{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 27 Solutions

# Chapter 27 Solutions - wright(daw2557 – Chapter 27 – de...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: wright (daw2557) – Chapter 27 – de – (14443) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four resistors are connected as shown in the figure. 97 V S 1 c d a b 1 8 Ω 47Ω 54Ω 7 1 Ω Find the resistance between points a and b . Correct answer: 12 . 5721 Ω. Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 18 Ω , R 2 = 47 Ω , R 3 = 54 Ω , R 4 = 71 Ω , and E = 97 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc- tions connected by zero resistance. E B a d b c R 1 R 2 R 3 R 4 The series connection of R 2 and R 3 gives the equivalent resistance R 23 = R 2 + R 3 = 47 Ω + 54 Ω = 101 Ω . The total resistance R ab between a and b can be obtained by calculating the resistance in the parallel combination of the resistors R 1 , R 4 , and R 23 ; i.e. , 1 R ab = 1 R 1 + 1 R 2 + R 3 + 1 R 4 = R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) R 1 R 4 ( R 2 + R 3 ) R ab = R 1 R 4 ( R 2 + R 3 ) R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) The denominator is R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) = (71 Ω)[47 Ω + 54 Ω] + (18 Ω) (71 Ω) + (18 Ω) [47 Ω + 54 Ω] = 10267 Ω 2 , so the equivalent resistance is R ab = (18 Ω) (71 Ω) [47 Ω + 54 Ω] (10267 Ω 2 ) = 12 . 5721 Ω ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

Chapter 27 Solutions - wright(daw2557 – Chapter 27 – de...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online