PES 1120 Spring 2014, Spendier Lecture 24/Page 1

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1 PES 1120 Spring 2014, Spendier Lecure 24/Page 1 Today: - circuis (ime-varying currens) - Elecric Eel Circuis Thus far, we deal only wih circuis in which he currens did no vary wih ime. Here we begin a discussion of ime-varying currens. Charging a capacior: We are now going o explore he ime dependence of he charge on a capacior. If we sar wih a circui wih a resisor and a capacior in series wih a emf, bu wih he circui broken by an open swich, he capacior is uncharged, since here is no poenial difference across i. A some ime, 0, we close he swich, le us now examine, using he loop rule: q( ) V i( ) R 0 C Bu his equaion relaes wo quaniies ha vary over ime: i() and q(). Bu, since he charge on he capacior, q(), comes from he curren, i()=dq/d, we can relae hese wo as well: q( ) q( ) R 0 d C q( ) 1 q( ) 0 d R This is a differenial equaion, you do no need o know how o solve. Bu you need o be able o apply he soluion. The soluion o his equaion is

2 PES 1120 Spring 2014, Spendier Lecure 24/Page 2 q( ) C 1 e Q0 1 e Le s also check our boundary values of q() a =0 and as = : 0 q(0) C 1 e C 11 0 q( ) C 1 e C 1 0 C Q0 Then we can ake he derivaive o solve for he curren: dq( ) i( ) e d R And do he same for he curren: 0 i(0) e 1 R R R i( ) e 0 0 R R Noe how he capacior acs a hese exremes of imes. A =0, immediaely afer he swich is closed, curren flows like he capacior is jus a wire. A very long imes afer he swich is closed, he capacior acs like a broken circui, so no curren passes hrough ha par of he circui.

3 PES 1120 Spring 2014, Spendier Lecure 24/Page 3 Using q = CV, we find ha he poenial difference V C () across he capacior during he charging process is q Vc ( ) 1 e C a = 0: V C () = 0 a = : V C () = ε Time Consan The facor in he denominaor of he exponenial is commonly called he ime consan, and deermines he rae a which he capacior charges (and discharges, as we will soon see). τ = q( ) Q0 1 e Example 1 A 10 M resisor is conneced in series wih a 1.0 μf capacior and a baery wih emf 12.0 V. Before he swich is closed a ime = 0, he capacior is uncharged. a) Wha is he ime consan? b) Wha fracion of he final charge Q 0 is on he capacior a = 46 s? c) Wha fracion of he iniial curren I 0 is sill flowing a = 46s?

4 PES 1120 Spring 2014, Spendier Lecure 24/Page 4 Discharging a capacior Now, le s sar wih a fully charged capacior in an open circui wih no baery. Iniially, here is charge q 0 on he capacior. For < 0, he swich is open: - he poenial difference across he capacior is given by V c = q/c. - he poenial difference across he resisor is zero because here is no curren flow, I=0. A some ime, =0, we will close he swich. The capacior will begin o discharge and charge will begin o flow around he circui. The charged capacior is now acing like a volage source o drive curren around he circui. When he capacior discharges (elecrons flow from he negaive plae hrough he wire o he posiive plae), he volage across he capacior decreases. The capacior is losing srengh as a volage source. Applying he loop rule by raversing he loop clockwise, he equaion ha describes he discharging process is given by q( ) V i( ) R 0 C The curren ha flows away from he posiive plae is proporional o he charge on he plae, I dq d The negaive sign in he equaion is due o he fac ha he charge on he posiive plae is decreasing as more posiive charges leave he posiive plae. Thus, charge saisfies a firs order differenial equaion: dq( ) 1 q( ) 0 d and he soluion o his equaion 0 0 q( ) Q e Q e

5 PES 1120 Spring 2014, Spendier Lecure 24/Page 5 To find he ime dependence of he curren, we jus have o ake he ime derivaive of his soluion: q( ) Q0 i( ) e d The curren mus be negaive since he posiive charge moves in he opposie direcion when a capacior is discharged hen when i is charged. Example 2: The resisor and capacior of example 1 are reconneced. The capacior has an iniial charge of 5.0 μc and is discharged by closing he swich a = 0. a) A wha ime will he charge be equal o 0.5 μc? b) Wha is he curren a his ime?

6 PES 1120 Spring 2014, Spendier Lecure 24/Page 6 Elecric Eel hp:// (old) The elecric eel (Elecrophorus elecricus), lives in rivers of Souh America. I lives on fish which he eel kills by elecric shocks. The elecric eel generaes he volage in special ses of cells called elecroplaques. How do elecric eels sun or kill fish wihou sunning or killing iself? The eels generae curren wih elecroplaques, biological cells ha are essenially baeries, wih an emf source and resisance. Each eel has approximaely 140 rows of hese cells ha run he lengh of heir body. Each row conains nearly 5000 elecroplaques. Each elecroplaque has an emf of 0.15V and a resisance of 0.25Ω. The resisance of he waer (and fish) beween he eel s head and ail o be 800Ω. There is a curren ha would pass hrough he prey (ouch) and a curren ha passes hrough he eel. Essenially hey form a complee circui- he curren leaves he eel, passes hrough he waer and fish and hen back o he eel. Bu, if we simply ake he eel as a single pahway, hen he curren would be he same hrough he predaor and prey, and ha s no good. The key is ha he eel has many rows over which he curren can pass. So, insead of a single pah or branch, here are acually 140. This means ha he curren divides among hese rows, hereby reducing he curren ha goes hrough any given par of he eel. (Also, I m going o guess ha much of he curren passes over he surface of he eel, raher han hrough is body. This would also reduce he risk of self-infliced injury.) We will rea he eel and fish sysem as an idealized circui. In each row, here are 5000 elecroplaques. Each one has a resisance and emf. Since he elecroplaques in a single row are in series we can replace he 10,000 componens wih simply wo- an equivalen emf source and resisance. The emf for each row is going o be V row = 5000* V elecroplaque = 5000* 0.15V= 750V For he resisance; R row = 5000* R elecroplaque = 5000* 0.25Ω= 1250Ω Once we recognize ha each of he 140 rows can be replaced wih an equivalen emf source and resisor, we end up wih somehing ha looks like his:

7 PES 1120 Spring 2014, Spendier Lecure 24/Page 7 (Okay, so I go lazy and didn draw all 140 rows; jus use your imaginaion.) This is now bordering on he doable. We sill have los of componens and branches, bu we can employ Kirchoff s rules o find he curren in each row as well as he curren hrough he prey. Le s redraw he circui, labeling he currens. Noice ha he currens in each row are equivalen. (Even if you do no see ha a his sage, and insead work wih a differen curren for each row, I1, I2, I3, ec., you ll ge he same resul in he end.) Bu, he curren hrough he fish is differen. Now we ll be able o wrie down wo equaions and solve for he wo unknowns- curren hrough he fish and he curren hrough each row along he eel. Curren/ juncion rule: 140 * I= I fish Nex we need o selec a loop and direcion of ravel:

8 PES 1120 Spring 2014, Spendier Lecure 24/Page 8 I ve picked one of he rows and he fish. Poenial/ loop rule: -1250Ω I + 750V-800Ω I fish =0 Wih hese wo equaions we can now solve for he currens. I fish 0.9A and I 7 x 10-3 A Does his make any sense? In an earlier lecure I menioned ha 100 ma is a lehal shock for humans. So, our curren hrough he fish does make sense. Also, he curren hrough he eel, is much lower. (I migh cause some pain o he eel, bu I m sure ha is more appealing han sarving o deah.) This is he beauy of having many separae rows hrough which he curren can ravel.