9_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

# 9_Ch 18 College Physics ProblemCH18 Direct-Current Circuits...

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Unformatted text preview: Direct-Current Circuits 18.9 Turn the circuit given in Figure P18.9 90° counterclockwise to observe that it is equivalent to that shown in Figure 1 below. This reduces, in stages, as shown in the following figures. b 25.0 V 10.0 W b I20 5.00 W 5.00 c W 20.0 W 10.0 W 25.0 V 10.0 W I20 5.00 W 10.0 W a Figure 1 25.0 W a Figure 2 b I 25.0 V 2.94 W 10.0 W I 25.0 V a Figure 3 12.9 W Figure 4 From Figure 4, I= ∆V 25.0 V = = 1.93 A R 12.9 Ω (b) From Figure 3, ( ∆V )ba = I Rba = ( 1.93 A ) ( 2 .94 Ω ) = 5.68 V (a) From Figures 1 and 2, the current through the 20.0 Ω resistor is I 20 = ( ∆V )ba Rbca = 5.68 V = 0.227 A 25.0 Ω 107 ...
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## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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