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Lecture18 - Physics 132 Introductory Physics Electricity...

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1 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 1 Physics 132 Introductory Physics: Electricity and Magnetism Winter Quarter 2008 Lecture 18 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 2 Writing Circuit Equations At an EMF device: Add + E going from - to + Add - E going from + to - At a Resistor: Add -IR going in direction of I Add +IR going opposite from I At a capacitor: Define the positive plate. Going from + to -: -Q/C Going from - to +: +Q/C Start at junctions. (Single loop circuits have no junctions.) Label each branch with an I. (Directions are arbitrary, but you need to put them on the diagram!) At each junction, Σ I = 0 (Kirchhoff’s Current Law) For each closed loop, Σ V = 0 (Kirchhoff’s Voltage Law) You get one equation for each junction and for each closed loop. You will get more than enough equations to solve for all the unknowns. Loop Rules
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2 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 3 Charging a Capacitor through a Resistor With switch open, Q=0 and V c =0. Close the switch. V c is still zero until current has time to charge C. Loop equation: E C R i Loop rule for a capacitor: Define the positive plate. Going from + to -: -Q/C Going from - to +: +Q/C E iR Q / C = 0 Initially Q=0 so i= E /R. Current flows until E =Q/C. Then i=0 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 4 Kirchhoff Analysis of an RC Circuit E iR Q / C = 0 E C R i i = dQ dt for positive plate on ‘upstream’ side of C E dQ dt R Q / C = 0 A differential equation Uniqueness theorems give us a ‘license to guess’.
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3 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 5 The Solution E dQ dt R Q / C = 0 C E dQ dt RC Q = 0 Guess: Q ( t ) = ae bt + d dQ dt = abe bt C E abe bt ( ) RC ae bt d = 0 CE d ( ) bRC + 1 ( ) ae bt = 0 Zero if both terms in parentheses are zero. d = C E b = 1 RC Q ( t ) = C E ae t RC Choose a so Q(t=0) is zero. Q ( t ) = C E C E e t RC Q ( t ) = C E 1 e t RC February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 6 The Solution Q ( t ) = C E 1 e t RC V c ( t ) = E 1 e t RC E C i t V c E Voltage and charge
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