{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online