barcode reader code in c# net In order to set a voltage source equal to zero, we replace it with a short circuit R1 vS + _ R1 in Software

Generation QR Code ISO/IEC18004 in Software In order to set a voltage source equal to zero, we replace it with a short circuit R1 vS + _ R1

1 In order to set a voltage source equal to zero, we replace it with a short circuit R1 vS + _ R1
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2 In order to set a current source equal to zero, we replace it with an open circuit R1 R1 vS + _ vS + _
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Figure 327 Zeroing voltage and current sources
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however, other methods result in a more ef cient solution Example 39 further illustrates the use of superposition to analyze a simple network The Check Your Understanding exercises at the end of the section illustrate the fact that superposition is often a cumbersome solution method
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EXAMPLE 39 Principle of Superposition
Problem
Determine the current i2 in the circuit of Figure 318 using the principle of superposition
Solution
Known Quantities: Source voltage and current values Resistor values Find: Unknown current i2 Given Data Figure 318 Assumptions: Assume reference node is at the bottom of the circuit Analysis: Part 1: Zero the current source Once the current source has been set to zero
(replaced by an open circuit), the resulting circuit is a simple series circuit; the current owing in this circuit, i2 V , is the current we seek Since the total series resistance is 5 + 2 + 4 = 11 , we nd that i2 V = 10/11 = 0909 A Part 2: Zero the voltage source After zeroing of the voltage source by replacing it with a short circuit, the resulting circuit consists of three parallel branches: On the left we have a single 5- resistor; in the center we have a 2-A current source (negative because the source current is shown to ow into the ground node); on the right we have a total resistance of 2 + 4 = 6 Using the current divider rule, we nd that the current owing in the right branch, i2 I , is given by: i2 I = 1 6 1 1 + 5 6 ( 2) = 0909 A
And, nally, the unknown current i2 is found to be i2 = i2-V + i2 I = 0 A The result is, of course, identical to that obtained by mesh analysis
Comments: Superposition may appear to be a very ef cient tool However, beginners
may nd it preferable to rely on more systematic methods, such as nodal analysis, to solve circuits Eventually, experience will suggest the preferred method for any given circuit
Check Your Understanding
39 Find the voltages va and vb for the circuits of Example 34 by superposition 310 Repeat Check Your Understanding Exercise 32, using superposition This exercise illustrates that superposition is not necessarily a computationally ef cient solution method
Part I
Circuits
311 Solve Example 35, using superposition 312 Solve Example 37, using superposition
ONE-PORT NETWORKS AND EQUIVALENT CIRCUITS
You may recall that, in the discussion of ideal sources in 2, the ow of energy from a source to a load was described in a very general form, by showing the connection of two black boxes labeled source and load (see Figure 210) In the same gure, two other descriptions were shown: a symbolic one, depicting an ideal voltage source and an ideal resistor; and a physical representation, in which the load was represented by a headlight and the source by an automotive battery Whatever the form chosen for source-load representation, each block source or load may be viewed as a two-terminal device, described by an i-v characteristic This general circuit representation is shown in Figure 328 This con guration is called a one-port network and is particularly useful for introducing the notion of equivalent circuits Note that the network of Figure 328 is completely described by its i-v characteristic; this point is best illustrated by the next example
i + v Linear network
Figure 328 One-port network
EXAMPLE 310 Equivalent Resistance Calculation
Problem
Determine the source (load) current i in the circuit of Figure 329 using equivalent resistance ideas
i + vS
Source
Load R1 R2 R3
Figure 329 Illustration of equivalent-circuit concept
Load circuit
Solution
Known Quantities: Source voltage, resistor values Find: Source current Given Data: Figures 329, 330 Assumptions: Assume reference node is at the bottom of the circuit
Equivalent load circuit
Figure 330 Equivalent load resistance concept
3
Resistive Network Analysis
Analysis: Insofar as the source is concerned, the three parallel resistors appear identical to a single equivalent resistance of value
REQ =
1 1 1 1 + + R1 R2 R3
Thus, we can replace the three load resistors with the single equivalent resistor REQ , as shown in Figure 330, and calculate i= vS REQ
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