12_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

12_Ch 18 College Physics ProblemCH18 Direct-Current...

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Unformatted text preview: 110 CHAPTER 18 18.13 The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of Rad = ( 63 11) Ω . 6.0 W 3.0 W a 3.0 W b I1 c d 3.0 W 6.0 W 4.0 W I2 3.0 W a I2 2.0 W I e 12 W I12 I1 3.0 W c d b e 3.0 W 2.0 W I 18 V Figure 1 Figure 2 18 V 6.0 W 3.0 W I1 a I I2 b 3W a d b 30  11 W d I 5.0 W 63  11 a W I 18 V Figure 4 18 V Figure 3 From Figure 5, I = ( ∆V )ad Rad Then, from Figure 4, = 18 V Figure 5 18 V = 3.14 A ( 63 11) Ω ( ∆V )bd = I Rbd = ( 3.14 A ) ( 30 11 Ω ) = 8.57 V Now, look at Figure 2 and observe that I2 = so ( ∆V )bd 3 .0 Ω + 2 .0 Ω = 8.57 V = 1.71 A 5.0 Ω ( ∆V )be = I 2 Rbe = ( 1.71 A ) ( 3.0 Ω ) = 5.14 V Finally, from Figure 1, I12 = ( ∆V ) b e R12 = 5.14 V = 0.43 A 12 Ω d ...
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