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8.022 (E&M) – Lecture 17 Topics: Discussion of Exam 2 and make-up exam Back to E&M: RCL circuits: recap undriven RCLs, driven RCLs, inductance Last time What happens when we put inductors in circuits? RL circuits: exponential solutions LC circuits: oscillatory solution RCL circuits: damped oscillation RCL circuits are particularly interesting Let’s see them in some more detail… G. Sciolla – MIT 8.022 – Lecture 17 4 1 Undriven RCL circuits: recap Kirchoff’s second rule: d 2Q dQ 1 L + R + Q =0 dt 2 dt C Does it look familiar? d 2x dx m + kf + ke x = 0 2 dt dt Mechanics: harmonic oscillator! RCL L d2Q/dt2 Mechanics ma=m Interpretation d2x/dt2 L ~ m: inertia term R dQ/dt kf v = kf dx/dt R ~ kf Æ friction (damping) term 1/C Q ke x 1/C ~ke Æ elastic term due to spring G. Sciolla – MIT 8.022 – Lecture 17 5 Undriven RCLs: solution Differential equation governing loop: d 2Q R dQ 1 Q =0 + + 2 dt L dt LC Solve using complex number notation: Q (t ) = e βt = e −αt e i ωt NB: β = −α + i ω is a complex number, with α and ω real e −αt = damping term, e i ωt = oscillatory term Throw this into the equation and we get a quadratic equation in β : β2 + β R 1 R R2 1 + =0 ⇒ β =− ± − 4L2 LC L LC 2L G. Sciolla – MIT 8.022 – Lecture 17 6 3 RCL circuits: solution ⎧ ⎪ 2 ⎪• β purely real: R − 1 > 0 ⇒ R>2 L ⇒ ⎪ 4L2 LC C ⎪ ⎪ ⎪• β purely imaginary: ⇒ R = 0 ⇒ undamped LC ⇒ R R2 1 ⎪ β =− ± :⎨ − 2L 4L2 LC ⎪ 2 ⎪• β truly complex: R>0 and R 2 − 1 < 0 ⇒ 4L LC ⎪ Q (t ) = e βt = e −αt e i ωt ⎪ 2 ⎪α= R and ω= 1 − R ⎪ 2L LC 4L2 ⎪ ⎩ 2 R 1 When 2 − = 0 critical damping (fastest way to damp an oscillator). 4L LC G. Sciolla – MIT 8.022 – Lecture 17 7 RCL in weak damping limit Initial conditions: Q(0)=Q0 =Acos(φ0 ) and I(0)=0=Aω0 sinφ0 ⇒ A = Q 0 ; φ0 = 0 R − t ⎧ 2L cos(ω 0t ) ⎪Q (t ) ~ Q 0e ⇒ ⎨ R − t ⎪ I (t ) ~ ω Q e 2 L sin(ω t ) 0 0 0 ⎩ Graphical representation of solution: I(t) t G. Sciolla – MIT 8.022 – Lecture 17 8 4 Energy Energy of the circuit in the weak damping limit: U C (t ) = U L (t ) = Q 2 (t ) Q 02 −Rt / L e cos 2 ω0t = 2C 2C Q2 1 1 LI (t )2 = ω02LQ 20 e −Rt / L sin2 ω0t = 0 e −Rt / L sin2 ω0t 2 2 2C ⇒ U (t ) = U L (t ) + U C (t ) = Q 02 −Rt / L Q2 (sin2 ω0t + cos 2 ω0t ) = 0 e −Rt / L e 2C 2C Since Q20/2C=total energy stored initially in the system Æ U decreases exponentially over time: as expected! G. Sciolla – MIT 8.022 – Lecture 17 9 Quality Factor Definition 1: the quality factor measures how many times the circuit oscillates before it loses a certain amount of energy Q2 U (t )= 0 e −Rt / L In the time τ =L/R the energy decreases by ∆U(t)=1/e 2C The oscillation is ωτ radians ⇒ Q = ωτ = Definition 2: the quality factor measures the ratio between energy stored (in C and L) and average power dissipated (in R) For an oscillation with frequency ω ⇒ Q = ω ωL R LI 2 / 2 ωL Energy stored = ω 02 = <Power> RI 0 / 2 R Q factor can be defined for any system that creates vibrations. Acoustics: Q of a tuning fork is much higher than the Q of a table… G. Sciolla – MIT 8.022 – Lecture 17 10 5 Today’s goal: Driven RCL circuits ~ is an AC e.m.f. AC voltage supplied to the circuit: Convenient assumption: emf (t ) = V 0 cos ωt V (t ) = Re ⎡⎣V (t ) ⎤⎦ with V (t ) = V 0e i ωt NB: V0 is purely real! How to solve this? Just generalize what we used for DC! Sum of voltage drops in loop is equal to emf (Kirchoff #2) Vemf (t ) = VR (t ) +VC (t ) +V L (t ) Vemf (t ) = VR (t ) +VC (t ) +V L (t ) The same current must pass through every circuit element I (t ) = I R (t ) = I C (t ) = I L (t ) G. Sciolla – MIT I (t ) = I R (t ) =8.022 I C (t –) Lecture = I L (t 17 ) 11 AC current Consider a B constant in magnitude and a loop rotating around its axis with angular velocity ω ω B θ ∫ If S is the area of the loop: B ida = BS cos θ = BS cos ωt S Faraday: 1 ∂ ω e .m .f . = (BS cos ωt ) = BS sin ωt c ∂t c This is how AC power is generated. In U.S.: ν=60 Hz Æ ω=377 G. Sciolla – MIT 8.022 – Lecture 17 12 6 AC emf + resistor R Ohm’s law holds for AC too: I V (t ) = VR (t ) = I (t )R V Let’s plot I(t) and V(t) on the same graph: ~ R V(t) --- I(t) __ V(t) t Æ In a resistor the voltage and the current are in phase (peak voltage occurs at the same time as peak current) G. Sciolla – MIT 8.022 – Lecture 17 13 Reminder: phasor notation Any complex number z = x + i y with i= -1 can always be represented as the product of a real number (magnitude) and a complex exponential: ⇒ z = re i θ (Phasor representation) y where magnitude r= x 2 +y 2 and phase θ =arctg x ⇒ z = r (cos θ + i sin θ ) and given Euler’s relation: e i θ = cos θ + i sin θ which can be easily proved using Maclaurin expansion G. Sciolla – MIT y 8.022 – Lecture 17 x z = x +i y y x 14 7 AC emf + R with phasors The same information can be represented with phasors in the complex plane: V (t ) = RI (t ) I ~ V R Æ In a resistor the voltage and the current are in phase In phase means that both phasors are at the same angle G. Sciolla – MIT 8.022 – Lecture 17 15 AC emf + capacitor C Connect AC emf across a capacitor C: Q (t ) V (t ) = VC (t ) = C ~ V C Since V(t)=V0cosωt and I(t)= dQ/dt: dQ (t ) π = −ωCV 0 sin ωt = ωCV 0 cos(ωt + ) I (t ) = dt 2 Æ I(t) LEADS V(t) by 90 deg / V(t) lags I(t) by 90 deg (maxima in I(t) occur before maxima in V(t)) V(t) t G. Sciolla – MIT 8.022 – Lecture 17 --- I(t) __ V(t) 16 8 Ohm’s law revisited and Impedance Relation between I(t) and V(t) becomes more obvious when using phasor notation: VC (t ) = V 0 cos ωt = Re ⎡⎣VC (t ) ⎤⎦ with V (t ) = V 0e i ωt For the current: π I (t ) = ωCV 0 cos(ωt + ) = Re ⎡⎣I C (t ) ⎤⎦ 2 with I (t ) = ωCV 0e π⎞ ⎛ i ⎜ ωt + ⎟ 2 ⎝ ⎠ i π = i ωCV 0e i ωt (remember: e 2 = i ) Combining complex currents and voltages we can write: V (t ) = I (t )Z C (complex equivalent of Ohm's law) where Z C is the impedance of a capacitor: Z C = G. Sciolla – MIT 8.022 – Lecture 17 1 iω C 17 AC emf + C: phasor representation Given V (t ) = V 0e i ωt and I (t ) = Z CV 0e i ωt =i ωCV 0e i ωt V(t) and I(t) can easily be represented in the complex plane: NB: I(t) is ahead of V(t) by 90 degrees: I(t) leads V(t) by 90 degrees G. Sciolla – MIT 8.022 – Lecture 17 18 9 AC emf + inductor L Connect AC emf across an inductor L: dI V (t ) = VL (t ) = L dt Since V(t)=V0cosωt: dI V 0 = cos ωt dt L ⇒ I (t ) = ~ L V V0 V π⎞ ⎛ sin ωt = 0 cos ⎜ ωt − ⎟ ωL ωL 2⎠ ⎝ V(t) t --- I(t) __ V(t) Æ I(t) LAGS V(t) by 90 degrees, or V(t) LEADS I(t) by 90 degrees (maxima in I(t) occur before maxima in V(t)) G. Sciolla – MIT 8.022 – Lecture 17 19 Impedance of inductors Using phasor notation: VC (t ) = V 0 cos ωt = Re ⎡⎣VL (t ) ⎤⎦ V (t ) = V 0e i ωt The current is: I (t ) = V0 π cos(ωt − ) = Re ⎡⎣I (t ) ⎤⎦ ωL 2 ⎛ with I (t ) = with π⎞ π -i V 0 i ⎜⎝ ωt − 2 ⎟⎠ V 0 i ωt −1 = e e (remember: e 2 = ( i ) = −i ) ωL i ωL Combining complex currents and voltages we can write: V (t ) = I (t )Z L (complex equivalent of Ohm's law) where ZL is the impedance of an inductor: ZL =iωL G. Sciolla – MIT 8.022 – Lecture 17 20 10 AC emf + L: phasor representation V 0 i ωt e i ωL V(t) and I(t) can easily be represented in the complex plane: Given V (t ) = V 0e i ωt and I (t ) = Z LV 0e i ωt = NB: I(t) is 90 degrees behind V(t): I(t) lags V(t) by 90 degrees G. Sciolla – MIT 8.022 – Lecture 17 21 Driven RCLs using inductance Inductance simplifies the study of driven RCL circuits Let’s work with complex numbers and use Ohm’s and Kirchoff’s extensions Vemf (t ) = VR (t ) +VC (t ) +V L (t ) ⎧ V R (t ) = RI (t ) ⎪ 1 ⎪ I (t ) Since ⎨VC (t ) = Z C I (t ) = i ω C ⎪ ⎪ V L (t ) = Z L I (t ) = i ωLI (t ) ⎩ ⎛ 1 ⎛ ⇒ V emf (t ) = I (t ) ⎜ R + i ⎜ ωL − ω C ⎝ ⎝ 1 ⎞ ⎛ where total impedance of the circuit is Z tot ≡ R + i ⎜ ωL − ω C ⎟⎠ ⎝ G. Sciolla – MIT 8.022 – Lecture 17 ⎞⎞ ⎟ ⎟ = I (t )Z tot ⎠⎠ 22 11 Driven RCLs: phasor notation The complex current can be written as I (t ) = V emf (t ) = Z tot V 0e i ωt This can be written as: I (t ) = ⎛ ⎝ V 0e i ωt V e i ωt = 0 * Z *tot = Z tot Z tot Z tot Remembering that e-iθ G. Sciolla – MIT 1 ⎞ ⎠ R + i ⎜ ωL − ωC ⎟ V 0e i ωt ⎡ 1 ⎞⎤ ⎛ i ωt − i φ ⎢R − i ⎜ ωL − ωC ⎟ ⎥ = I 0e e ⎝ ⎠ 1 ⎣ ⎦ ⎛ ⎞ R 2 + ⎜ ωL − ωC ⎟⎠ ⎝ V0 ⎧ ⎪ I0 = 2 1 ⎞ ⎛ ⎪ 2 R L ω + − ⎜ ⎟ ⎪ ωC ⎠ ⎝ = cos θ − i sin θ ⇒ ⎨ ⎪ 1 ωL − ⎪ C = ωL − 1 ω ⎪ tgφ = 8.022 – Lecture 17 23 R R ωRC ⎩ 2 Dependence of φ from ω φ(ω) ω tgφ = 1 ωL − R ωRC NB: I (t ) = I 0e i ωt e −i φ → high ω: I lags voltage by 90 o → low ω: I leads voltage by 90 o G. Sciolla – MIT 8.022 – Lecture 17 24 12 AC motor (H26) Coil 1 2 RL circuits driven by 60 Hz AC voltage Coil 2 What is the ∆φ between the 2 currents? ~ Coil 1: R=2.3 Ω, L=1.5mH Coil 2: R=2.5 Ω, L=31 mH 10-3 Z1=R1+iωL1=2.3+i 377 1.5 Z2=R2+iωL2=2.5+i 377 31 10-3 Æ ∆φ=64 degrees ~ The difference in phase will create a rotating B field Æ Eddie currents in the metal can will make it rotate! G. Sciolla – MIT 8.022 – Lecture 17 25 Dependence of I0 from ω I0 V0 I0 = ⎛ ⎝ R 2 + ⎜ ωL − 1 ⎞ ωC ⎟⎠ 2 Maximum current when ωL = ⇒ ω0 = ω0 G. Sciolla – MIT 1 LC 1 ωC resonance frequency ω 8.022 – Lecture 17 26 13 RCL resonance (Demo L8) RCL circuit driven with variable frequency ω L=50 mH C=0.3 µF scope Measure VR on scope and tune frequency to maximize VR What is the expect resonance frequency? ω0 = G. Sciolla – MIT 1 LC = 8.2 × 103 ⇒ ν = 1.3 kHz 8.022 – Lecture 17 27 Demo L8: part 2 Same RCL circuit driven with variable frequency ω Frequency is driven by a voltage Vin L=50 mH C=0.3 µF scope Display VR vs on the scope while sweeping Vin What do you expect to see? G. Sciolla – MIT 8.022 – Lecture 17 ω0=1.3 kHz 28 14 Resonant RCL with light bulb (L6) RCL circuit driven by AC voltage C can be adjusted using set of switches L can be adjusted moving the Fe core inside a solenoid For each setting of C we can find an L that turn on the light bulb What is that L? G. Sciolla – MIT L= 1 C ω2 8.022 – Lecture 17 29 Summary and outlook Today: Undriven RCL circuits Driven RCL AC circuits Energy stored and quality factor in weak damping limit Simple solution when introducing complex impedance Z ZR = R ZC = 1/(iωC) ZL = iωL Next Tuesday: More on driven RCLs: power, resonances, filters… G. Sciolla – MIT 8.022 – Lecture 17 30 15