{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

14. Complex circuits-solutions

# 14. Complex circuits-solutions - gilvin(jg47854 14 Complex...

This preview shows pages 1–4. Sign up to view the full content.

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 – fnd 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 ±owing in the direction in- dicated by the arrows. A negative current denotes ±ow opposite to the direction oF the arrow. Assume the batteries have zero inter- nal resistance. 9 . 9 Ω 19 . 4 Ω 4 V 15 . 4 V I I I ²ind 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 . R 1 R 2 E 1 E 2 I 1 I 2 I 3 At nodes, we have I 1 - I 2 - I 3 = 0 . (1) Pay attention to the sign oF the battery and the direction oF the current in the fgure. Us- ing the lower circuit in the fgure, we get E 2 + I 2 R 2 = 0 (2) so I 2 = -E 2 R 2 = - 15 . 4 V 19 . 4 Ω = - 0 . 793814 A . Then, For the upper circuit E 1 - I 2 R 2 - I 1 R 1 = 0 . (3) E 1 + E 2 - I 1 R 1 = 0 . 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 ²ind the current through the 19 . 4 Ω resistor in the center oF the circuit. Correct answer: - 0 . 793814 A. Explanation: ²rom 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
gilvin (jg47854) – 14. Complex circuits – meyers – (21235) 2 4 . 4 Ω 3 . 4 Ω 15 . 6 Ω 25 V 12 . 5 V Find the current through the 15 . 6 Ω lower- right resistor. Correct answer: 1 . 02459 A. Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I 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 . From the junction rule, I = i 1 + i 2 . Applying Kirchhof’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 ) Adding, 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) (3 . 4 Ω) (15 . 6 Ω) + (4 . 4 Ω) (15 . 6 Ω + 3 . 4 Ω) = 1 . 02459 A . 004 (part 1 of 2) 10.0 points An automobile battery has an em± o± 15 . 5 V and an internal resistance o± 0 . 116 Ω. The headlights have total constant resistance 4 . 95 Ω. What is the potential diference across the headlight bulbs when they are the only load on the battery? Correct answer: 15 . 1451 V. Explanation: Let : E = 15 . 5 V , R i = 0 . 116 Ω , and R h = 4 . 95 Ω . With the starter of, I h = E R i + R h = 15 . 5 V 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 What is the potential diference across the
gilvin (jg47854) – 14. Complex circuits – meyers – (21235) 3 headlight bulbs when the starter motor is op- erated, requiring an additional 46 A from the battery? Correct answer: 9 . 93127 V. Explanation: Let : I s = 46 A . R b R s R h The starter requires additional current, so I b = I h + I s .

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.

{[ snackBarMessage ]}

### Page1 / 10

14. Complex circuits-solutions - gilvin(jg47854 14 Complex...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online