Lect19 - Physics 212 Le cture19 LC and RLC Circuits -...

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Unformatted text preview: Physics 212 Le cture19 LC and RLC Circuits - Oscillation frequency - Energy - Damping 40 30 20 10 0 Physics 212 Le cture19, S 1 lide I 2 ddI Q VL==LL 2 dt dt L C Q Q VC = C but I = dQ dt d 2Q Q =- 2 dt LC d 2Q = - 2Q dt 2 = 1 LC Physics 212 Le cture19, S 2 lide d 2Q = - 2Q dt 2 = 1 LC L C d 2x = - 2 x dt 2 k = m k F = -kx a m x Same thing if we notice that k1 1 C and md L Physics 212 Le cture19, S 3 lide L C Physics 212 Le cture19, S 4 lide At time t = 0 the capacitor is fully charged with Qmax and the current through the circuit is 0. CD L C 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 30 20 10 0 Pendulum... Physics 212 Le cture19, S 5 lide At time t = 0 the capacitor is fully charged with Qmax and the current through the circuit is 0. CD 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 40 30 20 10 0 Physics 212 Le cture19, S 6 lide 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 = 500 radians/s. L = 4 x 10 H = 500 rad/s -3 L C What is the value of the capacitor C? A) B) C) C = 1 x 10-3 F C = 2 x 10-3 F C = 4 x 10-3 F 50 40 30 20 10 0 Physics 212 Le cture19, S 7 lide Which plot best represents the energy in the inductor as a function of time starting just after the switch is closed? 1 2 U L = LI 2 Can UL ever be negative? A) Yes B) No Energy proportional to I2 cant be negative Physics 212 Le cture19, S 8 lide Which plot best represents the energy in the inductor as a function of time starting just after the switch is closed? 1 2 U L = LI 2 Energy proportional to I2 cant be negative Current is changing so UL is not constant Initial current is zero 35 30 25 20 15 10 5 0 Physics 212 Le cture19, S 9 lide Just like LC circuit but the oscillations get smaller because of R Concept makes sense... ...but answer looks kind of complicated Physics 212 Le cture19, S 10 lide 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 Le cture19, S 11 lide C alculation Theswitch in thecircuit shown has be n e close for a long tim . At t = 0, theswitch is d e ope d. ne What is QMAX, them umchargeon the axim capacitor? once C ptual Analysis ne Onceswitch is ope d, wehavean LCcircuit urre que C nt will oscillatewith natural fre ncy 0 V R L C trate S gic Analysis De rm initial curre te ine nt De rm oscillation fre ncy 0 te ine que Find m umchargeon capacitor axim Maxim chargede rm s m umvoltage um te ine axim Physics 212 Le cture19, S 12 lide C alculation Theswitch in thecircuit shown has be n e close for a long tim . At t = 0, theswitch is d e ope d. ne I V R L L C What is I L, thecurre in theinductor, im e nt m diate AFTER theswitch is ope d? Take ly ne positivedire ction as shown. (A) I L < 0 (B) I L = 0 (C I L > 0 ) Curre through inductor im e nt m diate AFTER switch is ope d ly ne I STHE S AME AS thecurre through inductor im e nt m diate BEFORE switch is ope d ly ne BEFORE switch is ope d: ne all curre goe through inductor in dire nt s ction shown Physics 212 Le cture19, S 13 lide C alculation Theswitch in thecircuit shown has be n e close for a long tim . At t = 0, theswitch is d e ope d. ne I V R L V =0 L C C I L(t=0+) > 0 Thee rgy store in thecapacitor im e ne d m diate afte theswitch is ope d is ze ly r ne ro. (A) TRUE BEFORE switch is opened: dI L/dt ~ 0 VL = 0 BUT: VL = VC sincethe arein paralle y l VC= 0 (B) FALS E AFTER switch is opened: VC cannot change abruptly VC = 0 UC = CVC2 = 0 !! IMPORTANT: NOTE DIFFERENT CONSTRAINTS AFTER SWITCH OPENED CURRENT through INDUCTOR cannot change abruptly VOLTAGE across CAPACITOR cannot change abruptly Physics 212 Le cture19, S 14 lide C alculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. I V R IL(t=0+) > 0 VC(t=0+) = 0 L C L What is thedire ction of thecurre im e nt m diate afte theswitch is ope d? ly r ne (A) clockwise (B) counte rclockwise Curre through inductor im e nt m diate AFTER switch is ope d ly ne I STHE S AME AS thecurre through inductor im e nt m diate BEFORE switch is ope d ly ne BEFORE switch is opened: Current moves down through L AFTER switch is opened: Current continues to move down through L Physics 212 Le cture19, S 15 lide C alculation Theswitch in thecircuit shown has be n e close for a long tim . At t = 0, theswitch is d e ope d. ne V R I L(t=0+) > 0 VC (t=0+) = 0 L C What is them agnitudeof thecurre right afte theswitch is ope d? nt r ne (A) I 0 = V C (B) L I0 = V R (C ) 2 L C I0 (D)= V R I0 = 1V 2R Curre through inductor im e nt m diate AFTER switch is ope d ly ne I STHE S AME AS thecurre through inductor im e nt m diate BEFORE switch is ope d ly ne I BEFORE switch is ope d: ne V L I L L VL = 0 V =0 L C V = I LR Physics 212 Le cture19, S 16 lide I R C alculation Theswitch in thecircuit shown has be n e close for a long tim . At t = 0, theswitch is d e ope d. ne Hint: Ene is conse d rgy rve V R I L(t=0+) =V/R VC (t=0+) = 0 I L L C What is Qmax, them umchargeon thecapacitor during theoscillations? axim (A) Qmax = I L m ax V LC (B) R 1 Qmax = (C CV ) 2 Q Qmax (D) = CV Qmax = V R LC C L m ax C 2 1 2 1 Qmax LI max = 2 2 C Whe I is m n ax (and Q is 0) Whe Q is m n ax (and I is 0) 2 1 Qmax U= 2 C Qmax = I max LC = V LC R 1 2 U = LI max 2 Physics 212 Le cture19, S 17 lide ...
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