Electricity and Magnetism-RL,LC,LCR (DC) circuits

Topics: 􀂄 􀂄 RL circuits 􀂄 LC circuits 􀂄 8.022 (E&M) – Lecture 16 Inductors in circuits RCL circuits 2 Last time 􀂄 􀂄 􀂄 􀃆 􀂄 􀂄 Self and mutual inductance 􀂄 􀂄 Today is our 3rd G. Sciolla – MIT 8.022 – Lecture 16 Our second lecture on electromagnetic inductance 3 ways of creating emf using Faraday’s law: Change area of circuit S(t) Change angle between B and S AC generators Change B magnitude Energy stored in inductor Applications: transformers lecture on inductance: inductors in circuits 13 RL circuits: intuitive description 􀂄 At t=0, close S1 : 􀂄 ΦB 􀂄 Since ΦB 􀃆 􀂄 􀂄 􀂄 ΦB 􀂄 􀂄 i􀃆 the current will die exponentially G. Sciolla – MIT 8.022 – Lecture 16 Lentz’s law opposes change in through L (t=0)= 0, L will impede current flow I(0)=0 As time passes, I will start flowing saturating at I=V/R After a long time, simultaneously open S1 and close S2: Lentz’s law opposes change in through L Back emf will keep current flowing for a while R disspates power 4 RL circuits: quantitative description 􀂄 At t=0: close S1 􀂄 Kirchoff’s rule #2: 0dIV IR Ldt− − = /(1 )ln /RtL RtL V dI RI dtVR t LI R R V Vt I eVR VIL R R eR − − −+ = ⇒ − − ⇒ ⇒ − ⇒ − = − G. Sciolla – MIT 8.022 – Lecture 16 Rewrite as: LdI RdIV R = − = − = − 25 RL circuits: quantitative description(2) 􀂄 At t=t’: open S1 and close S2 􀂄 􀂄 Graphically: (1 ) RtLV eR −− 0dI dt− − = 0 () 0 0 iIln I RtL t II t RI dtRdt I L Rt IL V eR − = = = −= ⇒ = − =⇒ ⇒ ∫ ∫ I(t) t RtLV eR − G. Sciolla – MIT 8.022 – Lecture 16 Kirchoff’s rule #2: IR LRewrte as: II t LdI dI = − 6 RL circuits: interpretation of results 􀂄 􀂄 􀂄 􀂄 emf) 􀂄 􀂄 􀂄 (1 ) RtLV eR −− I(t) t RtLV eR − L t= i→⎧ ⎨ ∞ →⎩ G. Sciolla – MIT 8.022 – Lecture 16 How do we interpret these results? Inductors cause currents to have an “inertia” If no current flowing: L forces I to build up gradually If current is flowing: L will do what it takes to make it continue (backAsympptoti behavior when “charging” L At t=0, I=0, as if L were an open circuit At t=infinity, I=V/R, as if L did not exist t=0: open circuit : L short crcuit 37 RL circuits: time constant 􀂄 􀂄 τ=RC 􀂄 τ=L/R 􀂄 NB: ti􀂄 Check units 􀂄 2/cm)/(sec/cm)=sec 􀂄 Ω = (V sec/A)/(V/A) = sec (1 ) RtLV eR −− I(t) t RtLV eR − G. Sciolla – MIT 8.022 – Lecture 16 Results of RL circuit are exponentials, as in RC circuits RC circuit: time constant RL circuits: time constant me constant is the time it takes the exponential function to decrease (increase) to 1/e (1-1/e) of its original (final) value cgs: [L]/[R]=(secSI: [L]/[R]= H/8 LR time constant 􀂄 Consider the following circuit 75 Hz Vin I VR VL 􀂄 􀂄 V, VL, VRt t t t VL = L dI/dt R L G. Sciolla – MIT 8.022 – Lecture 16 On the oscilloscope: input , I in the circuit 49 L C LC circuits 􀂄 􀂄 􀂄 0Q dILC dt − = 2 2 dQ : 0dt dQ Q dt + = 0 0 0 0 2 2 2 2 0 02 1dQ A t B td Lt C ω ω ω ω ωω⇒ − ⇒ = 0 0() sinQt A t B tω ω= + G. Sciolla – MIT 8.022 – Lecture 16 Start with charged capacitor and close switch at t=0: Kirchoff’s second rule: How to solve this? Educated guess: Since I=-LC cos sin ( ) Qt = − = − cos 10 LC circuits: solution 􀂄 􀂄 l conditions: 􀂄 Q(t=0)=Q0= A cos(0) + B sin(0) 􀃆 0 􀂄 I(t=0)=0 = -ω0A sin(0) + ω0B cos(0) 􀃆 􀂄 Complete solution: 􀂄 0 0 2 2 2 () 1 1() () 1Qtdt C LL C ωω=− ⇒ − =− ⇒ = 0 0 0 0 0 0 ()() c()QdQ = tdt L os C C QQtV t tC CQt Q t ω ωω ⇒ == = G. Sciolla – MIT 8.022 – Lecture 16 Plug this in the differential equation: Determine constants A and B from initiaA=QB=0 NB: current and voltages are off by 90 degrees ( ) dQt Qt Qt LC cos I(t) = -sin 511 LC circuits: solution t I(t) 0 0 0 0 Q () = sin t cos LC C QV t tC ω ω⎧ =⎪⎪⎨ ⎪ ⎪⎩ NB: Q and I have a phase of 90 deg G. Sciolla – MIT 8.022 – Lecture 16 Graphical representation of the solution: V(t) I(t) 12 Energy conservation 􀂄 􀂄 􀂄 Total energy: 􀂄 􀂄 22 20 0 ()()() 2 2C QtQtUt tC Cω== 2 2 2 2 20 0 0 0 1 1() () sin sin2 2 2L Q QUt L t tLC Cωω=== 2 2 2 20 0 0 0() () () in2 2L C Q QUt t tC Cωω=+=+ G. Sciolla – MIT 8.022 – Lecture 16 Energy stored in the capacitor over time: Energy stored in the inductor: What is happening over time? Energy swings back and forth between C and L but at any moment in time the total energy is equal to the energy initially stored in the capacitor: Energy is conserved! cos LI t (scos ) = U t U t 613 RCL circuits 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 LC 􀃆 􀂄 R 􀃆 􀂄 0Q IR LC − − = 2 2 dQ R 1 00d L dQ Qdt dt LC⇒ + + = G. Sciolla – MIT 8.022 – Lecture 16 LC circuits don’t belong to this world: R is never exactly 0! So let’s concentrate on RCLs Start with a charged C Intuitively: oscillatory part: sin and cos solution dissipative part: exponential damping Rigorous solution: Use Kirchoff: dI dt Since I(t) =dQ 14 RCL circuits: solution 􀂄 􀂄 􀂄 Intuition tell􀂄 􀂄 􀂄 NB: a can be complex! 2 2 1 0dQ Qdt + + = ( )-/0 0()sinte A t B tτ ω ω= + 0() () ()i itφ α ⎡ ⎤= ⇒ = ⎣ ⎦ G. Sciolla – MIT 8.022 – Lecture 16 How to solve this equation? Educated guess! s us that the solution must have an oscillatory term and a damping term Strategy #1: exponential * sin/cos functions: Very heavy on algebra!!! Strategy #2: complex exponentials Idea: the solution is the real part of a complex solution Much easier algebra!!! R dQ L dt LC cos Qt Re Qt Ae e Qt Qt 7G. Sciolla – MIT 8.022 – Lecture 16 15 Complex number notation 􀂄 Complex number: number with both a real and an imaginary part 􀂄 Complex plane representation z=(x,y) 􀃆 􀂄 Another useful representation 􀂄 Given Euler’s relation: 􀂄 Prove it using Maclaurin expansion (see handout) with 1i= -z x i y= + x y z x i y= +x y 2 2 ySet magnitude r= x +y and phas (cos sie =arct )g x nz r iθ θθ ⇒= + cos sinie iθ θ θ= + (Phasor representation)iz reθ⇒ = See handout on complex number + sections next week G. Sciolla – MIT 8.022 – Lecture 16 16 RCL circuits: solution (cont) Plug expected solution into the differential equation 0 2 2 2 2 2 2 2 2 2 () 1 0 1; Q -0 1Simple quadratic equation: ­ This gives us 2 complex solutions for Q(t): 1 2 4 R iQt Ae e R dQ R dQ Qdt L dt LC dQ d Q RiQ Q idt dt L LC Ri L LC Ri L LC L φ α α α α α α α − + = + + = ⎛ ⎞ = =− ⇒ + + =⎜ ⎟⎝ ⎠ = ± −+ + = ⇒ 2 2 2 2 0 1 42 1 42 2 2 0 2 () 1() cos( ) =real part: w t 4ih Ri tt LC LL RR i tti LC LL RtL e Qt Ae e e RQt Ae t LC L φ ω φ ω − − −− − − = ⎧ ⎪⎪⎨ ⎪ ⎪⎩ ⇒ = ± + − 0() i itQt e eφ α= 817 The weak damping limit 􀃆 􀃆 several 2 0 0 0 2 2 02 () [ si) 2 im i i1 1 sii= ~ 4 RtL RtL dQ RIt t tdt L RIt Ae t LC t L L Q Q C ω ω φ ω φ ω ω ω ω ω − − =− = + + + − = ⇒ 2 0 0 0 2 0 0 0 0 0 ) si) 1 RtL RtL e t It t LC ω φ ω ω φ ω − − ⎧ +⎪ ⎪⎪ +⎨ ⎪ ⎪ = ⎪⎩ G. Sciolla – MIT 8.022 – Lecture 16 Weak damping limit: small Rthe damping is small oscillations occur before amplitude start decreasing in sizable way n( cos( ) ] W hen > > R/(2L) (dam ping lt), the second term can begnored and () ~ n( ) w th final solution for "w eak dam ping": () ~ Qe cos( () ~ n( Qe 18 RCL in weak damping limit 􀂄 􀂄 2 0 0 2 0 0 0 ) si) RtL RtL Q t t It t ω ω ω − − ⎧ ⎪ ⇒ ⎨ ⎪⎩ t I(t) 0 00 0 0 0sin ; 0AQφ ω φ φ=⇒ = Demo L2: Dumped RCL G. Sciolla – MIT 8.022 – Lecture 16 Initial conditions: Graphical representation of solution: () ~ cos( () ~ n( Qe Qe Q(0)=Q =Acos( ) and I(0)=0=A 919 Summary and outlook 􀂄 􀂄 RL circuits: exponential solutions 􀂄 􀂄 􀂄 Next Tuesday: 􀂄 Quiz # 2: good luck!!! G. Sciolla – MIT 8.022 – Lecture 16 Today: What happens when we put L in circuits? LC circuits: oscillatory solution RCL circuits: damped oscillation 10