Complex circuits-solutions

Complex - gilvin(jg47854 14 Complex circuits meyers(21235 This print-out should have 17 questions Multiple-choice questions may continue on the

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

R 2 = ?15.4 V1 gilvin (jg47854) - 14. Complex circuits - meyers - (21235) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. This homework is due Tuesday, March 29, at midnight Tucson time. 001 (part 1 of 2) 10.0 points The currents are flowing in the direction in- Pay attention to the sign of the battery and the direction of the current in the figure. Us- ing the lower circuit in the figure, we get E 2 + I 2 R 2 = 0 (2) so I 2 = −E 2 = −0.793814 A . dicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. Assume the batteries have zero inter- nal resistance. 4 V 9.4 Ω Then, for the upper circuit E 1 − I 2 R 2 − I 1 R 1 = 0 . (3) 9.9 Ω E 1 + E 2 − I 1 R 1 = 0 . I I 19.4 Ω 15.4 V I Find the current through the 9.9 Ω resistor and the 4 V battery at the top of the circuit. Correct answer: 1.9596 A. Explanation: Let : R 1 = 9.9 Ω , R 2 = 19.4 Ω , E 1 = 4 V , and E 2 = 15.4 V . E 1 I 1 = E 1 + E 2 R 1 = 4 V + 15.4 V 9.9 Ω = 1.9596 A . Alternate Solution: Using the outside loop −E 1 − E 2 + I 1 R 1 = 0 (4) I 1 = E 1 + E 2 R 1 002 (part 2 of 2) 10.0 points Find the current through the 19.4 Ω resistor in the center of the circuit. Correct answer: −0.793814 A. R 1 I 1 I 2 R 2 E 2 I 3 At nodes, we have I 1 − I 2 − I 3 = 0 . (1) Explanation: From Eq. (2) I 2 = − E 2 R 2 = − 15.4 V 19.4 Ω = −0.793814 A . 003 10.0 points

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

View Full Document
2 Adding, 25 V 12.5 V 4.4 Ω 3.4 Ω E 1 r 2 + E 2 r 1 = I [r 2 R + r 1 (R + r 2 )] I = E 1 r 2 + E 2 r 1 r 2 R + r 1 (R + r 2 ) (25 V) (3.4 Ω) + (12.5 V) (4.4 Ω) 15.6 Ω = (3.4 Ω) (15.6 Ω) + (4.4 Ω) (15.6 Ω + 3.4 Ω) = 1.02459 A . Find the current through the 15.6 Ω lower- right resistor. Correct answer: 1.02459 A. Explanation: E 1 004 (part 1 of 2) 10.0 points An automobile battery has an emf of 15.5 V and an internal resistance of 0.116 Ω. The headlights have total constant resistance 4.95 Ω. B E 2 C F r 1 What is the potential difference across the A headlight bulbs when they are the only load i 1 on the battery? r 2 D Correct answer: 15.1451 V. i 2 Explanation: R E I Let : E = 15.5 V , R i = 0.116 Ω , and Let : E 1 = 25 V , E 2 = 12.5 V , r 1 = 4.4 Ω , r 2 = 3.4 Ω , and R = 15.6 Ω . R h = 4.95 Ω . With the starter off, I h = E R i + R h 15.5 V From the junction rule, I = i 1 + i 2 . Applying Kirchhoff ’s loop rule, we obtain two equations: E 1 = i 1 r 1 + I R (1) E 2 = i 2 r 2 + I R = (I − i 1 ) r 2 + I R = −i 1 r 2 + I (R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = i 1 r 1 r 2 + r 2 I R E 2 r 1 = −i 1 r 1 r 2 + I r 1 (R + r 2 ) = 0.116 Ω + 4.95 Ω = 3.05961 A Thus the voltage across the headlight bulbs is V = I h R h = (3.05961 A) (4.95 Ω) = 15.1451 V . 005 (part 2 of 2) 10.0 points
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

Page1 / 14

Complex - gilvin(jg47854 14 Complex circuits meyers(21235 This print-out should have 17 questions Multiple-choice questions may continue on the

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

View Full Document
Ask a homework question - tutors are online