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barcode reader code in c# net Node 1 ib iS RS Rb Node 2 + in Software
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Creating UPCA In VS .NET Using Barcode creator for Reporting Service Control to generate, create UPCA image in Reporting Service applications. Generating Code128 In Java Using Barcode creation for Java Control to generate, create Code 128C image in Java applications. (326) RS v2 iS = Rb + R S RC which can be used to solve for v1 and v2 Note that, in this particular case, the two equations are independent of each other The following example illustrates a case in which the resulting equations are not independent Bar Code Encoder In ObjectiveC Using Barcode creation for iPhone Control to generate, create bar code image in iPhone applications. Creating Barcode In Java Using Barcode generator for BIRT Control to generate, create barcode image in Eclipse BIRT applications. EXAMPLE 38 Analysis with Dependent Sources
Creating Bar Code In Java Using Barcode drawer for Java Control to generate, create bar code image in Java applications. Making Bar Code In None Using Barcode creator for Online Control to generate, create barcode image in Online applications. Problem
Find the node voltages in the circuit of Figure 323
Solution
Known Quantities: Source current; resistor values; dependent voltage source
R1 v R2 + v3
relationship
Find: Unknown node voltage v
vx + _ I R3
R3 = 4 Schematics, Diagrams, Circuits, and Given Data: I = 05 A; R1 = 5
Dependent source relationship: vx = 2 v3
; R2 = 2
Assumptions: Assume reference node is at the bottom of the circuit Analysis: Applying KCL to node v we nd that Figure 323 v v3 vx v +I =0 R1 R2 Applying KCL to node v3 we nd v3 v v3 =0 R2 R3 If we substitute the dependent source relationship into the rst equation, we obtain a system of equations in the two unknowns v and v3 : 1 1 + R1 R2 1 R2 v+ v+ 1 2 R1 R2 v3 = I 1 1 + R2 R3
v3 = 0 Substituting numerical values, we obtain: 07v 09v3 = 05 05v + 075v3 = 0 Solution of the above equations yields v = 5 V; v3 = 333 V Comments: You will nd the solution to the same example computed using MathCad in
the electronic les that accompany this book
3
Resistive Network Analysis
Remarks on Node Voltage and Mesh Current Methods The techniques presented in this section and the two preceding sections nd use more generally than just in the analysis of resistive circuits These methods should be viewed as general techniques for the analysis of any linear circuit; they provide systematic and effective means of obtaining the minimum number of equations necessary to solve a network problem Since these methods are based on the fundamental laws of circuit analysis, KVL and KCL, they also apply to any electrical circuit, even circuits containing nonlinear circuit elements, such as those to be introduced later in this chapter You should master both methods as early as possible Pro ciency in these circuit analysis techniques will greatly simplify the learning process for more advanced concepts Check Your Understanding
37 The current source ix is related to the voltage vx in Figure 324 by the relation
vx 3 Find the voltage across the 8ix =
resistor by nodal analysis
+ vx 12 i12
6 8 6 6 + 15 V _ ix 15 V + _ Figure 325 Figure 324
38 Find the unknown current ix in Figure 325 using the mesh current method The dependent voltage source is related to the current i12 through the 12 resistor by vx = 2i12 THE PRINCIPLE OF SUPERPOSITION
This brief section discusses a concept that is frequently called upon in the analysis of linear circuits Rather than a precise analysis technique, like the mesh current and node voltage methods, the principle of superposition is a conceptual aid that can be very useful in visualizing the behavior of a circuit containing multiple sources The principle of superposition applies to any linear system and for a linear circuit may be stated as follows: In a linear circuit containing N sources, each branch voltage and current is the sum of N voltages and currents each of which may be computed by setting all but one source equal to zero and solving the circuit containing that single source Part I
Circuits
An elementary illustration of the concept may easily be obtained by simply considering a circuit with two sources connected in series, as shown in Figure 326 + vB2 _ + vB2 _ R vB1 + _ i
R vB1 + _ iB1
R iB2
The net current through R is the sum of the individual source currents: i = iB1 + iB2
Figure 326 The principle of superposition
The circuit of Figure 326 is more formally analyzed as follows The current, i , owing in the circuit on the lefthand side of Figure 326 may be expressed as: i= vB1 vB2 vB1 + vB2 = + = iB1 + iB2 R R R (327) Figure 326 also depicts the circuit as being equivalent to the combined effects of two circuits, each containing a single source In each of the two subcircuits, a short circuit has been substituted for the missing battery This should appear as a sensible procedure, since a short circuit by de nition will always see zero voltage across itself, and therefore this procedure is equivalent to zeroing the output of one of the voltage sources If, on the other hand, one wished to cancel the effects of a current source, it would stand to reason that an open circuit could be substituted for the current source, since an open circuit is by de nition a circuit element through which no current can ow (and which will therefore generate zero current) These basic principles are used frequently in the analysis of circuits, and are summarized in Figure 327 The principle of superposition can easily be applied to circuits containing multiple sources and is sometimes an effective solution technique More often,

