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Unformatted text preview: Solution for Homework 12 Kirchhoff’s Laws Solution to Homework Problem 12.1(Reduce Resistance Network) Problem: For the circuit to the right, with R 1 = 5 . 0Ω , R 2 = 10 . 0Ω , and R 3 = 15Ω , and Δ V = 6 . 0V . What is the current flowing through R 3 ? Select One of the Following: (aAnswer) . 22A (b) 8 . 2A (c) . 041A (d) 1 . 1A (e) . 73A Δ V R 1 R 3 R 2 Solution (a) Find the Equivalent Resistance: The combination R 2 and R 3 are in parallel and can be reduced to give the equivalent resistance, 1 R 23 = 1 R 2 + 1 R 3 = 1 10Ω + 1 15Ω giving R 23 = 6Ω . The circuit is redrawn to the right. The equiv alent resistor R 23 is in series with R 1 , so the total resistance of the circuit is R eq = R 1 + R 23 = 5Ω + 6Ω = 11Ω . R 1 R 23 Δ V (b) Find Current Delivered by the Battery: The current drawn by the circuit is given by Ohm’s Law, I = Δ V /R eq = 6V / 11Ω = 6 11 A . (c) Compute the Potential Difference Across R 23 : The current through by the battery flows through both R 1 and the equivalent resistor R 23 because they are in series. The potential difference across R 23 is then Δ V 23 = IR 23 = ( 6 11 A)(6Ω) = 36 11 V (d) Compute the current through R 3 : The potential difference across the parallel combination Δ V 23 is the same as the potential difference across R 3 , Δ V 3 . Use Ohm’s law to find the current, I 3 = Δ V 3 R 3 = ( 36 11 V)( 1 15Ω ) = 12 55 A = 0 . 22A 1 Total Points for Problem: 5 Points Solution to Homework Problem 12.2(Line Charge Potential) Problem: A uniform line charge of finite length occupies a length 2 L along the y axis from y = L to y = + L . The line charge has uniform linear charge density λ . Select the integral that would allow you to calculate the electric potential at the point vector r P = ( L,L, 0) Select One of the Following: (a) V ( L,L, 0) = integraltext L L kλdy y 2 (b) V ( L,L, 0) = integraltext L L kλdy √ L 2 + y 2 (c) V ( L,L, 0) = integraltext L L kλdy y (d) V ( L,L, 0) = integraltext L L kλdy y L (eAnswer) V ( L,L, 0) = integraltext L L kλdy √ L 2 +( L y ) 2 x y L LL P Solution Divide the line charge into small segments of length Δ y . The center of the i th segment is y i . The charge of the i th segment is q i = λ Δ y . The distance of the i th segment to the point P is d i = radicalbig L 2 + ( L y i ) 2 The potential at point P is the sum potentials of the segments...
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This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.
 Spring '10
 Stewat
 Current, Resistance, Work

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