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ANALYSIS METHODS
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Fig. 4-6 Fig. 4-7
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Thevenin too). The output resistance is found by dividing the open-circuited voltage to the shortcircuited current at the desired node. The short-circuited current is found in Section 4.6.
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A driving voltage in one part of a network results in currents in all the network branches. For example, a voltage source applied to a passive network results in an output current in that part of the network where a load resistance has been connected. In such a case the network has an overall transfer resistance. Consider the passive network suggested in Fig. 4-7, where the voltage source has been designated as Vr and the output current as Is . The mesh current equation for Is contains only one term, the one resulting from Vr in the numerator determinant:     rs Is 0 1s 0 Vr 0 R R The network transfer resistance is the ratio of Vr to Is : Rtransfer;rs R rs
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Because the resistance matrix is symmetric, rs sr , and so Rtransfer;rs Rtransfer;sr This expresses an important property of linear networks: If a certain voltage in mesh r gives rise to a certain current in mesh s, then the same voltage in mesh s produces the same current in mesh r. Consider now the more general situation of an n-mesh network containing a number of voltage sources. The solution for the current in mesh k can be rewritten in terms of input and transfer resistances [refer to (7), (8), and (9) of Example 4.4]: Ik V1 Vk 1 Vk Vk 1 Vn Rtransfer;1k Rtransfer; k 1 k Rinput;k Rtransfer; k 1 k Rtransfer;nk
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There is nothing new here mathematically, but in this form the current equation does illustrate the superposition principle very clearly, showing how the resistances control the e ects which the voltage sources have on a particular mesh current. A source far removed from mesh k will have a high transfer resistance into that mesh and will therefore contribute very little to Ik . Source Vk , and others in meshes adjacent to mesh k, will provide the greater part of Ik .
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The mesh current and node voltage methods are the principal techniques of circuit analysis. However, the equivalent resistance of series and parallel branches (Sections 3.4 and 3.5), combined with the voltage and current division rules, provide another method of analyzing a network. This method is tedious and usually requires the drawing of several additional circuits. Even so, the process of reducing
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ANALYSIS METHODS
the network provides a very clear picture of the overall functioning of the network in terms of voltages, currents, and power. The reduction begins with a scan of the network to pick out series and parallel combinations of resistors.
EXAMPLE 4.6 Obtain the total power supplied by the 60-V source and the power absorbed in each resistor in the network of Fig. 4-8. Rab 7 5 12  12 6 Rcd 4 12 6
Fig. 4-8 These two equivalents are in parallel (Fig. 4-9), giving Ref 4 12 3 4 12
Then this 3- equivalent is in series with the 7- resistor (Fig. 4-10), so that for the entire circuit, Req 7 3 10 
Fig. 4-9
Fig. 4-10
The total power absorbed, which equals the total power supplied by the source, can now be calculated as PT V2 60 2 360 W Req 10
This power is divided between Rge and Ref as follows: Pge P7 7 360 252 W 7 3 Pef 3 360 108 W 7 3
Power Pef is further divided between Rcd and Rab as follows: Pcd 12 108 81 W 4 12 Pab 4 108 27 W 4 12
ANALYSIS METHODS
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Finally, these powers are divided between the individual resistances as follows: P12 P6 6 81 27 W 12 6 12 81 54 W 12 6 P7 P5 7 27 15:75 W 7 5 5 27 11:25 W 7 5
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