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Resistive Network Analysis in Software
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QR Code 2d Barcode Generation In VB.NET Using Barcode maker for Visual Studio .NET Control to generate, create Quick Response Code image in .NET framework applications. Bar Code Printer In None Using Barcode drawer for Software Control to generate, create barcode image in Software applications. A comparison of this result with the analogous result obtained by the node voltage method reveals that we are using Ohm s law in conjunction with KVL (in contrast with the use of KCL in the node voltage method) to determine the minimum set of equations required to solve the circuit Mesh Analysis with Current Sources Bar Code Creation In None Using Barcode creation for Software Control to generate, create bar code image in Software applications. Drawing GS1  12 In None Using Barcode printer for Software Control to generate, create Universal Product Code version A image in Software applications. 5 10 V GS1128 Generation In None Using Barcode maker for Software Control to generate, create EAN128 image in Software applications. Paint Code 128B In None Using Barcode drawer for Software Control to generate, create Code 128 Code Set A image in Software applications. 2 + vx 2A i2 4
Code 93 Extended Generator In None Using Barcode drawer for Software Control to generate, create USS93 image in Software applications. Printing GS1  12 In Java Using Barcode printer for Java Control to generate, create UPC Symbol image in Java applications. Mesh analysis is particularly effective when applied to circuits containing voltage sources exclusively; however, it may be applied to mixed circuits, containing both voltage and current sources, if care is taken in identifying the proper current in each mesh The method is illustrated by solving the circuit shown in Figure 318 The rst observation in analyzing this circuit is that the presence of the current source requires that the following relationship hold true: i1 i 2 = 2 A (317) Create Code 39 In Visual Studio .NET Using Barcode encoder for .NET framework Control to generate, create Code 39 image in VS .NET applications. Painting Barcode In .NET Using Barcode drawer for Reporting Service Control to generate, create bar code image in Reporting Service applications. Figure 318 Mesh analysis with current sources
ECC200 Scanner In Visual Basic .NET Using Barcode decoder for .NET Control to read, scan read, scan image in VS .NET applications. UCC  12 Printer In None Using Barcode generation for Font Control to generate, create Universal Product Code version A image in Font applications. If the unknown voltage across the current source is labeled vx , application of KVL around mesh 1 yields: 10 5i1 vx = 0 while KVL around mesh 2 dictates that vx 2i2 4i2 = 0 (319) (318) Encode GS1 DataBar Expanded In .NET Framework Using Barcode maker for VS .NET Control to generate, create GS1 RSS image in Visual Studio .NET applications. Bar Code Creator In None Using Barcode generation for Font Control to generate, create barcode image in Font applications. Substituting equation 319 in equation 318, and using equation 317, we can then obtain the system of equations 5i1 + 6i2 = 10 i1 + i2 = 2 which we can solve to obtain i1 = 2 A i2 = 0 A (321) (320) Note also that the voltage across the current source may be found by using either equation 318 or equation 319; for example, using equation 319, vx = 6i2 = 0 V (322) The following example further illustrates the solution of this type of circuit
EXAMPLE 37 Mesh Analysis with Current Sources
Problem
Find the mesh currents in the circuit of Figure 319
Part I
Circuits
Solution
Known Quantities: Source current and voltage; resistor values Find: Mesh currents
R3 V
R2 = 8 ; R3 = 6
; R4 = 4
Assumptions: Assume clockwise mesh currents i1 , i2 , and i3 Analysis: Starting from mesh 1, we see immediately that the current source forces the Figure 319 mesh current to be equal to I : i1 = I There is no need to write any further equations around mesh 1, since we already know the value of the mesh current Now we turn to meshes 2 and 3 to obtain: R2 (i2 i1 ) R3 (i2 i3 ) + V = 0 R1 (i3 i1 ) R4 i3 R3 (i3 i2 ) = 0 mesh 2 mesh 3 Rearranging the equations and substituting the known value of i1 , we obtain a system of two equations in two unknowns: 14i2 6i3 = 10 6i2 + 13i3 = 15 which can be solved to obtain i2 = 095 A i3 = 055 A As usual, you should verify that the solution is correct by applying KVL
Comments: Note that the current source has actually simpli ed the problem by
constraining a mesh current to a xed value
Check Your Understanding
34 Find the unknown voltage, vx , by mesh current analysis in the circuit of Figure 320
6 5 60 15 V + _ 6 + vx
12 24 V + _ + 15 V _ 35 Find the unknown current, Ix , using mesh current methods in the circuit of Figure
Schematics, Diagrams, Circuits, and Given Data: I = 05 A; V = 6 V; R1 = 3
3
Resistive Network Analysis
36 Show that the equations given in Example 36 are correct, by applying KCL at each
node
NODAL AND MESH ANALYSIS WITH CONTROLLED SOURCES
The methods just described also apply, with relatively minor modi cations, in the presence of dependent (controlled) sources Solution methods that allow for the presence of controlled sources will be particularly useful in the study of transistor ampli ers in 8 Recall from the discussion in Section 23 that a dependent source is a source that generates a voltage or current that depends on the value of another voltage or current in the circuit When a dependent source is present in a circuit to be analyzed by node or mesh analysis, one can initially treat it as an ideal source and write the node or mesh equations accordingly In addition to the equation obtained in this fashion, there will also be an equation relating the dependent source to one of the circuit voltages or currents This constraint equation can then be substituted in the set of equations obtained by the techniques of nodal and mesh analysis, and the equations can subsequently be solved for the unknowns It is important to remark that once the constraint equation has been substituted in the initial system of equations, the number of unknowns remains unchanged Consider, for example, the circuit of Figure 322, which is a simpli ed model of a bipolar transistor ampli er (transistors will be introduced in 8) In the circuit of Figure 322, two nodes are easily recognized, and therefore nodal analysis is chosen as the preferred method Applying KCL at node 1, we obtain the following equation: iS = v1 1 1 + RS Rb (323) KCL applied at the second node yields: ib + v2 =0 RC (324) Next, it should be observed that the current ib can be determined by means of a simple current divider: ib = iS 1/Rb RS = iS 1/Rb + 1/RS Rb + R S

