solutions 10

solutions 10 - Patel, Kinal – Homework 10 – Due: Feb 26...

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

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.

Unformatted text preview: Patel, Kinal – Homework 10 – Due: Feb 26 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 10 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 5 . 9 Ω 4 . 9 Ω 21 . 6 Ω 22 . 4 V 11 . 2 V Find the current through the 21 . 6 Ω lower- right resistor. Correct answer: 0 . 670659 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 = 22 . 4 V , E 2 = 11 . 2 V , r 1 = 5 . 9 Ω , r 2 = 4 . 9 Ω , and R = 21 . 6 Ω . 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 ) 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 ) = (22 . 4 V) (4 . 9 Ω) + (11 . 2 V) (5 . 9 Ω) (4 . 9 Ω) (21 . 6 Ω) + (5 . 9 Ω) (21 . 6 Ω + 4 . 9 Ω) = . 670659 A . keywords: 002 (part 1 of 1) 10 points 2 Ω 5 Ω 7 Ω 8 Ω 11 Ω 2 Ω 2 Ω 1 Ω 13 V 22 V 38 V Find the magnitude of the current in the 13 V cell. Correct answer: 0 . 941545 A. Explanation: i 1 R 1 i 3 R 2 i 1 R 3 i 3 R 4 i 2 R 5 r 1 r 2 r 3 E 1 E 2 E 3 Patel, Kinal – Homework 10 – Due: Feb 26 2008, 7:00 pm – Inst: Weathers 2 Let : E 1 = 13 V , E 2 = 22 V , E 3 = 38 V , R 1 = 2 Ω , R 2 = 5 Ω , R 3 = 7 Ω , R 4 = 8 Ω , R 5 = 11 Ω , r 1 = 2 Ω , r 2 = 2 Ω , and r 3 = 1 Ω . Basic Concepts: Kirchhoff’s Laws: X V = 0 around a closed loop . X I = 0 at a circuit junction . Solution: Applying Kirchhoff’s law to the outside loop and the lower loop we get 3 equa- tions in 3 unknowns; i.e. , E 1-E 3 = ( R 3 + r 1 + R 1 ) i 1 + ( R 2 + r 3 + R 4 ) i 3 (1) E 2-E 3 = ( R 5 + r 2 ) i 2 + ( R 2 + r 3 + R 4 ) i 3 (2) 0 =- i 1- i 2 + i 3 . (3) Subtracting the first two equations, E 1-E 2 = ( R 3 + r 1 + R 1 ) i 1- ( R 5 + R 2 ) i 2 (4) Eliminating i 3 in equations (2) and (3), E 2-E 3 = ( R 2 + r 3 + R 4 ) i 1 + ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 (5) Multiply equation (4) by ( R 2 + r 3 + R 4 + R 5 + r 2 ) and equation (5) by ( R 5 + r 2 ): ( E 1-E 2 ) ( R 2 + r 3 + R 4 + R 5 + r 2 ) = ( R 3 + r 1 + R 1 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 1- ( R 5 + r 2 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 ( E 2-E 3 ) ( R 5 + r 2 ) = ( R 2 + r 3 + R 4 )( R 5 + r 2 ) i 1- ( R 5 + r 2 ) × ( R 2 + r 3 + R 4 + R 5 + r 2 ) i 2 Adding, E 1 ( R 2 + r 3 + R 4 + R 5 + r 2 )-E 2 ( R 2 + r 3 + R 4 )-E 3 ( R 5 + r 2 ) = [( R 3 + r 1 + R 1 )( R 2 + r 3 + R 4 + R 5 + r 2 ) + ( R 2 + r 3 + R 4 )( R 5 + r 2 )] i 1 Since a = R 2 + r 3 + R 4 + R 5 + r 2 = 5 Ω + 1 Ω + 8 Ω + 11 Ω + r 2 = 27 Ω , b = R 2 + r 3 + R 4 = 5 Ω + 1 Ω + 8 Ω = 14 Ω , c = R 5 + r 2 = 11 Ω + 2 Ω = 13 Ω , and d = ( R 3 + r 1 + R 1 ) a + b c = (7 Ω + 2 Ω + 2 Ω) (27 Ω) + (14 Ω) (13 Ω) = 479 Ω , we have...
View Full Document

This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

Page1 / 9

solutions 10 - Patel, Kinal – Homework 10 – Due: Feb 26...

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

View Full Document
Ask a homework question - tutors are online