{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

38_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

# 38_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: 136 CHAPTER 18 18.61 (a) Using the rules for combining resistors in series and parallel, the circuit reduces as shown below: 10.0 W 25.0 V + 10.0 W 25.0 V + I a 10.0 W 5.00 W 5.00 W 10.0 W 25.0 V + I b I1 10.0 W a I b 5.00 W 20.0 W I1 a 25.0 W Step 2 Step 1 b 2.94 W Step 3 From the figure of Step 3, observe that I= 25.0 V = 1.93 A 12.94 A ∆Vab = I ( 2.94 Ω ) = ( 1.93 A ) ( 2.94 Ω ) = 5.68 V and (b) From the figure of Step 1, observe that I1 = 18.62 ∆Vab 5.68 V = = 0.227 A 25.0 Ω 25.0 Ω (a) When the power supply is connected to points A and B, the circuit reduces as shown below to an equivalent resistance of Req = 0.099 9 Ω . 100 W R2 + R3 + 100 W = 111 W Req = 0.0999 W A R1 B R2 C R3 D A R1 = 0.100 W B I1 + + + 5.00 V 5.00 V From the center figure above, observe that I R1 = I1 = and I R2 = I R3 = I100 = 5.00 V 5.00 V = 50.0 A 0.100 Ω 5.00 V = 0.045 0 A = 45.0 mA 111 Ω ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online