19_InstSolManual_PDF_Part22

19_InstSolManual_PDF_Part22 - 19-22 Chapter 19 Solve: (a)...

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Solve: (a) Replacing series and parallel combinations of resistors by their equivalents gives the equivalent networks as shown in Figure 19.86. The equivalent resistance of the network is Figure 19.86 (b) The voltage across the resistor in Figure 19.86a is This is the voltage across the resistor in Figure 19.86b. The current through the resistor is 4.8 A. This is also the current through the resistor and the voltage across that resistor is 96.0 V. Therefore, the voltage between points x and y is This voltage is divided equally between the resistors in Figure 19.86b, so the voltage between points x and y is The voltmeter will read 57.6 V. 19.87. Set Up: Table 19.1 gives the resistivities of copper and aluminum to be and For the cables in series (end-to-end), For the cables in parallel the equivalent resistance is given by Note that in the two configurations the copper and aluminum sections have different lengths. And, for the parallel cables the cross sectional area of each cable is half what it is for the end to end configuration.
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