Lect19 - Physics 212 Lecture 19 LC and RLC Circuits -...

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Unformatted text preview: Physics 212 Lecture 19 LC and RLC Circuits - Oscillation frequency - Energy Energy - Damping Physics 212 Lecture 19, Slide 1 Music Who is the Artist? BB A) B) C) D) E) Jefferson Airplane Grateful Dead Moby Grape The Doors Cream Why? One reason that will be revealed on Thursday One (theme of the week) (theme San Francisco Giants are in the World Series Physics 212 Lecture 19, Slide 2 Your Comments “All of this stuff, I wish we could go slower but I know we cant.” TOO TRUE: Hang in there, we’ll do our best to work on the issues…. “Differential equations have taken Differential over my life. I'm not sure I approve.” over “How did yawl derive the differential equations?” “Why do capacitors start off with a Why charge of zero when the switch is opened? shouldn't they start off with charge? “ charge? It all depends on how the circuti was It circuti was started. You have to determine the initial conditions from the problem statement conditions Differential Equations Do Determine Differential Much Behavior in Physics. We will show corresponding equations in mechanics today mechanics 35 30 25 20 15 10 5 0 05 Confused Confident Physics 212 Lecture 19, Slide 3 Some Exam Stuff • Exam Tomorrow Night ( Oct 27) at 7:00 – – – – Covers material in Lectures 9 – 18 (through last Thursday’s lecture) Bring your ID: Rooms determined by discussion section (see link) Conflict exam at 5:15 – sign up in your gradebook before Mon (Oct 25) If you have conflicts with both of these, contact Prof Beck Physics 212 Lecture 19, Slide 4 LC Circuit - + I L dI VL = L dt C + Circuit Equation: dQ I= dt Q VC = Q C Q dI +L =0 C dt d 2Q Q =− LC dt 2 d 2Q dt 2 = −ω 2Q where ω= 1 LC Physics 212 Lecture 19, Slide 5 d 2Q dt 2 2 dx dt 2 = −ω Q 2 = −ω 2 x ω= 1 LC L k k ω= m C F = -kx a m x Same thing if we notice that k↔ 1 C and m↔L Physics 212 Lecture 19, Slide 6 Time Dependence I L ++ C -- Physics 212 Lecture 19, Slide 7 Preflight 2 At time t = 0 the capacitor is At the fully charged with Qmax and the and current through the circuit is 0. L BB C What is the potential difference across the inductor at t = 0 ? What A) VL = 0 B) VL = Qmax/C B) /C C) VL = Qmax/2C C) /2C since VL = VC 60 50 40 The voltage across the capacitor is Qmax/C Kirchhoff's Voltage Rule implies that must also be equal to the voltage across the inductor 30 20 10 0 Pendulum… Physics 212 Lecture 19, Slide 8 Preflight 4 At time t = 0 the capacitor is At the fully charged with Qmax and the and current through the circuit is 0. BB C L What is the potential difference across the What inductor when the current is maximum ? inductor 50 A) VL = 0 B) VL = Qmax/C B) /C C) VL = Qmax/2C C) /2C 40 30 20 10 dI/dt is zero when current is maximum 0 Physics 212 Lecture 19, Slide 9 Preflight 6 At time t = 0 the capacitor is At the fully charged with Qmax and the and current through the circuit is 0. BB C L How much energy is stored in the capacitor How when the current is a maximum ? when A) U = Qmax2/(2C) /(2C) B) U = Qmax2/(4C) B) /(4C) C) U = 0 C) Total Energy is constant ! ULmax = ½ LImax2 UCmax = Qmax2/2C I = max when Q = 0 50 40 30 20 10 0 Physics 212 Lecture 19, Slide 10 10 Preflight 8 BB The capacitor is charged such The that the top plate has a charge +Q0 and the bottom plate -Q0. +Q and At time t=0, the switch is closed t=0 the and the circuit oscillates with frequency ω = 500 radians/s. frequency L ++ C -- L = 4 x 10-3 H ω = 500 rad/s rad/s What is the value of the capacitor C? 50 A) C = 1 x 10-3 F B) C = 2 x 10-3 F B) C) C = 4 x 10-3 F C) ω= 1 LC C= 40 30 1 ωL 2 = 1 4 (25 ×10 )(4 ×10 −3 ) = 10−3 20 10 0 Physics 212 Lecture 19, Slide 11 11 Preflight 10 closed at t=0 L C +Q0 BB -Q0 Which plot best represents Which the energy in the inductor as a function of time starting just after the switch is closed? 12 U L = LI 2 35 30 Energy proportional to I fl C cannot be negative Energy cannot 2 Current is changing fl UL is not constant Current 25 20 15 10 5 Initial current is zero 0 Physics 212 Lecture 19, Slide 12 12 Preflight 12 When the energy stored in When the capacitor reaches its maximum again for the first time after t=0, how much time how charge is stored on the top plate of the capacitor? A) B) C) D) E) BB closed at t=0 L +Q0 C +Q0 +Q0 /2 0 -Q0/2 -Q0 -Q0 30 25 20 Q is maximum when current goes to zero I= dQ dt Current goes to zero twice during one cycle 15 10 5 0 Physics 212 Lecture 19, Slide 13 13 Add R: Damping Just like LC circuit but energy but the oscillations get smaller because of R Concept makes sense… …but answer looks kind of complicated Physics 212 Lecture 19, Slide 14 14 Physics Truth #1: Even though the answer sometimes looks complicated… Q(t ) = Qo cos(ωt − φ ) …the physics under the hood is still very simple !! d 2Q dt 2 = −ω 2Q Physics 212 Lecture 19, Slide 15 15 The elements of a circuit are very simple: dI VL = L dt VC = V = VL + VC + VR I= Q C dQ dt VR = IR This is all we need to know to solve for anything ! Physics 212 Lecture 19, Slide 16 16 A Different Approach Start with some initial V, I, Q, VL Now take a tiny time step dt dI = (1 ms) VL dt L dQ = Idt VC = Q C Repeat… VR = IR VL = V − VR − VC Physics 212 Lecture 19, Slide 17 17 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V What is QMAX, the maximum charge on the capacitor? L C R • Conceptual Analysis – – Once switch is opened, we have an LC circuit Current will oscillate with natural frequency ω0 • Strategic Analysis – – – Determine initial current Determine oscillation frequency ω0 Find maximum charge on capacitor Physics 212 Lecture 19, Slide 18 18 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. IL V L C R What is IL, the current in the inductor, immediately AFTER the switch is opened? Take positive direction as shown. (A) IL < 0 (B) IL = 0 (B) (C) IL > 0 (C) Current through inductor immediately AFTER switch is opened Current AFTER switch IS THE SAME AS IS the current through inductor immediately BEFORE switch is opened the BEFORE switch BEFORE switch is opened: all current goes through inductor in direction shown Physics 212 Lecture 19, Slide 19 19 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. IL V L VC=0 C R IL(t=0+) > 0 The energy stored in the capacitor immediately after the switch is opened is zero. (A) TRUE (B) FALSE (B) BEFORE switch is opened: AFTER switch is opened: AFTER dIL/dt ~ 0 ⇒ VL = 0 VC cannot change abruptly cannot VC = 0 BUT: VL = VC since they are in parallel since VC = 0 BB UC = ½ CVC2 = 0 !! IMPORTANT: NOTE DIFFERENT CONSTRAINTS AFTER SWITCH OPENED CURRENT through INDUCTOR cannot change abruptly through INDUCTOR VOLTAGE across CAPACITOR cannot change abruptly across CAPACITOR Physics 212 Lecture 19, Slide 20 20 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. IL V C L R IL(t=0+) > 0 VC(t=0+) = 0 What is the direction of the current immediately after the switch is opened? (A) clockwise (B) counterclockwise (B) Current through inductor immediately AFTER switch is opened Current AFTER switch IS THE SAME AS IS the current through inductor immediately BEFORE switch is opened the BEFORE switch BB BEFORE switch is opened: Current moves down through L BEFORE switch down AFTER switch is opened: Current continues to move down through L Physics 212 Lecture 19, Slide 21 21 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V C L R IL(t=0+) > 0 VC(t=0+) = 0 What is the magnitude of the current right after the switch is opened? (A) I o = V C L (B) I o = V R2 L C (D) I o = V (C) I o = V 2R R Current through inductor immediately AFTER switch is opened Current AFTER switch IS THE SAME AS IS the current through inductor immediately BEFORE switch is opened the BEFORE switch BEFORE switch is opened: IL V IL R VL = 0 IL L BB VL=0 C V = ILR Physics 212 Lecture 19, Slide 22 22 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. IL V C L R Hint: Energy is conserved IL(t=0+) =V/R VC(t=0+) = 0 What is Qmax, the maximum charge on the capacitor during the oscillations? V (A) Qmax = LC R Imax L 1 (B) Qmax = CV 2 Qmax C L C (C) Qmax = CV (D) Qmax = V R LC 2 1 2 1 Qmax LI = 2 2C BB When I is max When (and Q is 0) 1 U = LI 2 2 When Q is max When (and I is 0) Qmax = I max LC = V LC R 2 1 Qmax U= 2C Physics 212 Lecture 19, Slide 23 23 Follow-Up 1 The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V Is it possible for the maximum voltage on the capacitor to be greater than V? (A) YES Qmax = V LC R VL RC C L R Imax =V/R max =V/R (B) NO (B) Vmax = IL Qmax = Vmax can be greater than V IF: can IF V LC R L >R C We can rewrite this condition in terms of the resonant frequency: ω0 L > R OR 1 >R ω0C We will see these forms again when we study AC circuits!! BB Physics 212 Lecture 19, Slide 24 24 ...
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