This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: wright (daw2557) – Chapter 27 – de – (14443) 1 This printout should have 7 questions. Multiplechoice 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
 Spring '09
 SURESHSHARMA
 Physics, Resistor, SEPTA Regional Rail, Jaguar Racing, Electrical resistance, Wright

Click to edit the document details