{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW14Sol - Platt David Homework 14 Due 4:00 am Inst Ken Shih...

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

View Full Document Right Arrow Icon
Platt, David – Homework 14 – Due: Oct 10 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note: For 008, give your answer in units of muJ(E-6J); For 011,012 give your answers in units of kOhm(E+3Ohm) 001 (part 1 of 3) 10 points 7 . 3 V 2 . 1 V 4 . 8 V I 1 0 . 8 Ω 2 . 1 Ω I 2 7 . 2 Ω I 3 9 Ω Find the current I 1 in the 0 . 8 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 0 . 975845 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation of Charge) I 1 + I 2 - I 3 = 0 . (1) Kirchhoff’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoff’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 0 . 8 Ω , R B = 2 . 1 Ω , R C = 7 . 2 Ω , R D = 9 Ω , E 1 = 7 . 3 V , E 2 = 2 . 1 V , and E 3 = 4 . 8 V . Using determinants, I 1 = fl fl fl fl fl fl 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D fl fl fl fl fl fl fl fl fl fl fl fl 1 1 - 1 R A + R B 0 R D 0 R C R D fl fl fl fl fl fl Expanding along the first row, the numera- tor is D 1 = fl fl fl fl fl fl 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D fl fl fl fl fl fl = 0 - 1 fl fl fl fl E 1 + E 2 R D E 3 R D fl fl fl fl + ( - 1) fl fl fl fl E 1 + E 2 0 E 3 R C fl fl fl fl = - [( E 1 + E 2 ) R D - E 3 R D ] - [ R C ( E 1 + E 2 ) - 0] = R D ( E 3 - E 1 - E 2 ) - R C ( E 1 + E 2 ) = (9 Ω) (4 . 8 V - 7 . 3 V - 2 . 1 V) - (7 . 2 Ω) (7 . 3 V + 2 . 1 V) = - 109 . 08 V Ω . Expanding along the first column, the de- nominator is D = fl fl fl fl fl fl 1 1 - 1 R A + R B 0 R D 0 R C R D fl fl fl fl fl fl = 1 fl fl fl fl 0 R D R C R D fl fl fl fl - ( R A + R B ) fl fl fl fl 1 - 1 R C R D fl fl fl fl + 0 = 0 - R C R D - ( R A + R B ) ( R D + R C ) = (7 . 2 Ω) (9 Ω) - (0 . 8 Ω + 2 . 1 Ω) (9 Ω + 7 . 2 Ω) = - 111 . 78 Ω 2 , and
Background image of page 1

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

View Full Document Right Arrow Icon
Platt, David – Homework 14 – Due: Oct 10 2005, 4:00 am – Inst: Ken Shih 2 I 1 = D 1 D = - 109 . 08 V Ω - 111 . 78 Ω 2 = 0 . 975845 A .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}