8.022 (E&M) Lecture 16

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1 8. (E&M) ecure 16 Topics: Inducors in circuis circuis circuis circuis as ime Our second lecure on elecromagneic inducance 3 ways of creaing emf using Faraday s law: hange area of circui S() hange angle beween B and S hange B magniude Self and muual inducance Energy sored in inducor Applicaions: ransformers A generaors Today is our 3 rd lecure on inducance: inducors in circuis G. Sciolla MIT 8. ecure 16 1

2 circuis: inuiive descripion A =, close S1 : enz s law opposes change in Φ B hrough Since Φ B (=)=, will impede curren flow I()= As ime passes, I will sar flowing sauraing a I=V/ Afer a long ime, simulaneously open S1 and close S: enz s law opposes change in Φ B hrough Back emf will keep curren flowing for a while dissipaes power he curren will die exponenially G. Sciolla MIT 8. ecure 16 3 circuis: quaniaive descripion A =: close S1 Kirchoff s rule #: V I di = d V di di ewrie as: I + = = d d V I I V / V V ln = I = e V / G. Sciolla MIT 8. ecure 16 V I = (1 e ) 4

3 circuis: quaniaive descripion() A = : open S1 and close S Kirchoff s rule #: di I = d Graphically: I() ewrie as: I ln I I = I ( ) di di I = = d d I = I = V e I = I = V (1 e ) V e G. Sciolla MIT 8. ecure 16 5 circuis: inerpreaion of resuls I() How do we inerpre hese resuls? Inducors cause currens o have an ineria emf) Asympoic behavior when charging V (1 e ) V e G. Sciolla MIT 8. ecure 16 If no curren flowing: forces I o build up gradually If curren is flowing: will do wha i akes o make i coninue (back A =, I=, as if were an open circui A =infiniy, I=V/, as if did no exis =: open circui = : shor circui 6 3

4 circuis: ime consan I() V (1 e ) V e esuls of circui are exponenials, as in circuis circui: ime consan τ= circuis: ime consan τ=/ NB: ime consan is he ime i akes he exponenial funcion o decrease (increase) o 1/e (1-1/e) of is original (final) value heck unis cgs: []/[]=(sec /cm)/(sec/cm)=sec SI: []/[]= H/ Ω = (V sec/a)/(v/a) = sec G. Sciolla MIT 8. ecure 16 7 ime consan onsider he following circui On he oscilloscope: V inpu, V, V, I in he circui V in 75 Hz I V V = di/d V G. Sciolla MIT 8. ecure

5 circuis Sar wih charged capacior and close swich a =: di = d d d Since I=- : + = d d () = A cos ω + B sinω Kirchoff s second rule: How o solve his? Educaed guess: d = ω A cos ω ω B sin ω = ω ( ) ω = d G. Sciolla MIT 8. ecure circuis: soluion Plug his in he differenial equaion: d () = () ω () = () ω = d Deermine consans A and B from iniial condiions: (=)= = A cos() + B sin() A= I(=)= = -ω A sin() + ω B cos() B= omplee soluion: () () = cos ω V () = = cos ω d = I() = - sin ω d NB: curren and volages are off by 9 degrees G. Sciolla MIT 8. ecure

6 I() V() circuis: soluion Graphical represenaion of he soluion: V () I() = cos ω = sin ω NB: and I have a phase of 9 deg G. Sciolla MIT 8. ecure Energy conservaion Energy sored in he capacior over ime: () () U () = = cos ω Energy sored in he inducor: 1 1 U () = I () = sin ω = sinω Toal energy: U() = U () + U () = (sinω + cos ω) = Wha is happening over ime? Energy swings back and forh beween and bu a any momen in ime he oal energy is equal o he energy iniially sored in he capacior: Energy is conserved! G. Sciolla MIT 8. ecure

7 circuis circuis don belong o his world: is never exacly! So le s concenrae on s Sar wih a charged Inuiively: oscillaory par: sin and cos soluion dissipaive par: exponenial damping igorous soluion: Use Kirchoff: Since I() = I di d = d d d 1 d d d + + = G. Sciolla MIT 8. ecure circuis: soluion d d = d d Educaed guess! Inuiion ells us ha he soluion mus have an oscillaory erm and a damping erm Sraegy #1: exponenial * sin/cos funcions: τ () = e A cos ω + B sinω How o solve his equaion? Very heavy on algebra!!! - / ( ) Sraegy #: complex exponenials Idea: he soluion is he real par of a complex soluion i φ iα () = A e e () = e () Much easier algebra!!! NB: a can be complex! G. Sciolla MIT 8. ecure

8 See handou on complex number + secions nex week omplex number noaion omplex number: number wih boh a real and an imaginary par z = x + i y wih i= -1 y x z = x + i y omplex plane represenaion z=(x,y) y Anoher useful represenaion Se magniude r= x +y and phas e θ =arc g z = r (cos θ + i sinθ) x Given Euler s relaion: y e = cos θ + i sinθ Prove i using Maclaurin expansion (see handou) z = i re θ iθ (Phasor represenaion) G. Sciolla MIT 8. ecure x circuis: soluion (con) Plug expeced soluion () e i φ i α = e ino he differenial equaion d d = d d d d 1 = iα ; = α -α + iα + = d d Simple quadraic equaion: This gives us complex soluions for (): 1 1 α + i α + = α = i ± 4 i i φ ( ) = Ae e e + ( ) = Ae e e 1 real par: ( ) = Ae cos( ± ω + φ ) w ih ω = i i φ 4 G. Sciolla MIT 8. ecure

9 The weak damping limi Weak damping limi: small he damping is small several oscillaions occur before ampliude sar decreasing in sizable way d I ( ) = = e [ ω si n( ω + φ ) + cos( ω + φ ) ] d W hen ω > > /() (dam ping lim i ), he second erm can be ignored and 1 1 I ( ) ~ Ae ω sin( ω ) w i h ω = ~ = ω 4 ( ) ~ e cos( ω + φ ) final soluion for "w eak dam ping": I ( ) ~ ω e si n( ω + φ ) 1 ω = G. Sciolla MIT 8. ecure in weak damping limi Iniial condiions: ( ) ~ e cos( ω ) I ( ) ~ ω e si n( ω ) Graphical represenaion of soluion: I() ()= =Acos( φ ) and I()==A ω sin φ A = ; φ = G. Sciolla MIT 8. ecure 16 Demo : Dumped 18 9

10 Summary and oulook Today: Wha happens when we pu in circuis? circuis: exponenial soluions circuis: oscillaory soluion circuis: damped oscillaion Nex Tuesday: uiz # : good luck!!! G. Sciolla MIT 8. ecure

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