29_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

29_Ch 18 College Physics ProblemCH18 Direct-Current...

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Unformatted text preview: 127 Direct-Current Circuits 18.47 (a) The resistors combine to an equivalent resistance of Req = 15 Ω as shown. 6.0 W a I1 15 V I3 I2 + – 2.4 W c e I4 6.0 W 15 V + – f b 15 V + – I2 6.0 W a I3 c I1 + 15 V – 6.0 W a I1 6.0 W + 15 V – 3.0 W 6.0 W b 3.6 W Figure 2 c 6.0 W I3 d Figure 1 I1 e 6.0 W d a I2 6.0 W 9.0 W b 2.4 W c I1 I5 6.0 W 6.0 W 6.0 W a 15 W 6.0 W d Figure 3 b d Figure 4 (b) From Figure 5, I1 = b Figure 5 ∆Vab 15 V = = 1.0 A Req 15 Ω Then, from Figure 4, ∆Vac = ∆Vdb = I1 ( 6.0 Ω ) = 6.0 V and ∆Vcd = I1 ( 3.0 Ω ) = 3.0 V ∆Vcd 3.0 V = = 0.50 A 6.0 Ω 6.0 Ω From Figure 3, I2 = I3 = From Figure 2, ∆Ved = I 3 ( 3.6 Ω ) = 1.8 V I4 = Then, from Figure 1, and I5 = ∆Vfd 9.0 Ω = ∆Ved 1.8 V = = 0.30 A 6.0 Ω 6.0 Ω ∆Ved 1.8 V = = 0.20 A 9.0 Ω 9.0 Ω (c) From Figure 2, ∆Vce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences were calculated above in part (b). The results were ∆Vac = ∆Vdb = 6.0 V ; ∆Vcd = 3.0 V ; and ∆Vfd = ∆Ved = 1.8 V ...
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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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