Lectures_Lect19 - Physics 212 Lecture 19 LC and RLC Circuits P h y s ic s 2 1 2 Le c tu re 1 9 S lid e 1 LC C irc uit I L C Q dI VL = L dt Circuit

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 212 Lecture 19 LC and RLC Circuits P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 1 LC C irc uit - I L C Q + dI VL = L dt + Circuit Equation: VC = Q C - Q dI +L =0 C dt d 2Q Q =- 2 LC dt d 2Q dt 2 = - 2Q 1 LC dQ I= dt where = P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 2 d 2Q dt 2 = - Q 2 = 1 LC L C d x dt 2 2 = - 2 x k = m k F = kx a m x S a m e th ing if we no tic e th a t 1 k and C mL P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 3 T im e De p e nd e nc e I L C ++ - - P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 4 C h e c kp o int 1 a At tim e t = 0 th e c a p a c ito r is fully c h a rg e d with Q m a x a nd th e c urre nt th ro ug h th e c irc uit is 0 . L C Wh a t is th e p o te ntia l d iffe re nc e a c ro s s th e ind uc to r a t t = 0 ? A) VL = 0 s in c e V L = VC B) VL = Q m a x /C C ) VL = Q m a x /2 C T h e vo lta g e a c ro s s th e c a p a c ito r is Q m a x /C Kirc h h o ff's Vo lta g e R u le im p lie s th a t m u s t a ls o b e e q ua l to th e vo lta g e a c ro s s th e in d u c to r Pendulum... P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 5 C h e c kp o int 1 b At tim e t = 0 th e c a p a c ito r is fully c h a rg e d with Q m a x a nd th e c urre nt th ro ug h th e c irc uit is 0 . L C Wh a t is th e p o te ntia l d iffe re nc e a c ro s s th e ind uc to r wh e n th e c urre nt is m a xim um ? A) VL = 0 B) V L = Q m a x /C C ) VL = Q m a x /2 C d I/d t is ze ro wh e n c urre nt is m a xim um P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 6 C h e c kp o int 1 c At tim e t = 0 th e c a p a c ito r is fully c h a rg e d with Q m a x a nd th e c urre nt th ro ug h th e c irc uit is 0 . L C Ho w m uc h e ne rg y is s to re d in th e c a p a c ito r wh e n th e c urre nt is a m a xim um ? A) U = Q m a x 2 /(2 C ) B) U = Q m a x 2 /(4 C ) C ) U = 0 T o ta l Ene rg y is c o ns ta nt ! ULm a x = LIm a x 2 UC m a x = Q m a x 2 /2 C I = m a x wh e n Q = 0 P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 7 C h e c kp o int 2 a T h e c a p a c ito r is c h a rg e d s uc h th a t th e to p p la te h a s a c h a rg e +Q 0 a nd th e b o tto m p la te Q 0 . At tim e t=0 , th e s witc h is c lo s e d a nd th e c irc uit o s c illa te s with fre q ue nc y = 5 0 0 ra d ia ns /s . Wh a t is th e va lue o f th e c a p a c ito r C ? A) C = 1 x 1 0 3 F B) C = 2 x 1 0 3 F C ) C = 4 x 1 0 3 F = 1 LC L C ++ - - L = 4 x 1 0 3 H = 5 0 0 ra d /s C= 1 2L = 1 (25 10 4 )( 4 10- 3 ) = 10 - 3 P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 8 closed at t=0 C h e c kp o int 2 b +Q0 -Q0 L C Wh ic h p lo t b e s t re p re s e nts th e e ne rg y in th e ind uc to r a s a func tio n o f tim e s ta rting jus t a fte r th e s witc h is c lo s e d ? 1 2 U L = LI 2 Ene rg y p ro p o rtio na l to I2 U c a nno t b e ne g a tive C urre nt is c h a ng ing UL is no t c o ns ta nt Initia l c urre nt is ze ro P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 9 C h e c kp o int 2 c Wh e n th e e ne rg y s to re d in th e c a p a c ito r re a c h e s its m a xim um a g a in fo r th e firs t tim e a fte r t=0 , h o w m uc h c h a rg e is s to re d o n th e to p p la te o f th e c a p a c ito r? A) +Q 0 B) +Q 0 /2 C ) 0 D) Q 0 /2 E) Q 0 Q is maximum when current goes to zero I= dQ dt P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 1 0 closed at t=0 L C +Q0 -Q0 Current goes to zero twice during one cycle Ad d re s is ta nc e R I(t) L C VL + VC + VR = 0 dI Q L + + I R=0 dt C d Q Q dQ R + + =0 2 dt LC dt L 2 Use I = dQ/dt and divide by L: Q ( t ) = Q0 e - R t 2L cos ( 't + ) Da m p e d o s c illa tio ns P h y s ic s 2 1 2 Le c tu re 1 9 , S lid e 1 2 T h e s witc h in th e c irc uit s h o wn h a s b e e n c lo s e d fo r a lo ng tim e . At t = 0 , th e s witc h is o p e ne d . C a lc ula tio n V R L C Wh a t is Q MAX , th e m a xim um c h a rg e o n th e c a p a c ito r? Conceptual Analysis Once switch is opened, we have an LC circuit Current will oscillate with natural frequency 0 Strategic Analysis Determine initial current Determine oscillation frequency 0 Find maximum charge on capacitor Physics 212 Lecture 19, Slide 13 Calculation The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. V R I L 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 (B) (C) 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 14 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 I L L VC=0 C The energy stored in the capacitor immediately after the switch is opened is zero. (A) TRUE (B) FALSE (B) BEFORE switch is opened: dIL/dt ~ 0 VL = 0 BUT: VL = VC since they are in parallel VC = 0 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 Lecture 19, Slide 15 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 I L L C 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 IS THE SAME AS the current through inductor immediately BEFORE switch is opened BEFORE switch is opened: Current moves down through L AFTER switch is opened: Current continues to move down through L Physics 212 Lecture 19, Slide 16 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? C (A) (B) (C) (D) = V Io = Io Io = V 2 L R V L C R Io = V 2R Current through inductor immediately AFTER switch is opened IS THE SAME AS the current through inductor immediately BEFORE switch is opened V I I L BEFORE switch is opened: I L L VL = 0 VL=0 C V = ILR Physics 212 Lecture 19, Slide 17 R 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 I L L C What is Qmax, the maximum charge on the capacitor during the oscillations? 1 V Q Qmax = CV (A) (B) (C) (D)max = CV Qmax = LC 2 R Qmax = V R LC Imax L When I is max (and Q is 0) 1 U = LI 2 2 C L Qmax C 2 1 2 1 Qmax LI = 2 2 C When Q is max (and I is 0) 2 1 Qmax U= 2 C Qmax = I max LC = V LC R Physics 212 Lecture 19, Slide 18 The switch in the circuit shown has been closed for a long time. At t = 0, the switch is opened. Is it possible for the maximum voltage on the capacitor to be greater than V? FollowUp 1 V R Imax =V/R I L L C Qmax = (A) YES (B) NO V LC R Qmax = V LC R Vmax V L = R C Vmax can be greater than V IF: L >R C We can rewrite this condition in terms of the resonant frequency: 0 L > R OR 1 >R 0 C We will see these forms again when we study AC circuits!! Physics 212 Lecture 19, Slide 19 ...
View Full Document

This document was uploaded on 03/08/2012.

Ask a homework question - tutors are online