12_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online