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EXAMPLE 32 Nodal Analysis
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Write the nodal equations and solve for the node voltages in the circuit of Figure 36
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Solution
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Known Quantities: Source currents, resistor values Find: All node voltages and branch currents Figure 36
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va i1 ia R1 R2 i2 R3 i3 ib vb i4 R4
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R2 = 500 circuit
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Schematics, Diagrams, Circuits, and Given Data: ia = 1 mA; ib = 2 mA; R1 = 1 k ;
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; R3 = 22 k ; R4 = 47 k
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Assumptions: The reference (ground) node is chosen to be the node at the bottom of the
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Analysis: To write the node equations, we start by selecting the reference node (step 1)
Figure 37 illustrates that two nodes remain after the selection of the reference node Let us label these a and b and de ne voltages va and vb (step 2)
3
Resistive Network Analysis
Next, we apply KCL at each of the nodes, a and b (step 3): ia va va vb =0 R1 R2 (node a) (node b)
vb vb va vb + ib =0 R2 R3 R4 1 1 + R1 R2 1 R2 va + 1 R2
and rewrite the equations to obtain a linear system: va + v b = ia v b = ib
1 1 1 + + R2 R3 R4
Substituting the numerical values in these equations, we get 3 10 3 va 2 10 3 vb = 1 10 3 2 10 3 va + 267 10 3 vb = 2 10 3 or 3va 2vb = 1 2va + 267vb = 2 The solution va = 1667 V, vb = 2 V may then be obtained by solving the system of equations
EXAMPLE 33 Solution of Linear System of Equations Using Cramer s Rule
Problem
Solve the circuit equations obtained in Example 32 using Cramer s rule (see Appendix A)
Solution
Known Quantities: Linear system of equations Find: Node voltages Analysis: The system of equations generated in Example 32 may also be solved by using
linear algebra methods, by recognizing that the system of equations can be written as: 3 2 2 267 va vb = 1V 2V
By using Cramer s rule (see Appendix A), the solution for the two unknown variables, va and vb , can be written as follows: va = 1 2 3 2 3 2 3 2 2 267 2 267 1 2 2 267 = 667 (1)(267) ( 2)(2) = = 1667 V (3)(267) ( 2)( 2) 4
vb =
8 (3)(2) ( 2)(1) = =2V (3)(267) ( 2)( 2) 4
The result is the same as in Example 32
Part I
Circuits
Comments: While Cramer s rule is an ef cient solution method for simple circuits (eg, two nodes), it is customary to use computer-aided methods for larger circuits Once the nodal equations have been set in the general form presented in equation 39, a variety of computer aids may be employed to compute the solution You will nd the solution to the same example computed using MathCad in the electronic les that accompany this book
Nodal Analysis with Voltage Sources It would appear from the examples just shown that the node voltage method is very easily applied when current sources are present in a circuit This is, in fact, the case, since current sources are directly accounted for by KCL Some confusion occasionally arises, however, when voltage sources are present in a circuit analyzed by the node voltage method In fact, the presence of voltage sources actually simpli es the calculations To further illustrate this point, consider the circuit of Figure 38 Note immediately that one of the node voltages is known already! The voltage at node a is forced to be equal to that of the voltage source; that is, va = vS Thus, only two nodal equations will be needed, at nodes b and c: vS vb vb vb vc =0 R1 R2 R3 vb v c vc + iS =0 R3 R4 Rewriting these equations, we obtain: 1 1 1 + + R1 R2 R3 1 R3 vb + 1 R3 vc = vS R1 (node b) (310) (node c)
Figure 38 Nodal analysis with voltage sources
+ v _ S
R4 iS
vb +
1 1 + R3 R4
(311)
vc = iS
Note how the term vS /R1 on the right-hand side of the rst equation is really a current, as is dimensionally required by the nature of the node equations
EXAMPLE 34 Nodal Analysis with Voltage Sources
Problem
Find the node voltages in the circuit of Figure 39
va Node a
vb Node b V + _
Solution
Known Quantities: Source current and voltage; resistor values Find: Node voltages
R1 R3
R 2 = 2 k ; R3 = 3 k
Schematics, Diagrams, Circuits, and Given data: I = 2 mA; V = 3 V; R1 = 1 k ; Figure 39 Assumptions: Place the reference node at the bottom of the circuit
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