28_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

# 28_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: 126 CHAPTER 18 18.44 From Q = Qmax ( 1 − e −t τ ) , the ratio Q Qmax at t = 2τ is found to be Q 1 = 1 − e −2τ τ = 1 − 2 = 0.865 , or Q is 86.5% of Qmax at t = 2τ Qmax e 18.45 The resistive network between a an b reduces, in the stages shown below, to an equivalent resistance of Req = 7.5 Ω . 2.4 W 5.1 W a 18.46 a a 1.8 W b 2.4 W 2.4 W 3.6 W 8.6 W 1.8 W 3.5 W 1.5 W 3.6 W b a 3.6 W b b (a) The circuit reduces as shown below to an equivalent resistance of Req = 14 Ω . 5.0 W 28 V 3.0 W 10 W 3.0 W 10 W 4.0 W 5.0 W 4.0 W 2.0 W 28 V 3.0 W 10 W 3.0 W 10 W 4.0 W 5.0 W 3.0 W 10 W 5.0 W 4.0 W 2.0 W 28 V 10 W 2.0 W Figure 1 28 V 7.5 W Figure 2 5.0 W 10 W 28 V 10 W Figure 3 Figure 5 Figure 4 (b) The power dissipated in the circuit is P = 5.0 W 4.0 W 4.0 W ( ∆V ) Req 2 = I 5.0 W ( 28 V ) 14 Ω 28 V 14 W Figure 6 2 = 56 W (c) The current in the original 5.0 -Ω resistor (in Figure 1) is the total current supplied by the battery. From Figure 6, this is I= ∆V 28 V = = 2.0 A Req 14 Ω ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online