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AC Circuits w i m wL im wC m im R iron V1 V2 N1 (primary) N2 (secondary) Physics 1304: Lecture 18, Pg 1 Lecture Outline Driven Series LCR Circuit: • General solution • Resonance condition » Resonant frequency » “Sharpness of resonance” = Q • Power considerations » Power factor depends on impedance Transformers • Voltage changes • Faraday’s Law in action gives induced primary current. • Power considerations Text Reference: Chapter 33.4-6 Physics 1304: Lecture 18, Pg 2 Phasors i R m sin wt R Q m sin wt C i C wC m cos wt di L m sin wt dt • R: V in phase with i VR Ri R m sin wt • C: V lags i by 90 VC • L: V leads i by 90 VL L iL m cos wt wL A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2-d plane with angular velocity w. Recall uniform circular motion: x r coswt y r sinwt The projections of r (on the vertical y axis) execute sinusoidal oscillation. y y w x Physics 1304: Lecture 18, Pg 3 Suppose: xx 00,, r1 r1 nn .... r1 r1 Phasors for L,C,R 11 V i R V f(f(xx))000 VR Ri m sinwt xx 00 22 44 66 1 i m cos wt wC x xx f( x ) 00 V r1 0 , .. r1 n 1 wt C V 2 4 V i f( x ) 0 0 V 0 C 66 x 1.01 1 w i 0 1.01 1 VL wLi m cos wt R t i VC w wt r1 00,, r1 .... r1 r1 n 11 11 i L w i wt L 2 4 6 Physics 1304: Lecture 18, Pg 4 Series LCR AC Circuit R • Back to the original problem: the loop equation gives: L d 2Q dt 2 Q dQ R m sin wt C dt C L • Assume a solution of the form: Here all unknowns, (im,) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: m sinwt. • To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram. i i m sin(wt ) Physics 1304: Lecture 18, Pg 5 Phasors: LCR • Given: m sin wt • Assume: R C L w i m wL From these equations, we can draw the phasor diagram to the right. This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time. im wC m im R Phasors: LCR w i m XL R C m L i m XC • im R The phasor diagram has been relabeled in terms of the reactances defined from: X L wL XC 1 wC The unknowns (im,) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf . Physics 1304: Lecture 18, Pg 7 Lecture 20, ACT 3 A driven RLC circuit is connected as shown. For what frequencies w of the voltage source is the current through the resistor largest? (a) w small (b) w large (c) w R w 1 LC L C Physics 1304: Lecture 18, Pg 8 Conceptual Question A driven RLC circuit is connected as shown. For what frequencies w of the voltage source is the current through the resistor largest? (a) w small (b) w large (c) w R w 1 LC C • This is NOT a series RLC circuit. We cannot blindly apply our techniques for solving the circuit. We must think a little bit. • However, we can use the frequency dependence of the impedances (reactances) to answer this question. • The reactance of an inductor = XL = wL. • The reactance of a capacitor = XC = 1/(wC). • Therefore, • in the low frequency limit, XL 0 and XC . • Therefore, as w 0, the current will flow mostly through the inductor; the current through the capacitor approaches 0. • in the high frequency limit, XL and XC 0 . • Therefore, as w , the current will flow mostly through the capacitor, approaching a maximum imax = /R. L Phasors:LCR i m XL m i m XC X L wL im R i m (XL-X C) m X L XC tan R im R XC 1 wC 2m i 2m R 2 X L XC 2 Z R 2 X L XC 2 im m m 2 Z R 2 X L XC Physics 1304: Lecture 18, Pg 10 Phasors:Tips i m XL y • This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis. • Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0). i m XC imXL m m imR imXC “Full Phasor Diagram” x im R From this diagram, we can also create a triangle which allows us to calculate the impedance Z: Z X L XC R “ Impedance Triangle” Phasors:LCR We have found the general solution for the driven LCR circuit: i i m sin(wt ) imXL m w the loop eqn tan i mZ im m Z imR imXC XL Z X L wL 1 XC wC Z R 2 X L XC X L XC R 2 XL - XC R XC Physics 1304: Lecture 18, Pg 12 Lagging & Leading The phase between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. X L wL 1 XC wC X XC tan L R im m Z XL Z XL XL R XC XL > XC >0 current LAGS applied voltage Z R Z XC XL < XC <0 current LEADS applied voltage R XC XL = XC =0 current IN PHASE applied voltage Conceptual Question The series LCR circuit shown is driven by a generator with voltage = msinwt. The time dependence of the current i which flows in the circuit is shown in the plot. How should w be changed to 1A bring the current and driving voltage into phase? (b) decrease w (a) increase w (c) impossible 1B • Which of the following phasors represents the current i at t=0? (c) (b) i (a) i i Physics 1304: Lecture 18, Pg 14 Lecture 21, ACT 1 The series LCR circuit shown is driven by a generator with voltage = msinwt. The time dependence of the current i which flows in the circuit is shown in the plot. xx r1 .... r1 nn im 11 1.01 R C How should w be changed to 1A bring the current and driving voltage into phase? (a) increase w 00, L f( f(ixx)) 00 0 1.01-i 1 m 1 00 .53 (b) decrease w io 2 2 t 4 4 xx 6 6 6.81 (c) impossible • From the plot, it is clear that the current is LEADING the applied voltage. XL Therefore, the phasor diagram must look like this: i XC Therefore, XC X L • To bring the current into phase with the applied voltage, we need to increase XL and decrease XC. • Increasing w will do both!! Physics 1304: Lecture 18, Pg 15 Lecture 21, ACT 1 x The series LCR circuit shown is driven by a generator with voltage = msinwt. The time dependence of the current i which flows in the circuit is shown in the plot. How should w be changed to 1A bring the current and driving voltage into phase? (a) increase w 0, r1 .. r1 n i1 1.01 m R C (b) decrease w io L f(ix ) 00 1.01-i 1 m 0 .53 2 t 4 x (c) impossible 1B • Which of the following phasors represents the current i at t=0? w (c) (b) i (a) i w i • The projection of i along the vertical axis is negative here. • no way jose • The sign of i is correct at t=0. • However, it soon will become negative! • nope • This one looks just right!! • = -30 Physics 1304: Lecture 18, Pg 16 6 6.81 Resonance For fixed R,C,L the current im will be a maximum at the resonant frequency w0 which makes the impedance Z purely resistive. ie: m im Z m R 2 X L XC reaches a maximum when: 2 X L XC the frequency at which this condition is obtained is given from: 1 1 woL wo w oC LC • Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! • At this frequency, the current and the driving voltage are in phase! X XC tan L 0 R Physics 1304: Lecture 18, Pg 17 Resonance The current in an LCR circuit depends on the values of the elements and on the driving frequency through the relation m im Z m R 2 X L XC 1 im m m cos R 1 tan 2 R Suppose you plot the current versus w, the source voltage frequency, you would get: 2 x 0.0 , r1 .. r1 n m 1 R0 R=Ro f( x ) im0.5 g( x ) R=2Ro 00 00 1 wx 2w2o Physics 1304: Lecture 18, Pg 18 Power in LCR Circuit • The power supplied by the emf in a series LCR circuit depends on the frequency w. It will turn out that the maximum power is supplied at the resonant frequency w0. The instantaneous power (for some frequency, w) delivered at time t is given by: Remember what this stands for • The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle. • To evaluate the average on the right, we first expand the sin(wt) term. Physics 1304: Lecture 18, Pg 19 Power in LCR Circuit • Expanding, • Taking the averages, 1 1.01+1 sinwtcoswt (Product of even and odd function = 0) • Generally: h( x ) 0 0 1.01 1 -1 • x Putting it all back together again, 1/2 0 00 2 r1 0.0 ,0 .. r1 n 1 +1 wt x 4 2p 6 6.28 sin2wt h( x ) 0 0 -11 2p wt Physics 1304: Lecture 18, Pg 20 x 00 2 4 6 Power in LCR Circuit This result is often rewritten in terms of rms values: rms 1 m 2 i rms 1 im 2 P( t ) rmsi rms cos Power delivered depends on the phase, , the “power factor” phase depends on the values of L, C, R, and w therefore... Physics 1304: Lecture 18, Pg 21 Power in RLC P( t ) rmsi rms cos Power, as well as current, peaks at w = w 0. The sharpness of the resonance depends on the values of the components. Recall: • Therefore, We can write this in the following manner (you can do the algebra): x …introducing the curious factors Q and x... w wo Q Lw o R Physics 1304: Lecture 18, Pg 22 The Q factor A parameter “Q” is often defined to describe the sharpness of resonance peaks in both mechanical and electrical oscillating systems. “Q” is defined as where Umax is max energy stored in the system and DU is the energy dissipated in one cycle period For RLC circuit, Umax is (e.g.) And losses only in R, namely This gives And for completeness, note Physics 1304: Lecture 18, Pg 23 Power in RLC For Q > few, x 0.0 , r1 2 rms 1 R0 wres Q fwhm .. r1 n Q=3 FWHM R=Ro f( x ) <P>0.5 g( x ) R=2Ro 00 00 w 1 2wo2 x Physics 1304: Lecture 18, Pg 24 Conceptual Question 2 Consider the two circuits shown where CII = 2 CI. What is the relation between the quality factors, QI and QII , of the two circuits? 2A (a) QII < QI (b) QII = QI R R CI L CII L (c) QII > QI 2B • What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency? (a) PII < PI (b) PII = PI (c) PII > PI Physics 1304: Lecture 18, Pg 25 Lecture 21, ACT 2 Consider the two circuits shown where CII = 2 CI. What is the relation between the quality factors, QI and QII , of the two circuits? 2A (a) QII < QI (b) QII = QI • We know the definition of Q: Q R R CI L CII L (c) QII > QI w 0L R • At first glance, it looks like Q is independent of C. • At second glance, we see this cannot be true, since the resonant frequency wo depends on C! w0 1 LC Doubling C decreases wo by sqrt(2)! \ Doubling C decreases Q by sqrt(2)! Doubling C increases the width of the resonance! Physics 1304: Lecture 18, Pg 26 Lecture 21, ACT 2 Consider the two circuits shown where CII = 2 CI. What is the relation between the quality factors, QI and QII , of the two circuits? 2A (a) QII < QI (b) QII = QI R R CI L CII L (c) QII > QI 2B • What is the relation between PI and PII , the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency? (a) PII < PI (b) PII = PI (c) PII > PI • At the resonant frequency, the impedance of the circuit is purely resistive. • Since the resistances in each circuit are the same, the impedances at the resonant frequency for each circuit are equal. • Therefore, The power delivered by the generator to each circuit is identical. Physics 1304: Lecture 18, Pg 27 Power Transmission How can we transport power from power stations to homes? Why do we use “high tension” lines? At home, the AC voltage obtained from outlets in this country is 120V at 60Hz. Transmission of power is typically at very high voltages ( eg ~500 kV) Transformers are used to raise the voltage for transmission and lower the voltage for use. We’ll describe these next. But why? Calculate ohmic losses in the transmission lines: Define efficiency of transmission: Pout iVin i 2 R iR 1 Pin iVin Vin Vin Vin Pin R 1 2 V in Keep R small Make Vin big – Note for fixed input power and line resistance, the inefficiency 1/V2 Example: Quebec to Montreal 1000 km R= 220W suppose Pin = 500 MW With Vin=735kV, = 80%. The efficiency goes to zero quickly if Vin were lowered! Physics 1304: Lecture 18, Pg 28 Physics 1304: Lecture 18, Pg 29 Transformers • AC voltages can be stepped up or stepped down by the use of transformers. The AC current in the primary circuit creates a time-varying magnetic field in the iron This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils. • iron V1 V2 N1 (primary) N2 (secondary) The iron is used to maximize the mutual inductance. We assume that the entire flux produced by each turn of the primary is trapped in the iron. Physics 1304: Lecture 18, Pg 30 Ideal No resistance losses Transformers (no load) All flux contained in iron Nothing connected on secondary The primary circuit is just an AC voltage source in series with an inductor. The change in flux produced in each turn is given by: d turn V1 dt N1 iron • The change in flux per turn in the secondary coil is the same as the change in flux per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by: V2 N 2 V1 V2 N1 (primary) N2 (secondary) d turn N 2 V1 dt N1 • Therefore, • N2 > N1 secondary V2 is larger than primary V1 (step-up) • N1 > N2 secondary V2 is smaller than primary V1 (step-down) • Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small! Physics 1304: Lecture 18, Pg 31 Ideal Transformers with a Load iron What happens when we connect a resistive load to the secondary coil? Flux produced by primary coil induces an emf in secondary emf in secondary produces current i2 V i2 2 R V1 V2 N1 (primary) R N2 (secondary) This current produces a flux in the secondary coil N2i2, which opposes the original flux -- Lenz’s law This changing flux appears in the primary circuit as well; the sense of it is to reduce the emf in the primary... However, V1 is a voltage source. Therefore, there must be an increased current i1 (supplied by the voltage source) in the primary which produces a flux N1i1 which exactly cancels the flux produced by i2. N2 i1 i2 N1 Physics 1304: Lecture 18, Pg 32 Transformers with a Load iron With a resistive load in the secondary, the primary current is given by: V1 V2 N1 (primary) R N2 (secondary) V1 N12 Req R 2 i1 N2 This is the equivalent resistance seen by the source. Physics 1304: Lecture 18, Pg 33 Lecture 21, ACT 3 iron The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. 3A If V1 = 120 V, what is the potential drop across the resistor R ? V1 V2 N1 (primary) (a) 30 V 3B (b) 120 V N2 (secondary) (c) 480 V If 960 W are dissipated in the resistor R, what is the current in the primary ? (a) 8 A (b) 16 A (c) 32 A Physics 1304: Lecture 18, Pg 34 R Lecture 21, ACT 3 iron The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. 3A If V1 = 120 V, what is the potential drop across the resistor R ? V1 V2 N1 (primary) (a) 30 V (b) 120 V N2 (secondary) (c) 480 V The ratio of turns, (N2/N1) = (200/50) = 4 The ratio of secondary voltage to primary voltage is equal to the ratio of turns, (V2/V1) = (N2/N1) Therefore, (V2/V1) = 480 V Physics 1304: Lecture 18, Pg 35 R iron Lecture 21, ACT 3 The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. If V1 = 120 V, what is the potential drop across the resistor R ? V1 V2 N1 3A (primary) (a) 30 V N2 (secondary) (c) 480 V (b) 120 V The ratio of turns, (N2/N1) = (200/50) = 4 The ratio (V2/V1) = (N2/N1). Therefore, (V2/V1) = 480 V 3B If 960 W are dissipated in the resistor R, what is the current in the primary ? (a) 8 A (c) 32 A (b) 16 A Gee, we didn’t talk about power yet…. But, let’s assume energy is conserved…since it usually is around here Therefore, 960 W should be produced in the primary P1 = V1 I1 implies that 960W/120V = 8 A Physics 1304: Lecture 18, Pg 36 R Transformers with a Load iron • To get that last ACT, you had to use a general philosophy -- energy conservation. • An expression for the RMS power flow looks like this: Prms V1rmsi1rms V1 V2 N1 (primary) R N2 (secondary) N1 N2 V2 rms i 2 rms V2 rmsi 2 rms N2 N1 Note: This equation simply says that all power delivered by the generator is dissipated in the resistor ! Energy conservation!! Physics 1304: Lecture 18, Pg 37 Physics 1304: Lecture 18, Pg 38