in a matrix form 1 2 2 1 1 2 2 2 3 2 2 G G G v I I G G G v I + - -   =   - + And can be solved using MATLAB Let us solve a circuit using different methods. The example we will solve is the circuit in the figure 6 7 2 1A 4A 1 2 i1 i2 i3
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Method 1: Variable replacement technique At node 1: 1 2 1 A i i = + [1] At node 2:
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This note was uploaded on 10/28/2008 for the course ECE 103 taught by Professor Marconi during the Spring '08 term at Colorado State.

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Lecture_3 - ECE103 Spring 2008 Lecture 3 Methods of...

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