Physics 1304: Lecture 18, Pg 1 AC Circuits 

Physics 1304: Lecture 18, Pg 2 Lecture Outline l Driven Series LCR Circuit: General solution Resonance condition »Resonant frequency »“Sharpness of resonance” = Q Power considerations »Power factor depends on impedance l Transformers Voltage changes Faraday’s Law in action gives induced primary current. Power considerations Text Reference: Chapter

Physics 1304: Lecture 18, Pg 3 Phasors 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 . Recall uniform circular motion: The projections of r (on the vertical y axis) execute sinusoidal oscillation.    R: V in phase with i C: V lags i by 90  L: V leads i by 90   x y y

Physics 1304: Lecture 18, Pg 4 Phasors for L,C,R i tt  i tt  i tt   Suppose: t i 0 0 i i 0

Physics 1304: Lecture 18, Pg 5 Series LCR AC Circuit Back to the original problem: the loop equation gives: Here all unknowns, (i m,  ), must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be:  m sin  t. To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram. L C   R Assume a solution of the form:

Phasors: LCR Assume: l From these equations, we can draw the phasor diagram to the right.   L C   R Given:  l 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.

Physics 1304: Lecture 18, Pg 7 Phasors: LCR The phasor diagram has been relabeled in terms of the reactances defined from:  L C   R The unknowns (i m,  ) can now be solved for graphically since the vector sum of the voltages V L + V C + V R must sum to the driving emf .

Physics 1304: Lecture 18, Pg 8 Lecture 20, ACT 3 l A driven RLC circuit is connected as shown.  For what frequencies  of the voltage source is the current through the resistor largest? (a)  small (b)  large (c) 

Conceptual Question l A driven RLC circuit is connected as shown.  For what frequencies  of the voltage source is the current through the resistor largest? (a)  small (b)  large (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 = X L =  L. The reactance of a capacitor = X C = 1/(  C). Therefore, in the low frequency limit, X L  0 and X C  . Therefore, as  0, the current will flow mostly through the inductor; the current through the capacitor approaches 0. in the high frequency limit, X L  and X C  0 . Therefore, as , the current will flow mostly through the capacitor, approaching a maximum i max =  /R.

Physics 1304: Lecture 18, Pg 10 Phasors:LCR  

Phasors:Tips This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis. y x  imRimR imXLimXL imXCimXC mm “Full Phasor Diagram” From this diagram, we can also create a triangle which allows us to calculate the impedance Z: “ Impedance Triangle” 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).

Physics 1304: Lecture 18, Pg 12 Phasors:LCR We have found the general solution for the driven LCR circuit: R XLXL XCXC   the loop eqn X L - X C  imRimR imXLimXL imXCimXC mm 

Lagging & Leading The phase  between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances.  R XLXL XCXC  X L > X C  > 0 current LAGS applied voltage  R XLXL XCXC  X L < X C  < 0 current LEADS applied voltage X L = X C  = 0 current IN PHASE applied voltage R XLXL XCXC 

Physics 1304: Lecture 18, Pg 14 Conceptual Question The series LCR circuit shown is driven by a generator with voltage  =  m sin  t. The time dependence of the current i which flows in the circuit is shown in the plot.  How should  be changed to bring the current and driving voltage into phase? (a) increase  (b) decrease  (c) impossible Which of the following phasors represents the current i at t=0? 1B 1A (a) (b) (c) i i i

Physics 1304: Lecture 18, Pg 15 Lecture 21, ACT 1 The series LCR circuit shown is driven by a generator with voltage  =  m sin  t. The time dependence of the current i which flows in the circuit is shown in the plot.  How should  be changed to bring the current and driving voltage into phase? (a) increase  (b) decrease  (c) impossible 1A From the plot, it is clear that the current is LEADING the applied voltage.  i  XCXC XLXL Therefore, the phasor diagram must look like this: Therefore, To bring the current into phase with the applied voltage, we need to increase X L and decrease X C. Increasing  will do both!!

Physics 1304: Lecture 18, Pg 16 Lecture 21, ACT 1 The series LCR circuit shown is driven by a generator with voltage  =  m sin  t. The time dependence of the current i which flows in the circuit is shown in the plot.  How should  be changed to bring the current and driving voltage into phase? (a) increase  (b) decrease  (c) impossible Which of the following phasors represents the current i at t=0? (a) (b) (c) 1A1B i i 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 17 Resonance For fixed R,C,L the current i m will be a maximum at the resonant frequency  0 which makes the impedance Z purely resistive. the frequency at which this condition is obtained is given from:  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! ie: reaches a maximum when:

Physics 1304: Lecture 18, Pg 18 Resonance The current in an LCR circuit depends on the values of the elements and on the driving frequency through the relation imim 0 0  o  R=R o R=2R o Suppose you plot the current versus , the source voltage frequency, you would get:

Physics 1304: Lecture 18, Pg 19 Power in LCR Circuit The power supplied by the emf in a series LCR circuit depends on the frequency . It will turn out that the maximum power is supplied at the resonant frequency  0. The instantaneous power (for some frequency,  ) delivered at time t is given by: 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(  t-  ) term. Remember what this stands for

Physics 1304: Lecture 18, Pg 20 Power in LCR Circuit Expanding, Taking the averages, Generally: sin  tcos  t tt 0  0 +1 sin 2  t tt 0  0 +1 Putting it all back together again, (Product of even and odd function = 0) 0 1/2

Physics 1304: Lecture 18, Pg 21 Power in LCR Circuit l This result is often rewritten in terms of rms values:  Power delivered depends on the phase,  the  “power factor” phase depends on the values of L, C, R, and  l therefore...

Physics 1304: Lecture 18, Pg 22 Power in RLC Power, as well as current, peaks at  =  0. The sharpness of the resonance depends on the values of the components. l Recall: Therefore, We can write this in the following manner (you can do the algebra): …introducing the curious factors Q and x...

Physics 1304: Lecture 18, Pg 23 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 U max is max energy stored in the system and  U is the energy dissipated in one cycle For RLC circuit, U max is (e.g.) And losses only in R, namely period This gives And for completeness, note

Physics 1304: Lecture 18, Pg 24 Power in RLC 0 0  o  R=R o R=2R o Q=3 FWHM For Q > few,

Physics 1304: Lecture 18, Pg 25 Conceptual Question 2 Consider the two circuits shown where C II = 2 C I.  What is the relation between the quality factors, Q I and Q II, of the two circuits? (a) Q II < Q I (b) Q II = Q I (c) Q II > Q I (a) P II < P I (b) P II = P I (c) P II > P I What is the relation between P I and P II, the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency? 2B 2A I II

Physics 1304: Lecture 18, Pg 26 Lecture 21, ACT 2 Consider the two circuits shown where C II = 2 C I.  What is the relation between the quality factors, Q I and Q II, of the two circuits? (a) Q II < Q I (b) Q II = Q I (c) Q II > Q I 2A I II We know the definition of Q: At first glance, it looks like Q is independent of C. At second glance, we see this cannot be true, since the resonant frequency  o depends on C! Doubling C decreases  o by sqrt(2)! Doubling C decreases Q by sqrt(2)!  Doubling C increases the width of the resonance!

Physics 1304: Lecture 18, Pg 27 Lecture 21, ACT 2 Consider the two circuits shown where C II = 2 C I.  What is the relation between the quality factors, Q I and Q II, of the two circuits? (a) Q II < Q I (b) Q II = Q I (c) Q II > Q I What is the relation between P I and P II, the power delivered by the generator to the circuit when each circuit is operated at its resonant frequency? (a) P II < P I (b) P II = P I (c) P II > P I 2A2B I II 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 28 Power Transmission l How can we transport power from power stations to homes? Why do we use “high tension” lines? çAt home, the AC voltage obtained from outlets in this country is 120V at 60Hz. çTransmission of power is typically at very high voltages ( eg ~500 kV) çTransformers are used to raise the voltage for transmission and lower the voltage for use. We’ll describe these next. l But why? çCalculate ohmic losses in the transmission lines: çDefine efficiency of transmission: – Note for fixed input power and line resistance, the inefficiency  1/V 2 Example: Quebec to Montreal 1000 km  R= 220   suppose P in = 500 MW With V in =735kV,  = 80%. The efficiency goes to zero quickly if V in were lowered! Keep R small Make V in big

Physics 1304: Lecture 18, Pg 29

Physics 1304: Lecture 18, Pg 30 Transformers   (primary) (secondary) AC voltages can be stepped up or stepped down by the use of transformers. iron l The AC current in the primary circuit creates a time-varying magnetic field in the iron 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. V2V2 V1V1 l This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils.

Physics 1304: Lecture 18, Pg 31 Ideal Transformers (no load) l 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: 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: Therefore, N 2 > N 1  secondary V 2 is larger than primary V 1 (step-up) N 1 > N 2  secondary V 2 is smaller than primary V 1 (step-down) Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small! No resistance losses All flux contained in iron Nothing connected on secondary

Physics 1304: Lecture 18, Pg 32 Ideal Transformers l What happens when we connect a resistive load to the secondary coil? çFlux produced by primary coil induces an emf in secondary R with a Load çThis changing flux appears in the primary circuit as well; the sense of it is to reduce the emf in the primary... çHowever, V 1 is a voltage source.  Therefore, there must be an increased current i 1 (supplied by the voltage source) in the primary which produces a flux  N 1 i 1 which exactly cancels the flux produced by i 2. çemf in secondary produces current i 2  This current produces a flux in the secondary coil  N 2 i 2  which opposes the original flux -- Lenz’s law

Physics 1304: Lecture 18, Pg 33 Transformers with a Load l With a resistive load in the secondary, the primary current is given by: R This is the equivalent resistance seen by the source.

Physics 1304: Lecture 18, Pg 34 Lecture 21, ACT 3 l 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. çIf V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 3A R 3B 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 35 Lecture 21, ACT 3 l 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. çIf V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 3A R The ratio of turns, (N 2 /N 1 ) = (200/50) = 4 The ratio of secondary voltage to primary voltage is equal to the ratio of turns, (V 2 /V 1 ) = (N 2 /N 1 ) Therefore, (V 2 /V 1 ) = 480 V

Physics 1304: Lecture 18, Pg 36 Lecture 21, ACT 3 l 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. çIf V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 3A R The ratio of turns, (N 2 /N 1 ) = (200/50) = 4 The ratio (V 2 /V 1 ) = (N 2 /N 1 ). Therefore, (V 2 /V 1 ) = 480 V çIf 960 W are dissipated in the resistor R, what is the current in the primary ? (a) 8 A (b) 16 A (c) 32 A 3B 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 P 1 = V 1 I 1 implies that 960W/120V = 8 A

Physics 1304: Lecture 18, Pg 37 Transformers with a Load R To get that last ACT, you had to use a general philosophy -- energy conservation. An expression for the RMS power flow looks like this: Note: This equation simply says that all power delivered by the generator is dissipated in the resistor ! Energy conservation!!

Physics 1304: Lecture 18, Pg 38