# 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 - Damping 50 40 30 20 10 0 Confused Avg = 2.8 Confident Physics 212 Lecture 19, Slide 1 Physics Announcements • Due to the review session on Tue, the Homework due dates are a little out of resonance (pardon the pun). • Due dates for the remaining homeworks are on Thursday. • This means that Homework #10 (the next one) is due Thu April 10, and so on … Physics 212 Lecture 19, Slide 2 Physics LC Circuit I 2 ddI Q VL==LL 2 dt dt L C Q Q VC = C but I = dQ dt d 2Q Q + =0 2 dt LC d 2Q + ω 2Q = 0 dt 2 ω= 1 LC Physics 212 Lecture 19, Slide 3 Physics d 2Q + ω 2Q = 0 dt 2 ω= 1 LC L C d 2x + ω2x = 0 dt 2 k ω= m k a F = -kx m x Same thing if we notice that k↔ 1 C and m↔L Physics 212 Lecture 19, Slide 4 Physics Time Dependence L C Physics 212 Lecture 19, Slide 5 Physics Preflight 2 At time t = 0 the capacitor is fully charged with Qmax and the current through the circuit is 0. L C BB What is the potential difference across the inductor at t = 0 ? A) VL = 0 B) VL = Qmax/C C) VL = Qmax/2C VC = VL 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 6 Physics Preflight 4 At time t = 0 the capacitor is fully charged with Qmax and the current through the circuit is 0. BB L C What is the potential difference across the inductor when the current is maximum ? 50 A) VL = 0 B) VL = Qmax/C C) VL = Qmax/2C dI/dt is zero when current is maximum 40 30 20 10 0 Physics 212 Lecture 19, Slide 7 Physics Preflight 6 At time t = 0 the capacitor is fully charged with Qmax and the current through the circuit is 0. BB L C How much energy is stored in the capacitor when the current is a maximum ? A) U = Qmax2/(2C) B) U = Qmax2/(4C) C) U = 0 Total Energy is constant ! ULmax = ½ LI2 UCmax = (Qmax)2/2C I is max when Q = 0 50 40 30 20 10 0 Physics 212 Lecture 19, Slide 8 Physics Preflight 8 The capacitor is charged such that the top plate has a charge +Q0 and the bottom plate -Q0. At time t=0, the switch is closed and the circuit oscillates with frequency ω = 500 radians/s. What is the value of the capacitor C? A) C = 1 x 10-3 F B) C = 2 x 10-3 F C) C = 4 x 10-3 F ω= 1 LC C= 1 50 40 30 20 10 0 BB L = 4 x 10-3 H ω = 500 rad/s 500 L C ω 2L = 1 = 10−3 ( 25 × 104 )(4 × 10−3 ) Physics 212 Lecture 19, Slide 9 Physics closed at t=0 Preflight 10 +Q0 -Q0 BB L C Which plot best represents the energy in the inductor as a function of time starting just after the switch is closed? 12 U L = LI 2 Energy proportional to I ﬂ C cannot be negative 2 30 25 20 15 10 5 0 Current is changing ﬂ UL is not constant Initial current is zero Physics 212 Lecture 19, Slide 10 Physics Preflight 12 When the energy stored in the capacitor reaches its maximum again for the first time after t=0, how much charge is stored on the top plate of the capacitor? A) B) C) D) E) +Q0 +Q0 /2 0 -Q0/2 -Q 0 dQ I= dt closed at t=0 BB L C +Q0 -Q0 35 30 25 20 15 10 5 0 Q is maximum when current goes to zero Current goes to zero twice during one cycle Physics 212 Lecture 19, Slide 11 Physics Just like LC circuit but energy but the oscillations get smaller because of R Add R: Get Damping Concept makes sense… …but answer looks kind of complicated Physics 212 Lecture 19, Slide 12 Physics The elements of a circuit are very simple: dI VL = L dt V = VL + VC + VR Q VC = C dQ I= dt VR = IR This is all we need to know to solve for anything ! Physics 212 Lecture 19, Slide 13 Physics A Different Approach Start with some initial V, I, Q, VL Now take a tiny time step dt dI = VL dt L (1 ms) dQ = Idt VC = Q C Repeat… VR = IR VL = V − VR − VC Physics 212 Lecture 19, Slide 14 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. What is QMAX, the maximum charge on the capacitor? • Conceptual Analysis – – Once switch is opened, we have an LC circuit Current will oscillate with natural frequency ω0 V R L C • Strategic Analysis – – – Determine initial current Determine oscillation frequency ω0 Find maximum charge on capacitor Physics 212 Lecture 19, Slide 15 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V R IL L C 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 (C) IL > 0 Current through inductor immediately AFTER switch is opened IS THE SAME AS the current through inductor immediately BEFORE switch is opened BEFORE switch is opened: all current goes through inductor in direction shown Physics 212 Lecture 19, Slide 16 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V R IL(t=0+) > 0 IL L VC=0 C The energy stored in the capacitor immediately after the switch is opened is zero. (A) TRUE (B) FALSE BEFORE switch is opened: BUT: VL = VC since they are in parallel VC = 0 dIL/dt ~ 0 ⇒ VL = 0 AFTER switch is opened: AFTER VC cannot change abruptly VC = 0 UC = ½ CVC2 = 0 !! BB IMPORTANT: NOTE DIFFERENT CONSTRAINTS AFTER SWITCH OPENED CURRENT through INDUCTOR cannot change abruptly VOLTAGE across CAPACITOR cannot change abruptly Physics 212 Lecture 19, Slide 17 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V R IL(t=0+) > 0 VC(t=0+) = 0 IL L C What is the direction of the current immediately after the switch is opened? (A) clockwise (B) counterclockwise Current through inductor immediately AFTER switch is opened IS THE SAME AS the current through inductor immediately BEFORE switch is opened BEFORE switch is opened: Current moves down through L BEFORE AFTER switch is opened: Current continues to move down through L BB Physics 212 Lecture 19, Slide 18 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V R IL(t=0+) > 0 VC(t=0+) = 0 L C What is the magnitude of the current right after the switch is opened? (A) I 0 = V C L (B) I 0 = V R 2 L C (C) I 0 = V R (D) I 0 = 1V 2R Current through inductor immediately AFTER switch is opened IS THE SAME AS the current through inductor immediately BEFORE switch is opened V IL IL R IL L VL = 0 VL=0 C V = ILR BB BEFORE switch is opened: Physics 212 Lecture 19, Slide 19 Physics Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. Hint: Energy is conserved V R IL(t=0+) =V/R VC(t=0+) = 0 IL L C What is Qmax, the maximum charge on the capacitor during the oscillations? (A) Qmax = Imax L C When I is max (and Q is 0) V LC R 1 Qmax = CV (B) 2 Qmax C (C) Qmax = CV (D) Qmax = V R LC L 2 12 1 Qmax LI max = 2 2C When Q is max (and I is 0) 2 1 Qmax U= 2C Qmax = I max LC = V LC R BB 12 U = LI max 2 Physics 212 Lecture 19, Slide 20 Physics The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. Follow-Up 1 V R Imax =V/R =V/R IL L C Is it possible for the maximum voltage on the capacitor to be greater than V? (A) YES (B) NO Qmax = V LC R Qmax = V LC R Vmax = VL RC Vmax can be greater than V IF: L >R C We can rewrite this condition in terms of the resonant frequency: ω0 L > R OR 1 >R ω0 C BB We will see these forms again when we study AC circuits!! Physics 212 Lecture 19, Slide 21 Physics ...
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