Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits. Copyright 2009 Pearson Education, Inc.

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1 Chapter 30 Inductance, Electromagnetic Oscillations, and AC Circuits

2 30-1 Mutual Inductance Mutual inductance: a changing current in one coil will induce a current in a second coil: Coil 1 produces a flux at coil 2= F 21 Faraday s law says the Emf at coil 2 ~change in flux there Which is due to the current in 1 Total flux at coil 2 =N 2 F 21 a I 1 or N 2 F 21 =M 21 I 1 M 21 =Mutual inductance constant N 2 F 21 =M 21 I 1 /N 2 F 21 Faraday-> E 2emf = -N 2 df 21 /dt Unit of inductance, M: the henry, H: 1 H = 1 V s/a = 1 Ω s. =Weber/A=Tm 2 /A A transformer is an example of mutual inductance.

3 M depends on size,shape, N,iron core? Distance between coils, but not on I The reason we call it mutual inductance Is that the emf generated in coil 1 from 2 will be obtain by changes in the Current in 2 and M 12 =M 21 : Proof that They are equal is not shown di E M 2 and E M 1emf 2emf dt di dt 1

4 30-1 Mutual Inductance Example 30-1: Solenoid and coil. A long thin solenoid of length l (l)and cross-sectional area A contains N 1 closely packed turns of wire. Wrapped around it is an insulated coil of N 2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance. To get the flux inside we ask What is the B inside the solenoid Recall B =m 0 N 1 I 1 /l eq 28-4 F 21 BA m 0 N 1 I 1 A /l (l) M=M 21 = m 0 N 1 N 2 I 1 A /I

5 30-1 Mutual Inductance Conceptual Example 30-2: Reversing the coils. How would Example 30 1 change if the coil with turns was inside the solenoid rather than outside the solenoid? Same formula but A would be smaller!

6 30-2 Self-Inductance A changing current in a coil will also induce an emf in itself: Lenz s law applies and the EMF opposes the change in flux. I increases than Emf that forms will try to retard the increasing flux, So we define like M-> L=NF B /I &F B =LI/N Here, L is called the self-inductance:

7 30-2 Self-Inductance Example 30-3: Solenoid inductance. CLASS (a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length (l) and whose cross-sectional area is A. (b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm 2, and the solenoid is air filled. a. Solenoid B can be written B =m 0 ni n=n/l F B BA m 0 NIA/l and L=NF B /I = m 0 N 2 A/l b. We have N=100, l =5.0 x 10-2 m A= 3.0 x 10-5 m 2 m 0 =4p x 10-7 Tm/A plug in gives L=7.5mH

8 HW 30.07

9 Inductance some key points 1. Magnitude of L depends on geometry like last example 2. A changing current (a.c.) through a coil changes the flux. The induced EMF above opposes the change in flux (Lenz s Law). we note if df/dt is increasing ie. + then the emf is Or the induced emf opposes the current to oppose the increasing flux, that is resisting the incoming changing current. Or the emf is opposite the Driving voltage over the inductor. And df/dt is negative than emf is negative or supports The current and the Driving voltage since the flux is going down, so By Lenz s law the induced EMF needs to bolster up the flux.

10 30-2 Self-Inductance Conceptual Example 30-4: Direction of emf in inductor Lenz s law again. Note:+ CURRENT Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf? Loosely the Situation + ~ Feels like an Opposite Reaction To changes

11 30-2 Self-Inductance Example 30-5: Coaxial cable inductance. Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r 1 and the outer conductor has a radius r 2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions.

12 With the assumptions we are looking for the L between The two conductors.. That is for r with r 1 <r<r 2 To get L we need F since L=NF /I To get F We need B and in this region Since F=BA Ampere s law for r 2prB=m 0 I-> B=m 0 I/2pr The field B is circles around the inner conductor so we need To apply the definition of flux to the region in between the Two conducting cylinders. To apply the latter we Need to see the picture NEXT

13 Cross section of Coaxial cable L I B m l r 0 ln 2p r We have to sum up the flux in the hollow between The two conductors. The area element da=ldr so we have a df B =BdA=Bldr ie for A value of r B is a constant (in a circle radius r!) From before B=m 0 I/2pr Thus we have to integrate To get the total flux from r= r 1 to r 2 d B BdA Bldr r2 m0il dr m0il r r ln 2p 2p 1 B Or inductance per unit length L l m r 0 ln 2p r 2 1 r r 2

14 30-3 Energy Stored in a Magnetic Field Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor. An inductor stores energy analogous to a capacitor. Recall potential energy U =1/2CV 2 =1/2 e 0 A/d (Ed) 2 =1/2 e 0 E 2 Ad Ad= volume so u=energy density U/Vol or u =1/2 e 0 E 2 for electric field in general Energy in magnetic field =power going in=p=ie emf =ILdI/dt since Work being stored in the inductor is Then dw=pdt=lidi Or integrating W=U=1/2LI 2 (similar to) 1/2CV 2

15 Energy in the Magnetic field B Since we have U=1/2LI 2 for an inductor s storage of energy we can use a solenoid As a model analogous to using the capacitor storage for the Electric field. The solenoid gave us L=m 0 N 2 A/l and B=m 0 NI/l (solve for I) and solving For U=(1/2m 0 ) B 2 Al vol =Al So u=u/vol =energy density in the Magnetic field becomes analogous to the Electric field u=(1/2)e 0 E 2

16 HW 30.14

17 30-4 LR Circuits A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor.

18 30-4 LR Circuits / exp growth and decay The sum of potential differences around the loop gives Integrating gives the current as a function of time: I V di IR t dt L 1 V ln R IR V I t L. The time constant of an LR circuit is..

19 Start here Where is the current coming from in the Next slide?

20 30-4 LR Circuits If the circuit is then shorted across the battery, the current will gradually decay away: LdI/dt+RI=0 I di I t R dt ln L I I I R L t.

21 30-4 LR Circuits Example 30-6: An LR circuit. At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor s magnetic field?

22 At t-0 We see all the givens in the diagram (a) What is the current at t = 0?. t=0 I=0 not instantaneous (b) What is the time constant?. 220mH/30W 7.3ms (c) What is the maximum current? t->infinity Gives V 0/R= 12V/30W 0.40A then di/dt=0 (d) How long will it take the current to reach half its maximum possible value? I= V 0 /2R=V 0 /R(1-e-t/t) ½=1-e -t/t t=tln2=5.0ms (e) At this instant, at what rate is energy being delivered by the battery, P=I max/2 V=0.40A/2 x 12V=2.4W and (f) at what rate is energy being stored in the inductor s magnetic field? U=1/2LI 2 in the field but rate is du/dt=lidi/dt=? CLASS? Tricky Long way and short way?

23 HW 30.23

24 30-5 LC Circuits and Electromagnetic Oscillations An LC circuit is a charged capacitor shorted through an inductor.close S Just as circuit is closed By Kirchoffs loop rule Q/C LdI/dt=0 Since I=dQ/dt Then di/dt=d 2 Q/dt 2

25 30-5 LC Circuits and Electromagnetic Oscillations As we saw Summing the potential drops around the circuit gives a differential equation for Q: This is the same equation for simple harmonic motion, and has solutions. We see that dq/dt =-Q 0 sin(wt+ )w And d 2 Q/dt 2 = -Q 0 cos(wt+ )w 2 The differential equation becomes w 2 Q 0 cos(wt+ ) +(1/LC)Q 0 cos(wt+ ) 0. Or (-w 2 +1/LC)cos(wt+ )=0 to be zero for all t means w 2 +1/LC 0 thus

26 30-5 LC Circuits and Electromagnetic Oscillations The charge and current are both sinusoidal, but with different phases. We have. The current in the inductor is sinusoidal as well: Note the maximum Values of Q and I

27 30-5 LC Circuits and Electromagnetic Oscillations The total energy in the circuit is constant; it oscillates between the capacitor and the inductor:. I wq 0 sin(wt+ )->I 2 w 2 Q 02 sin(wt+ Recall w 2 1/LC and T=1/f=2p/w the total energy becomes

28 30-5 LC Circuits and Electromagnetic Oscillations Example 30-7: LC circuit. A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system. Straight forward solutions here: Given C=1200pF, V=500V, L=75mH. a. Q 0 =CV b. I max =wq 0 c. f=w/2p 1/{ (2p)(LC) 1/2 } And T=1/f d. U total =Q 02 /2C see text for numbers

29 30-6 LC Oscillations with Resistance (LRC Circuit) Any real (nonsuperconducting) circuit will have resistance. Kirchoffs loop LdI/dt IR+Q/C=0 And noting I=-dQ/dt ie charge on C is dropping Voltage drops Q/C ; -LdI/dt=Ld 2 Q/dt 2 (I=-dQ/dt) And -IR= RdQ/dt or we get K loop

30 30-6 LC Oscillations with Resistance voltage drops gave us (LRC Circuit) Same equation as md 2 x/dt 2 +bdx/dt +kx=0 ie. Eq This is the same equation for a damped Harmonic Oscillator found in chapter 14 section 14-7 see text Discussion page

31 30-6 LC Oscillations with Resistance (LRC Circuit). The detailed solutions to this equation are damped harmonic oscillations which we will not go into now. See chapter 14 is you are interested. The system will be underdamped for R 2 < 4L/C, and overdamped for R 2 > 4L/C. Critical damping will occur when R 2 = 4L/C. This figure shows the three cases of A: underdamping C: overdamping, and B: critical damping. For case A.

32 30-6 LC Oscillations with Resistance (LRC Circuit) Example 30-8: Damped oscillations. At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μf. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating? SEE TEXT SOLUTION!

33 30-7 AC Circuits with AC Source Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit. We examine an AC generator hooked to each one at a time Recall sect 25-7 AC voltage = V 0 sin(wt)=v 0 sin(2pft) Current (through a resistor for example I=V/R=V 0 /R sin(2pft) or I =I 0 sin(2pft) which is the general type of current produced by an AC generator. Recall to measure power we use P avg =I rms V rms Where I rms =0.707I 0 and V rms =0.707V =1/( 2) 1/2 see 25-7 So we can also writ I rms =I 0 /(2) 1/2 and V rms =V 0 /(2) 1/2

34 AC with a Resistor The current through a resistor is in phase with the voltage. AC Here we will assume I=I 0 cos wt Thus V=IR=I 0 Rcoswt=V 0 cos wt Both plotted to the right See 25-7 Energy in the Restitor converts to Heat P avg =I rms V rms = I 2 rms R =V 2 rms/r In phase!!

35 30-7 AC Source with Inductor The voltage across the inductor is given by LdI/dt K loop gives V-LdI/dt=0 or With I=I 0 coswt Or since sinwt=cos(wt+90 0 ) IE cos(a+b)=cosacosb-sinasinb. Therefore, the current through an inductor lags the voltage by 90. NOTE: V 0 =wli 0 is the Peak voltage

36 AC and Inductor We say Voltage is ahead of current By 90 0 or Current lags behind voltage by By 90 0 or Current reaches peak a quarter cycle After the voltage We note P=IV on the average is 0 so no heating of this ideal inductor. As we have see the inductor impedes The flow of charge since for a resistor V 0 =I 0 R with R being a measure of how a Resistor impedes charge: we define X L from V 0 =I o X L since V 0 =I 0 wl X L =wl=2pfl (inductive reactance!) Units are in OHMS higher X L the more it impedes, ie higher f &/or L

37 30-7 AC Source with Inductor In SUM: The voltage across the inductor is related to the current through it:. Ie divide both sides by (2) 1/2. V 0 /(2) 1/2 = I 0 /(2) 1/2 X L or V rms =I rms X L Since V rms = V 0 /(2) 1/2 and I rms = I 0 /(2) 1/2 definitions from sec 25-7 also For a resistor V=IR and R is for all times t. X L only defined For max and rms values The quantity X L is called the inductive reactance, and has units of ohms:

38 30-7 AC Circuits with AC Source Example 30-9: Reactance of a coil. A coil has a resistance R = 1.00 Ω and an inductance of H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied. Given R=1.00 W; L=0.300H a. V=120V DC b. V rms =120ac f=60hz a. X L =0 dc f=0 I=V/R=120A b. We know V rms =I rms X L and X L =2pf=113W I rms =V rms /X L =1.06A

39 HW 30.40

40 30-7 AC Source with Capacitor AC: charge flows to plates, one + and the other and when the Voltage reverses charge flows in the opposite directions. So current is continuous and as Before I=I 0 coswt =dq/dt K rule V-Q/C=0 or V=Q/C How does Q vary with time dq=i 0 coswtdt or integrating 0 to t Q=I 0 /w sinwt and V=Q/C=(I 0 /wc)sinwt V 0 = I 0 /wc or V=V 0 sinwt IE cos(a-b)=cosacosb+sinasinb-> cos(wt-90 0 )= Therefore, in a capacitor, the current leads the voltage by 90..

41 30-7 AC Source capacitor We sawthe voltage across the capacitor is related to the current through it:. As with the inductor we can write V 0 =I 0 X C or V rms =I rms X C The quantity X C is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:

42 30-7 AC Circuits with AC Source Example 30-10: Capacitor reactance. What is the rms current in the circuit shown if C = 1.0 μf and V rms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 10 5 Hz. Given: C=1.0mf, V rms =120V I rms =? a. And b. V rms =I rms X C X c =1/2pfC =2.7kW for f=60hz I rms =V rms /X C =44mA b. X C =0.27W for f=6 x 10 5 Hz I rms = V rms /X C =440A

43 30-7 AC Source + current (signal) filters This figure shows a high-pass filter (allows an ac signal to pass but blocks a dc voltage) and a low-pass filter (allows a dc voltage to be maintained but blocks higher-frequency fluctuations). Capacitively coupled circuit High f AC can go from A to B but not DC!= High Pass filter DC can go from A to B but high f AC is Grounded out! Low pass filter Some low f AC can get through Tweeter has C to impede low f Smaller f means higher X C =1/2pfC thus I is small for low f I =V/X C Woofer has L X L =2pfL to impede High f Ie higher X L lower currents I=V/X L

44 Analyzing the LRC series AC circuit is complicated, as the voltages are not in phase this means we cannot simply add them. Furthermore, the reactances depend on the frequency. V=V R +V L +V C 30-8 LRC Series AC Circuit We could solve this by the differential equation V=IR +LdI/dt+Q/C-> V =Ld 2 Q/dt 2 +RdQ/dt+Q/C Difficult to do and less insight than the PHASOR DIAGRAMS TO FOLLOW Kirchhoff s rule does Apply but only for some Time t, all the voltages Do not reach their peak At the same time CURRENT IS THE SAME ON ALL ELEMENTS I=I 0 coswt at the same time not VOLTAGES

45 30-8 LRC Series AC Circuit We calculate the voltage (and current) using what are called phasors these are vectors representing the individual voltages. RECALL Current and voltage on the Resistor V R are in phase (reach max at the same time) V L is 90 0 ahead of I and V R while V C is 90 0 behind Here, at t = 0, the current and voltage are both at a maximum. As time goes on, the phasors will rotate counterclockwise. At an angular rate of w. So at time t they all rotate an amount wt!

46 30-8 LRC Series AC Circuit Some time t later, the phasors have rotated. Note the magnitude of the vectors are the peak values as listed.(also rms versions exist as before)

47 30-8 LRC Series AC Circuit The voltages across each device are given by the x-component of each, and the current by its x- component. The current is the same throughout the circuit. IE. We had The V s as functions that had cosine which Is the x component in this phasor diagram form. See the equations below for the V s At time t rotation by wt

48 30-8 LRC Series AC Circuit The Total VOLTAGE at time t THE TOTAL MAX VOLTAGE V 0 AT ANY TIME t (Vectors rotated wt) IS THE MAGNITUDE OF THE VECTOR SUM OF THE PHASOR VOLTAGES. Note sum of V L0 - V C0 The final Sum with Magnitude V 0 is at some angle to V R0 and current Vector magnitude I 0 V L0 V C0 So total voltage at any time t Is x component of vector V 0 V=V 0 cos(wt+ ) see fig above

49 30-8 LRC Series AC Circuit The Total IMPEDANCE As with each of the L,R and C we define the impedance Z Of the total voltage as V rms =I rms Z As with the other voltages we have V 0 =I 0 Z We now use the Phasor diagram to get the magnitude V 0 from the Other voltages NOTE Vectors V R0 and V L0 -V C0 are at rt angles and we Can use the Pythagorean theorem to get magnitude of vector V 0 the hypotenuse of the right triangle. V VR0 + ( VL0 VC 0) Recall V R0 =I 0 R V L0 =I 0 X L V C0 =I 0 X C V I0 R + ( X L X C ) Z R 2 + ( X ) L X C 2 R ( wl ) wc 2

50 30-8 LRC Series AC Circuit In Sum: We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by The phase angle between the voltage and the current is derived From the diagram below and is given by The factor cos φ is called the power factor of the circuit. For pure R cos 1,for only L or only C 90 in general P avg =I 2 rmsr=i 2 rmszcos I rms V rms cos

51 30-8 LRC Series AC Circuit or The factor cos φ is called the power factor of the circuit.

52 Example 30-11: LRC circuit LRC Series AC Circuit Suppose R = 25.0 Ω, L = 30.0 mh, and C = 12.0 μf, and they are connected in series to a 90.0-V ac (rms) 500-Hz source. Calculate (a) the current in the circuit, (b) the voltmeter readings (rms) across each element, (c) the phase angle, and (d) the power dissipated in the circuit. STUDY text for more details on this problem a. To get I rms we need Z since I rms = V rms /Z so we need the impedances; R=25.0W Is known, at f=500hz X L =2pfL=94.2W C C =1/2pfC =26.5W Z=[R 2 +(X L -X C ) 2 ] 1/2 =72.2W so I rms =V rms /Z = 1.25A b. The rms voltagesv Rrms =I rms R =31.2V V Lrms =I rms X L =118V and V Crms =I rms X C =33.1V c. Cos =R/Z= and is positive see text d. P avg = I rms V rms cos =39.0W

53 30-9 Resonance in AC Circuits The rms current in an ac circuit is Clearly, I rms depends on the frequency. WHEN is the I rms What w? a Maximum or for wl-1/wc =0 that is X L =X C

54 30-9 Resonance in AC Circuits We see that I rms will be a maximum when X C = X L ; the frequency at which this occurs is f 0 = ω 0 /2π is called the resonant frequency.

55 HW 30.59

56 Skip the rest

57 30-10 Impedance Matching When one electrical circuit is connected to another, maximum power is transmitted when the output impedance of the first equals the input impedance of the second. The power delivered to the circuit will be a minimum when dp/dt = 0; this occurs when R 1 = R 2.

58 30-11 Three-Phase AC Transmission lines usually transmit threephase ac power, with the phases being separated by 120. This makes the power flow much smoother than if a single phase were used.

59 30-11 Three-Phase AC Example 30-12: Three-phase circuit. In a three-phase circuit, 266 V rms exists between line 1 and ground. What is the rms voltage between lines 2 and 3?

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