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Time Varying Circuits 2008 Induction 1 A look into the future We have one more week after today (+ one day) Time Varying Circuits Including AC Some additional topics leading to waves A bit of review if there is time. There will be one more Friday morning quiz. I hope to be able to return the exams on Monday at which time we will briefly review the solutions. Induction 2 The Final Exam 8-10 Problems similar to (or exactly) Web- Assignments Covers the entire semester’s work May contain some short answer questions. Induction 3 Max Current Rate of VR=iR increase = max emf ~current Induction 4 E i (1 e Rt / L ) R L (time constant) R Induction 5 We also showed that 1 uinductor 2 B 20 1 0 E 2 ucapacitor 2 Induction 6 At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen?? Induction 7 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 2 0 dt C dt Solution : Rt Q Qm axe 2L cos(d t ) where 2 1/ 2 1 R d LC 2 L Induction 8 The Graph of that LR (no emf) circuit .. Rt 2L I e Induction 9 Induction 10 Mass on a Spring Result Energy will swap back and forth. Add friction Oscillation will slow down Not a perfect analogy Induction 11 Induction 12 LC Circuit High Low High Q/C Low Induction 13 The Math Solution (R=0): LC Induction 14 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. Induction 15 Variable Emf Applied 1.5 1 0.5 emf DC Volts 0 0 1 2 3 4 5 6 7 8 9 10 -0.5 -1 -1.5 Sinusoidal Tim e Induction 16 Sinusoidal Stuff emf A sin(t ) “Angle” Phase Angle Induction 17 Same Frequency with PHASE SHIFT Induction 18 Different Frequencies Induction 19 Note – Power is delivered to our homes as an oscillating source (AC) Induction 20 Producing AC Generator xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx Induction 21 The Real World Induction 22 A Induction 23 Induction 24 The Flux: B A BA cos t emf BA sin t emf i A sin t Rbulb Induction 25 problems … Induction 26 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 27 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 28 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 29 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 30 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 31 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 32 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 33 Source Voltage: emf V V0 sin(t ) Induction 34 Average value of anything: T h T f (t )dt 0 h T 1 h f (t )dt T 0 T Area under the curve = area under in the average box Induction 35 Average Value T 1 V V (t )dt T0 For AC: T V V0 sin t dt 0 1 T 0 Induction 36 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 Induction 37 Average Value T 1 V V (t )dt 0 T 0 Vrms V 2 Induction 38 RMS 2 2 T 1 Vrms V02 Sin2t V0 Sin ( T t )dt T 0 1 T 2 2 2 T Vrms V0 Sin ( t )d t T 2 0 T T 2 V0 V0 Sin ( )d 2 2 Vrms 2 0 V0 Vrms 2 Induction 39 Usually Written as: V peak Vrms 2 V peak Vrms 2 Induction 40 Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit: R E ~ Induction 41 Power V V0 sin(t ) V V0 i sin(t ) R R 2 V0 V02 P (t ) i R sin(t ) R 2 sin t 2 R R Induction 42 More Power - Details V02 V02 P Sin2t Sin2t R R V02 1 T T 0 P Sin2 (t )dt R V02 T1 2 0 P Sin2 (t )dt R V02 1 2 V02 1 0 Sin ( )d R 2 2 P R 2 V02 1 1 V0 V0 Vrms 2 P R 2 R 2 2 R Induction 43 Resistive Circuit We apply an AC voltage to the circuit. Ohm’s Law Applies Induction 44 e iR emf Consider this circuit i R CURRENT AND VOLTAGE IN PHASE Induction 45 Induction 46 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 = VP sin (t - v ) 2 t I = IP sin (t - I ) v -Vp is the angular frequency (angular speed) [radians per second]. Sometimes instead of we use the frequency f [cycles per second] 2 f Induction 47 Frequency f [cycles per second, or Hertz (Hz)] V = Term Phase VP sin (t - v ) V(t) Vp 2 t v -Vp Induction 48 Alternating Current Circuits V = VP sin (t - v ) I = IP sin (t - I ) I(t) V(t) Ip Vp Vrms Irms 2 t t v I/ -Vp -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 V and I are zero). Usually we’re free to set v=0 (but not I). Induction 49 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 50 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 51 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 52 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 53 Review: Resistors in AC Circuits R EMF (and also voltage across resistor): E 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 V and I 2 t “In-phase” Induction 54 Capacitors in AC Circuits C Start from: q = C V [V=Vpsin(t)] E Take derivative: dq/dt = C dV/dt ~ So I = C dV/dt = C VP cos (t) I = C VP sin (t + /2) V This looks like IP=VP/R for a resistor (except for the phase change). I So we call Xc = 1/(C) the Capacitive Reactance 2 t 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” by 90º. I leads V by 90º. Induction 55 I Leads V??? What the **(&@ does that mean?? 2 V 1 I Phase= -(/2) I = C VP sin (t + /2) Current reaches it’s maximum at an earlier time Induction than the voltage! 56 Capacitor Example A 100 nF capacitor is C connected to an AC supply E of peak voltage 170V and frequency 60 Hz. ~ What is the peak current? What is the phase of the current? 2f 2 60 3.77 rad/sec C 3.77 107 1 I=V/XC XC 2.65M C Also, the current leads the voltage by 90o (phase difference). Induction 57 Inductors in AC Circuits V = VP sin (t) ~ L Loop law: V +VL= 0 where VL = -L dI/dt Hence: dI/dt = (VP/L) sin(t). Integrate: I = - (VP / L cos (t) or I = [VP /(L)] sin (t - /2) V Again this looks like IP=VP/R for a resistor (except for the phase change). I t So we call XL = L 2 the Inductive Reactance Here the current lags the voltage by 90o. V and I “out of phase” by 90º. I lags V by 90º. Induction 58 Induction 59 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 60 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 61 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 Ip t Induction Vp 62 Steady State Solution for AC Current (2)cos d t m sin d t I m d L cos d I m R sin d t Im d C • Expand sin & cos expressions sin d t sin d t cos cos d t sin High school trig! cos d t cos d t cos sin d t sin • Collect sindt & cosdt terms separately d L 1/ d C cos R sin 0 cosdt terms I m d L 1/ d C sin I m R cos m sindt terms • These equations can be solved for Im and 63 Induction (next slide) Steady State Solution for AC Current (2)cos d t m sin d t I m d L cos d I m R sin d t Im d C • Expand sin & cos expressions sin d t sin d t cos cos d t sin High school trig! cos d t cos d t cos sin d t sin • Collect sindt & cosdt terms separately d L 1/ d C cos R sin 0 cosdt terms I m d L 1/ d C sin I m R cos m sindt terms • These equations can be solved for Im and 64 Induction (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 m • Solve for and Im in terms of d L 1/ d C X XC m tan L Im R R 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 65 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 Ip t t t Induction Vp 66 Reactance - Phasor Diagrams Resistor Capacitor Inductor Vp Ip Vp Ip Ip t t t Induction Vp 67 “Impedance” of an AC Circuit R L ~ C The impedance, Z, of a circuit relates peak current to peak voltage: Vp Ip (Units: OHMS) Z Induction 68 “Impedance” of an AC Circuit R L ~ C The impedance, Z, of a circuit relates peak current to peak voltage: Vp Ip (Units: OHMS) Z (This is the AC equivalent of Ohm’s law.) Induction 69 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 70 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 71 Phasors for a Series RLC Circuit Ip VRp VLp VP (VCp- VLp) VCp Induction 72 Phasors for a Series RLC Circuit Ip VRp VLp VP (VCp- VLp) VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] Induction 73 Phasors for a Series RLC Circuit Ip VRp VLp VP (VCp- VLp) VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] = Ip2 R2 +InductionXC - Ip XL) 2 (Ip 74 Impedance of an RLC Circuit R Solve for the current: L ~ C Vp Vp Ip R2 (X c X L )2 Z Induction 75 Impedance of an RLC Circuit R Solve for the current: L ~ C Vp Vp Ip R2 (X c X L )2 Z 1 2 Impedance: Z R 2 L C Induction 76 Impedance of an RLC Circuit Vp Ip The current’s magnitude depends on Z the driving frequency. When Z is a 1 2 L minimum, the current is a maximum. Z R 2 C This happens at a resonance frequency: 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 L=1mH R =10 C=10F The current dies away R = 1 0 0 at both low and high r frequencies. 0 1 0 2 1 0 3 1 0 4 Induction 5 1 0 77 Phase in an RLC Circuit Ip We can also find the phase: VRp VLp tan = (VCp - VLp)/ VRp VP or; tan = (XC-XL)/R. or (VCp- VLp) VCp tan = (1/C - L) / R Induction 78 Phase in an RLC Circuit Ip We can also find the phase: VRp VLp tan = (VCp - VLp)/ VRp VP or; tan = (XC-XL)/R. or (VCp- VLp) VCp 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 79 Power in an AC Circuit V = 0 V(t) = VP sin (t) I I(t) = IP sin (t) 2 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 80 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 P( t )dt (T=1/f ) T 0 Induction 81 Power in an AC Circuit Now: sin(t ) sin(t )cos cos(t )sin Induction 82 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 83 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 1 Use: sin ( t ) 2 2 and: sin( t ) cos( t ) 0 1 So P I PV P cos 2 Induction 84 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 1 Use: sin ( t ) 2 2 and: sin( t ) cos( t ) 0 1 So P I PV P cos 2 P which we usually write as Induction IrmsVrms cos 85 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 86 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 87 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 88 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 89 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 90

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