Lect18 - Physics 212 Le cture18 50 40 30 20 10 0 Confused...

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Unformatted text preview: Physics 212 Le cture18 50 40 30 20 10 0 Confused Avg = 2.7 Confident Physics 212 Le cture18, S 1 lide Music Who is the Artist? A) B) C) D) E) Stephane Grappelli Dave Brubeck Itzhak Perlman & Oscar Peterson Hot Club of Cowtown Regina Carter Theme of the week Jazz by violin players from other genres… Pretty cool !! Physics 212 Le cture18, S 2 lide Physics 212 Le cture18 50 40 30 20 10 0 Confused Avg = 2.7 Confident Physics 212 Le cture18, S 3 lide Fromthepre cture S lf I nductance le :e Wrap a wire into a coil to make an “inductor”… ε = -L dI dt Physics 212 Le cture18, S 4 lide What this re m ans: ally e emf induced across L tries to keep I constant ε L = -L dI dt L curre I nt Inductors prevent discontinuous current changes ! Inductors It’s like inertia! Physics 212 Le cture18, S 5 lide Pre flight 2 Two sole ade e ctional are and total num r of turns. a be Two noids arem with thesam cross se I nductor B is twiceas long as inductor A t wice L = µ0 n 2π r 2 z (1/2) 2 2 Compare the inductance of the two solenoids Compare 50 A) LA = 4 LB A) B) LA = 2 LB B) C) LA = LB C) D) LA = (1/2) LB D) E) LA = (1/4) LB E) 40 30 20 10 0 Physics 212 Le cture18, S 6 lide WHAT ARE I NDUC TORSAND C APAC TORSGOOD FOR? I I nsideyour i-clicke r Physics 212 Le cture18, S 7 lide How to think about RL circuits Episode1: How Episode Whe no curre is flowing initially: n nt VL I =0 L R τ = L/R L/R I =V/R L R I V V τ = L/R L/R At t = 0: I =0 VL = VBATT VR = 0 (L is likea giant re sistor) At t >> L/R: At VL = 0 VR = VBATT I = VBATT/R (L is likea short circuit) Physics 212 Le cture18, S 8 lide Pre flight 4 I n thecircuit above theswitch has be n , e ope for a long tim , and thecurre is n e nt ze e rywhe . ro ve re At tim t=0 theswitch is close d. At e I BB What is thecurre I through theve nt rtical What re sistor im e m ly r d im diate afte theswitch is close ? (+ is in thedire ction of thearrow) 40 I IL=0 A) I = V/R V/R B) I = V/2R V/2R C I =0 ) D) I = -V/2R -V/2R E) I = -V/R -V/R Before: IL = 0 After: IL = 0 I = + V/2R 30 20 10 0 Physics 212 Le cture18, S 9 lide RL Circuit (Long Tim ) e What is thecurre I through theve nt rtical re sistor afte theswitch has be n close for a r e d What long tim ? long e (+ is in thedire ction of thearrow) (+ A) I = V/R V/R B) I = V/2R V/2R C I =0 ) D) I = -V/2R -V/2R E) I = -V/R -V/R BB - + + - Afte a long tim in any static circuit: VL = 0 r e KVR: V + IR= 0 Physics 212 Le cture18, S 10 lide 10 L V How to think about RL circuits Episode2: How Episode Whe ste curre is flowing initially: n ady nt VL τ = L/R L/R I =0 R L I =V/R R L R τ = L/R L/R At t = 0: At I = VBATT/R VR = I R VL = VR At t >> L/R: At I =0 VL = 0 VR = 0 Physics 212 Le cture18, S 11 lide 11 Pre flight 6 Afte a long tim , theswitch is ope d, r e ne abruptly disconne cting thebatte fromthe ry circuit. circuit. What is thecurre I through theve nt rtical What re sistor im e m ly r ne im diate afte theswitch is ope d? (+ is in thedire ction of thearrow) (+ A) I = V/R V/R B) I = V/2R V/2R C I =0 ) D) I = -V/2R -V/2R E) I = -V/R -V/R circuit when circuit switch opened switch 50 BB 40 L I =V/R R 30 20 10 0 Current through inductor Current cannot change DISCONTINUOUSLY DISCONTINUOUSLY L Physics 212 Le cture18, S 12 lide 12 Why is the e re xpone ntial be havior ? + VL − I L R τ = L/R L/R V=L dI dt V = IR − τ = L/R L/R + dI L + IR = 0 dt I (t ) = I 0e −tR / L = I 0e −t / τ L where τ = R Physics 212 Le cture18, S 13 lide 13 I L R VL VBATT τ = L/R L/R Le cture : Pre cture le : Did wem ss up?? e No: The resistance is simply twice as big in one case. Physics 212 Le cture18, S 14 lide 14 Pre flight 8 After long time at 0, moved to 1 After long time at 0, moved to 2 BB 60 After switch moved, which case After has larger time constant? has A) Case 1 B) Case 2 C) The same 50 40 30 20 L τ1 = 2R L τ2 = 3R 10 0 Physics 212 Le cture18, S 15 lide 15 Pre flight 10 After long time at 0, moved to 1 After long time at 0, moved to 2 BB Immediately after switch moved, Immediately in which case is the voltage across the inductor larger? across A) Case 1 After switch moved: B) Case 2 V VL1 = 2 R C) The same R Before switch moved: I = 60 50 40 30 20 10 0 V R VL 2 = V 3R R Physics 212 Le cture18, S 16 lide 16 Pre flight 12 After long time at 0, moved to 1 After long time at 0, moved to 2 BB After switch moved for finite time, in After which case is the current through the inductor larger? the A) Case 1 After awhile B) Case 2 I1 = Ie −t / τ C) The same 1 50 40 30 20 10 0 Immediately after: I1 = I 2 I 2 = Ie −t / τ 2 τ1 > τ 2 Physics 212 Le cture18, S 17 lide 17 C alculation Theswitch in thecircuit shown has be n ope for a e n long tim . At t = 0, theswitch is close e d. What is dI L/dt, thetim rateof changeof the e curre through theinductor im e nt m diate afte ly r switch is close d V R1 R2 L R3 once • C ptual Analysis d, nts – Onceswitch is close curre will flow through this 2-loop circuit. d te ine nts e – KVR and KCR can beuse to de rm curre as a function of tim . trate • S gic Analysis te ine nts m diate afte switch is close ly r d. – De rm curre im e te ine m diate afte switch is close ly r d. – De rm voltageacross inductor im e te ine m diate afte switch is close ly r d. – De rm dI L/dt im e Physics 212 Le cture18, S 18 lide 18 C alculation Theswitch in thecircuit shown has be n ope for a e n long tim . At t = 0, theswitch is close e d. V R1 R2 BB L I =0 R3 What is I L, thecurre in theinductor, im e nt m diate afte theswitch is close ly r d? (A) I L =V/R1 up (B) I L =V/R1 down (C I L = 0 ) I NDUC TORS C nt cannot changediscontinuously ! : urre Curre through inductor im e nt m diate AFTER switch is close ly AFTER d Curre I STHE S AME AS t hecurre through inductor im e nt m diate BEFORE switch is close ly BEFORE d the Im e m diate be switch is close I = 0 sinceno batte in loop ly fore d: ry Physics 212 Le cture18, S 19 lide 19 C alculation Theswitch in thecircuit shown has be n ope for a e n long tim . At t = 0, theswitch is close e d. V R1 R2 BB L R3 I L(t=0+) = 0 What is them agnitudeof I 2, thecurre in R2, im e nt m diate afte theswitch is close ly r d? (A) I2 = V (B) (B) R1 V I 2 = (C ) R2 + R3 V I 2 = (D) R1 + R2 + R3 R1 I2 = VR2 R3 R2 + R3 R2 Weknow I L = 0 im e m diate afte switch is close ly r d I Im e m ly r d, Im diate afte switch is close circuit looks V like : like I= V R1 + R2 + R3 R3 Physics 212 Le cture18, S 20 lide 20 C alculation Theswitch in thecircuit shown has be n ope for a e n long tim . At t = 0, theswitch is close e d. V R1 R2 I 2 BB L R3 I L(t=0+) = 0 I 2(t=0+) = V/(R1+R2+R3) What is them agnitudeof VL, thevoltageacross theinductor, im e m diate afte theswitch is ly r close d? (A) VL = V R2 + R3 (B) (B) R1 V(C) V L= V (D) L = 0 VL(E) =V R2 R3 R1 ( R2 + R3 ) VL = V R2 + R3 R1 + R2 + R3 Kirchhoff’s VoltageLaw, VL-I 2 R2 -I 2 R3 =0 VL = I 2 (R2+R3) VL = V ( R2 + R3 ) R1 + R2 + R3 Physics 212 Le cture18, S 21 lide 21 C alculation Theswitch in thecircuit shown has be n ope for a e n long tim . At t = 0, theswitch is close e d. What is dI L/dt, thetim rateof changeof the e curre through theinductor im e nt m diate afte ly r switch is close d V R1 R2 BB L R3 VL(t=0+) = V(R2+R3)/(R1+R2+R3) dI V R2 + R3 (A) L = (B) (B) dt L R1 dI L (C= 0 ) dt dI L V R2 + R3 = (D) dt L R1 + R2 + R3 dI L V = dt L Thetim rateof changeof curre through theinductor (dI L / dt) = VL / L e nt dI L V R2 + R3 = dt L R1 + R2 + R3 Physics 212 Le cture18, S 22 lide 22 Follow Up Theswitch in thecircuit shown has be n closed e f or a long tim . e What is I 2, thecurre through R2 ? (Positive nt value indicatecurre flows to theright) s nt V R1 R2 BB L R3 (A) I 2 = + V (B) R2 + R3 V ( R2 R3 ) I2 = + (C ) R1 + R2 + R3 (D) I 2 = 0 I2 = − V R2 + R3 Afte a long tim , dI /dt = 0 r e The fore thevoltageacross L = 0 re , The forethevoltageacross R2 + R3 = 0 re The forethecurre through R2 + R3 m beze !! re nt ust ro Physics 212 Le cture18, S 23 lide 23 Follow Up 2 Theswitch in thecircuit shown has be n closed e f or a long tim at which point, theswitch is e opened. V What is I 2, thecurre through R2 im e nt m diate ly afte switch is ope d ? (Positivevalue indicate r ne s curre flows to theright) nt R1 IL L R3 R2 I2 BB (A) I2 = + V (B) (B) R1 + R2 + R3 V I2 = + (C) R1 (D) I2 = 0 (E) I2 = − V R1 I2 = − V R1 + R2 + R3 Curre through inductor im e nt m diate AFTER switch is ope d ly AFTER ne Curre I STHE S AME AS t hecurre through inductor im e nt m diate BEFORE switch is ope d ly BEFORE ne the Im e m diate BEFORE switch is ope d: I L = V/R1 ly BEFORE ne Im e m diate AFTER switch is ope d: I L f lows in right loop ly AFTER ne The fore I L = -V/R1 re , Physics 212 Le cture18, S 24 lide 24 ...
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This note was uploaded on 03/05/2011 for the course PHYS 212 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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