1178864748Xuj

1178864748Xuj - Alvarado, Patrick – Homework 7 – Due:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Alvarado, Patrick – Homework 7 – Due: Mar 9 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The emf of a battery is E = 11 V . When the battery delivers a current of 0 . 8 A to a load, the potential difference between the terminals of the battery is 9 V volts. Find the internal resistance of the battery. Correct answer: 2 . 5 Ω. Explanation: Given : E = 11 V , V load = 9 V , and I = 0 . 8 A . The potential difference across the internal resistance is E - V load , so the internal resis- tance is given by r = E - V load I = 11 V- 9 V . 8 A = 2 . 5 Ω . keywords: 002 (part 1 of 1) 10 points Consider the circuit 23 . 4 Ω 46 . 8Ω 4 6 . 8 Ω 23 . 4Ω 46 . 8 Ω a b What is the equivalent resistance between the points a and b ? Correct answer: 40 . 95 Ω. Explanation: R 1 R 2 R 3 R 4 R 5 a b c d Let : R 1 = R = 23 . 4 Ω , R 2 = 2 R = 46 . 8 Ω , R 3 = 2 R = 46 . 8 Ω , R 4 = R = 23 . 4 Ω , and R 5 = 2 R = 46 . 8 Ω . The circuit is redrawn below. R 1 = R R 3 = 2 R R 2 = 2 R R 4 = R R 5 = 2 R a b c d Basic Concepts: R series = R 1 + R 2 + R 3 + · · · 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + · · · Solution: This circuit can be analyzed by straight-forward series/parallel analysis: R 4 and R 5 are in series, so R 45 = R + 2 R = 3 R . R 2 and R 3 are in parallel with R 45 , so 1 R 2345 = 1 2 R + 1 2 R + 1 3 R = 8 6 R R 2345 = 3 R 4 . R 1 and R 2345 are in series, so R eq = R + 3 R 4 = 7 R 4 = 7(23 . 4 Ω) 4 = 40 . 95 Ω . Alvarado, Patrick – Homework 7 – Due: Mar 9 2007, 11:00 pm – Inst: Opyrchal, H 2 keywords: 003 (part 1 of 1) 10 points The following diagram shows part of an elec- trical circuit. 11 Ω 3 Ω 6 Ω 7 Ω 2Ω 4Ω 5Ω A B Find the equivalent resistance R eq between points A and B of the resistor network. Correct answer: 4 Ω. Explanation: R 5 R 2 R 6 R 3 R 1 R 4 R 7 A B Let : R 1 = 2 Ω , R 2 = 3 Ω , R 3 = 7 Ω , R 4 = 4 Ω , R 5 = 11 Ω , R 6 = 6 Ω , and R 7 = 5 Ω . Start from the right-hand side in determining the equivalent resistances. Step 1: R 1 , R 2 , and R 3 are in series, so R 123 = R 1 + R 2 + R 3 = 2 Ω + 3 Ω + 7 Ω = 12 Ω . Step 2: R 123 is now parallel with R 4 , so 1 R 1234 = 1 R 4 + 1 R 123 = R 123 + R 4 R 123 R 4 R 1234 = R 123 R 4 R 123 + R 4 = (12 Ω)(4 Ω) 12 Ω + 4 Ω = 3 Ω . Step 3: R 1234 is in series with R 5 and R 6 , so R s = R 1234 + R 5 + R 6 = 3 Ω + 11 Ω + 6 Ω = 20 Ω . Step 4: Finally, R s is parallel with R 7 , so 1 R eq = 1 R s + 1 R 7 = R 7 + R 5 R 5 R 7 R eq = R s R 7 R s + R 7 = (20 Ω)(5 Ω) 20 Ω + 5 Ω = 4 Ω ....
View Full Document

This note was uploaded on 06/02/2010 for the course MGMT 121 taught by Professor Morozova during the Spring '10 term at NJIT.

Page1 / 9

1178864748Xuj - Alvarado, Patrick – Homework 7 – Due:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online