Lecture18 - 1 February 18, 2008 Physics 132-Winter 2008...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 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 = 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 = E C R i i = dQ dt for positive plate on ‘upstream’ side of C E − dQ dt R − Q / C = A differential equation Uniqueness theorems give us a ‘license to guess’. 3 February 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 5 The Solution E − dQ dt R − Q / C = C E − dQ dt RC − Q = Guess: Q ( t ) = ae bt + d dQ dt = abe bt C E − abe bt ( ) RC − ae bt − d = CE − d ( ) − bRC + 1 ( ) ae bt = 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....
View Full Document

This note was uploaded on 04/18/2008 for the course PHYS 132 taught by Professor Beatty during the Winter '08 term at Ohio State.

Page1 / 11

Lecture18 - 1 February 18, 2008 Physics 132-Winter 2008...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online