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

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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...
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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.

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

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