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Transcript
Time Varying Circuits 2009 Induction 1 The Final Exam Approacheth 8-10 Problems similar to Web-Assignments Covers the entire semester’s work May contain some short answer (multiple choice) questions. Induction 2 Spring 2009 Final Exam Schedule Tuesday, April 28 - Monday, May 4 No Exams on Sunday EXAM TIMES Class Meeting Times EXAM DAY 1 TUES 4/28 EXAM DAY 2 WED 4/29 EXAM DAY 3 THURS 4/30 EXAM DAY 4 FRI 5/1 EXAM DAY 5 SAT 5/2 7:00 a.m. 9:50 a.m. 7:30-10:20 F 9:007:30-10:20 W 8:3010:15 M/ 7:30-10:20 T 9:007:30-8:45 TR 7:309:20 MWF 9:007:30-8:45 F 9:3010:15 TR (all a.m.) 10:20 R (all a.m.) 10:15 MW (all a.m.) 10:20 MWF (all a.m.) 10:00 a.m. 12:50 p.m. 10:30-11:45 TR 10:30-1:20 T 1:00 p.m. 3:50 p.m. 1:30-4:20 W 2:301:30-2:45 TR 1:30- 3:20 MWF 3:001:30-4:20 R 3:004:20 T 4:15 WF 3:00-4:15 4:15 TR MW 4:00 p.m. 6:50 p.m. 6:00-7:15 TR 6:00-7:15 MW 4:30-5:45 TR 4:30-7:15 F FREE PERIOD and Alternate Time and Alternate Time and Alternate Time and Alternate Time 7:00 p.m. 9:50 p.m. 6:00-8:50 T 7:3010:20 T (all p.m.) 10:30-1:20 W 11:30-12:20 MWF 12:00-1:15 MW 12:00-1:15 WF 6:00-8:50 W 7:007:50 MWF 7:308:45 MW 7:3010:20 W (all p.m.) 10:30-1:20 F 12:3010:30-1:20 R 12:00- 1:20 MWF 12:001:15 TR 1:15 M/ 10:30-11:45 F 1:30-4:20 F 3:304:20 MWF 3:004:15 M/ 1:30-2:45 F EXAM DAY 6 MON 5/4 Finals For Saturday 7:30-8:20 MWF Classes Are Held 7:30-8:45 MW 7:30During Regular 10:20 M (all a.m.) Class Meeting Times Finals For Saturday 10:30-11:20 MWF Classes Are Held 10:30-11:45 MW During Regular 10:30-1:20 M Class Meeting Times FREE PERIOD 6:00-8:50 R 7:306:00-8:50 F 8:008:45 TR 7:30-10:20 FREE PERIOD Induction 8:50 MWF (all p.m.) R (all p.m.) 1:30-2:20 MWF 1:30-2:45 MW 1:304:20 M 4:30-5:45 MW and Alternate Time 6:00-6:50 MWF 6:00-8:50 M 7:303 10:20 M (all p.m.) HowjaDo?? A. B. C. D. 4 I done good I done ok I done not so ok I screwed up major Induction The Test Itself Was A. B. C. D. 5 Fair Not so fair. Really Unfair. The worst kind of unfair in the entire universe. Induction Sort of like RC circuit issues. Back to Circuits for a bit …. Definition Current in loop produces a magnetic field in the coil and consequently a magnetic flux. If we attempt to change the current, an emf will be induced in the loops which will tend to oppose the change in current. This this acts like a “resistor” for changes in current! Remember Faraday’s Law d emf V E ds dt Lentz Look at the following circuit: Switch is open NO current flows in the circuit. All is at peace! Close the circuit… After the circuit has been close for a long time, the current settles down. Since the current is constant, the flux through the coil is constant and there is no . Emf Current is simply E/R (Ohm’s Law) Close the circuit… When switch is first closed, current begins to flow rapidly. The flux through the inductor changes rapidly. An emf is created in the coil that opposes the increase in current. The net potential difference across the resistor is the battery emf opposed by the emf of the coil. Close the circuit… d emf dt Ebattery V (notation) d V iR 0 dt Moving right along … Ebattery V (notation) d V iR 0 dt The flux is proportion al to the current as well as to the number of turns, N. For a solonoid, i Li N B d di L dt dt di V iR L 0 dt Definition of Inductance L N B L i UNIT of Inductance = 1 henry = 1 T- m2/A B is the flux near the center of one of the coils making the inductor Consider a Solenoid l B ds i 0 enclosed Bl 0 nli or n turns per unit length B 0 ni So…. N B nlBA nl 0 niA L i i i or L 0 n 2 Al or inductance 2 L/l n A unit length Depends only on geometry just like C and is independent of current. Inductive Circuit i Switch to “a”. Inductor seems like a short so current rises quickly. Field increases in L and reverse emf is generated. Eventually, i maxes out and back emf ceases. Steady State Current after this. THE BIG INDUCTION As we begin to increase the current in the coil The current in the first coil produces a magnetic field in the second coil Which tries to create a current which will reduce the field it is experiences And so resists the increase in current. Back to the real world… Switch to “a” i sum of voltage drops 0 : di E iR L 0 dt same form as the capacitor equation q dq E R 0 C dt Solution (See textbook) E Rt / L i (1 e ) R time constant L R Switch position “b” E0 di L iR 0 dt E t / i e R VR=iR ~current Max Current Rate of increase = max emf E (1 e Rt / L ) R L (time constant) R i IMPORTANT QUESTION Switch closes. No emf Current flows for a while It flows through R Energy is conserved (i2R) WHERE DOES THE ENERGY COME FROM?? For an answer Return to the Big C E=e0A/d +dq +q -q We move a charge dq from the (-) plate to the (+) one. The (-) plate becomes more (-) The (+) plate becomes more (+). dW=Fd=dq x E x d The calc q dW (dq) Ed (dq) d (dq) d e0 e0 A d d q2 W qdq e0 A e0 A 2 or 1 2 Ad 1 2 1 2 W (A) e 0 2 Ad e 0 E Ad 2e 0 A 2 e0 2 e0 2 d 2 energy 1 2 u e0 E unit volum e 2 The energy is in the FIELD !!! What about POWER?? di E L iR dt i : di 2 iE Li i R dt power to circuit Must be dWL/dt power dissipated by resistor So dWL di Li dt dt 1 2 WL L idi Li 2 1 2 WC CV 2 Energy stored in the Capacitor WHERE is the energy?? l B ds i 0 enclosed 0l Bl 0 nil B 0 ni or B 0 Ni l BA 0 Ni l A Remember the Inductor?? N L i N Number of turns in inductor i current. Φ Magnetic flux throu gh one turn. So … N L i N i L 1 2 1 2 N 1 W Li i N i 2 2 i 2 0 NiA l 1 0 NiA 1 2 2 2 A W Ni 0 N i 2 l 2 0 l 1 A W N i 2 0 l 2 0 2 2 From before : B 0 Ni l 1 A 1 2 W Bl B V (volume) 2 0 l 2 0 2 2 or W 1 2 u B V 2 0 ENERGY IN THE FIELD TOO! IMPORTANT CONCLUSION A region of space that contains either a magnetic or an electric field contains electromagnetic energy. The energy density of either is proportional to the square of the field strength. Solution (From Before) E Rt / L i (1 e ) R time constant L R At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen?? 35 Induction The math … For an RLC circuit with no driving potential (AC or DC source): Q di iR L 0 C dt dQ Q d 2Q R L 0 2 dt C dt Solution : Q Qmax e Rt 2L cos( d t ) where 1 R d LC 2 L 36 2 1/ 2 Induction The Graph of that LR (no emf) circuit I 37 e Rt 2L Induction 38 Induction Mass on a Spring Result Energy will swap back and forth. Add friction 39 Oscillation will slow down Not a perfect analogy Induction 40 Induction LC Circuit Low High Q/C High Low 41 Induction The Math Solution (R=0): LC Induction 42 New Feature of Circuits with L and C These circuits produce oscillations in the currents and voltages Without a resistance, the oscillations would continue in an un-driven circuit. With resistance, the current would eventually die out. 43 Induction Variable Emf Applied 1.5 1 Volts emf 0.5 DC 0 0 1 2 3 4 5 6 7 8 -0.5 -1 Sinusoidal -1.5 Tim e 44 Induction 9 10 Sinusoidal Stuff emf A sin( t ) “Angle” Phase Angle 45 Induction Same Frequency with PHASE SHIFT 46 Induction Different Frequencies 47 Induction Note – Power is delivered to our homes as an oscillating source (AC) 48 Induction Producing AC Generator xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx 49 Induction The Real World 50 Induction A 51 Induction 52 Induction The Flux: B A BA cos t emf BA sin t emf i A sin t Rbulb 53 Induction problems … 54 Induction 14. Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of its final value in 3.00 s. Induction 55 18. In the circuit shown in Figure P32.17, let L = 7.00 H, R = 9.00 Ω, and ε = 120 V. What is the self-induced emf 0.200 s after the switch is closed? Induction 56 32. At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a resistance of 30.0 Ω. (a) Find the energy stored in the magnetic field when the current reaches half its maximum value. (b) After the emf is connected, how long does it take the current to reach this value? Induction 57 16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0. Induction 58 17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value? Induction 59 27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA? Induction 60 52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s? Induction 61 Source Voltage: emf V V0 sin( t ) 62 Induction Average value of anything: T h T f (t )dt 0 h 1 h T T f (t )dt 0 T Area under the curve = area under in the average box 63 Induction Average Value T 1 V V (t )dt T0 For AC: 64 T 1 V V0 sin t dt 0 T0 Induction So … Average value of current will be zero. Power is proportional to i2R and is ONLY dissipated in the resistor, The average value of i2 is NOT zero because it is always POSITIVE 65 Induction Average Value T 1 V V (t )dt 0 T0 Vrms 66 V 2 Induction RMS Vrms V02 Sin 2t V0 1 2 2 Sin ( t )dt T 0 T T 1 T 2 2 2 Sin ( t )d t T 2 0 T T T Vrms V0 67 Vrms V0 2 Vrms V0 2 2 V0 0 Sin ( )d 2 2 Induction Usually Written as: Vrms V peak 2 V peak Vrms 2 68 Induction in the circuit: R E ~ 69 Induction Power V V0 sin( t ) V V0 i sin( t ) R R 2 2 V V 0 2 2 0 P(t ) i R sin( t ) R sin t R R 70 Induction More Power - Details 2 V02 V P Sin 2t 0 Sin 2t R R P P P P 71 V02 R 2 V0 R V02 R V02 R 1 T 2 T Sin 2 (t )dt 0 T 1 0 Sin 2 (t )dt 2 V 1 2 2 0 1 Sin ( )d 2 0 R 2 2 1 1 V0 V0 Vrms 2 R 2 2 R Induction Resistive Circuit We apply an AC voltage to the circuit. Ohm’s Law Applies Induction 72 Consider this circuit 73 e iR emf i R CURRENT AND VOLTAGE IN PHASE Induction 74 Induction Alternating Current Circuits An “AC” circuit is one in which the driving voltage and hence the current are sinusoidal in time. V(t) Vp v 2 t V = VP sin (t - v ) I = IP sin (t - I ) -Vp is the angular frequency (angular speed) [radians per second]. Sometimes instead of we use the frequency f [cycles per second] Induction Frequency f [cycles per second, or Hertz (Hz)] 2 f 75 Phase Term V= V P sin (t - v ) V(t) Vp 2 t v -Vp Induction 76 Alternating Current Circuits V = VP sin (t - v ) I = IP sin (t - I ) I(t) V(t) Ip Vp Irms Vrms v -Vp 2 t I/ t -Ip Vp and Ip are the peak current and voltage. We also use the “root-mean-square” values: Vrms = Vp / 2 and Irms=Ip / 2 v and I are called phase differences (these determine when Induction 77 V and I are zero). Usually we’re free to set v=0 (but not I). Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. Induction 78 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V. Induction 79 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V. This 60 Hz is the frequency f: so =2 f=377 s -1. Induction 80 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so Vp=Vrms 2 = 170 V. This 60 Hz is the frequency f: so =2 f=377 s -1. So V(t) = 170 sin(377t + v). Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t). Induction 81 Review: Resistors in AC Circuits R E ~ EMF (and also voltage across resistor): V = VP sin (t) Hence by Ohm’s law, I=V/R: I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R) V I 2 t Induction V and I “In-phase” 82 Capacitors in AC Circuits C Start from: q = C V [V=Vpsin(t)] Take derivative: dq/dt = C dV/dt So I = C dV/dt = C VP cos (t) E ~ I = C VP sin (t + /2) V I 2 t This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference). V and I “out of phase”Induction by 90º. I leads V by 90º. 83 I Leads V??? What the **(&@ does that mean?? 2 V I 1 Phase= -(/2) I = C VP sin (t + /2) Induction Current reaches it’s maximum at an earlier time than the voltage! 84 Capacitor Example A 100 nF capacitor is connected to an AC supply of peak voltage 170V and frequency 60 Hz. C E ~ What is the peak current? What is the phase of the current? 2f 2 60 3.77 rad/sec C 3.77 107 1 XC 2.65M C I=V/XC 85 Also, the current leadsInduction the voltage by 90o (phase difference). Inductors in AC Circuits ~ L V = VP sin (t) Loop law: V +VL= 0 where VL = -L dI/dt Hence: dI/dt = (VP/L) sin(t). Integrate: I = - (VP / L cos (t) or V Again this looks like IP=VP/R for a resistor (except for the phase change). I I = [VP /(L)] sin (t - /2) 2 t So we call the XL = L Inductive Reactance Here the current lags the voltage by 90o. V and I “out of phase”Induction by 90º. I lags V by 90º. 86 Induction 87 Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Vp Ip t Induction 88 Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Capacitor Vp Ip Ip t t Induction Vp 89 Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. Resistor Capacitor Inductor Vp Ip Vp Ip t Induction Vp Ip 90 Steady State Solution for AC Im Current (2) I m d L cos d I m R sin d t cos d t e m sin d t d C • Expand sin & cos expressions sin d t sin d t cos cos d t sin cos d t cos d t cos sin d t sin High school trig! • Collect sindt & cosdt terms separately cosdt terms d L 1/ d C cos R sin 0 sindt terms I m d L 1/ d C sin I m R cos e m • These equations can be solved for I and m Induction 91 (next slide) Steady State Solution for AC Im Current (2) I m d L cos d I m R sin d t cos d t e m sin d t d C • Expand sin & cos expressions sin d t sin d t cos cos d t sin cos d t cos d t cos sin d t sin High school trig! • Collect sindt & cosdt terms separately cosdt terms d L 1/ d C cos R sin 0 sindt terms I m d L 1/ d C sin I m R cos e m • These equations can be solved for I and m Induction 92 (next slide) Steady State Solution for AC Current (3) d L 1/ d C cos R sin 0 I m d L 1/ d C sin I m R cos e m • Solve for and Im in terms of tan d L 1/ d C R X XC L R Im em Z • R, XL, XC and Z have dimensions of resistance X L d L Inductive “reactance” X C 1/ d C Capacitive “reactance” Z R2 X L X C 2 Total “impedance” • Let’s try to understand this solution using “phasors” Induction 93 REMEMBER Phasor Diagrams? A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. Resistor Capacitor Inductor Vp Ip Vp Ip t Ip t t Induction Vp 94 Reactance - Phasor Diagrams Resistor Capacitor Inductor Vp Ip Vp Ip t Ip t t Induction Vp 95 “Impedance” of an AC Circuit R L ~ C The impedance, Z, of a circuit relates peak current to peak voltage: Ip Vp Z Induction (Units: OHMS) 96 “Impedance” of an AC Circuit R L ~ C The impedance, Z, of a circuit relates peak current to peak voltage: Ip Vp Z (Units: OHMS) (This is the AC equivalent of Ohm’s law.) Induction 97 Impedance of an RLC Circuit R E ~ L C As in DC circuits, we can use the loop method: E - V R - VC - VL = 0 I is same through all components. Induction 98 Impedance of an RLC Circuit R E ~ L C As in DC circuits, we can use the loop method: E - V R - VC - VL = 0 I is same through all components. BUT: Voltages have different PHASES they add as PHASORS. Induction 99 Phasors for a Series RLC Circuit Ip VLp VRp (VCp- VLp) Induction VP VCp 100 Phasors for a Series RLC Circuit Ip VLp VRp (VCp- VLp) VP VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] Induction 101 Phasors for a Series RLC Circuit Ip VLp VRp (VCp- VLp) VP VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] = Ip2 R2 +Induction (Ip XC - Ip XL) 2 102 Impedance of an RLC Circuit R Solve for the current: Ip ~ L C Vp Vp Z R2 (X c X L )2 Induction 103 Impedance of an RLC Circuit R Solve for the current: Ip ~ L C Vp Z R2 (X c X L )2 Impedance: Vp Z 1 R L C 2 2 Induction 104 Impedance of an RLC Circuit Vp Ip Z 1 R L C Z The current’s magnitude depends on the driving frequency. When Z is a minimum, the current is a maximum. This happens at a resonance frequency: 2 2 The circuit hits resonance when 1/C-L=0: r=1/ LC When this happens the capacitor and inductor cancel each other and the circuit behaves purely resistively: IP=VP/R. IP R =10 L=1mH C=10F R = 1 0 0 0 1 0 r 2 1 0 3 1 0 4 Induction 5 1 0 The current dies away at both low and high frequencies. 105 Phase in an RLC Circuit Ip VLp We can also find the phase: VRp (VCp- VLp) VP tan = (VCp - VLp)/ VRp or; or VCp Induction tan = (XC-XL)/R. tan = (1/C - L) / R 106 Phase in an RLC Circuit Ip VLp We can also find the phase: VRp (VCp- VLp) VP tan = (VCp - VLp)/ VRp or; or VCp tan = (XC-XL)/R. tan = (1/C - L) / R More generally, in terms of impedance: cos R/Z At resonance the phase goes to zero (when the circuit becomes purely resistive, the current and voltage are in phase). Induction 107 Power in an AC Circuit V = 0 I 2 t V(t) = VP sin (t) I(t) = IP sin (t) (This is for a purely resistive circuit.) P P(t) = IV = IP VP sin 2(t) Note this oscillates twice as fast. 2 t Induction 108 Power in an AC Circuit The power is P=IV. Since both I and V vary in time, so does the power: P is a function of time. Use, V = VP sin (t) and I = IP sin ( t+ ) : P(t) = IpVpsin(t) sin ( t+ ) This wiggles in time, usually very fast. What we usually care about is the time average of this: 1 T P 0 P( t )dt T Induction (T=1/f ) 109 Power in an AC Circuit Now: sin( t ) sin( t )cos cos(t )sin Induction 110 Power in an AC Circuit Now: sin( t ) sin( t )cos cos(t )sin P( t ) I PVP sin( t )sin( t ) I PVP sin 2( t )cos sin( t )cos( t )sin Induction 111 Power in an AC Circuit Now: sin( t ) sin( t )cos cos(t )sin P( t ) I PVP sin( t )sin( t ) I PVP sin 2( t )cos sin( t )cos( t )sin Use: and: So sin ( t ) 2 1 2 sin( t ) cos( t ) 0 P 1 2 I PV P cos Induction 112 Power in an AC Circuit Now: sin( t ) sin( t )cos cos(t )sin P( t ) I PVP sin( t )sin( t ) I PVP sin 2( t )cos sin( t )cos( t )sin Use: and: So sin ( t ) 2 1 2 sin( t ) cos( t ) 0 P 1 2 I PV P cos which we usually write as InductionP IrmsVrms cos 113 Power in an AC Circuit P IrmsVrms cos goes from -900 to 900, so the average power is positive) cos( is called the power factor. For a purely resistive circuit the power factor is 1. When R=0, cos()=0 (energy is traded but not dissipated). Usually the power factor depends on frequency. Induction 114 16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0. Induction 115 17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value? Induction 116 27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA? Induction 117 52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s? Induction 118