Chapter 12 Inductors and AC Circuits

Transcription

1 hapter Inductors and A rcuts awrence B. ees 6. You may make a sngle copy of ths document for personal use wthout wrtten permsson. Hstory oncepts from prevous physcs and math courses that you wll need for ths chapter: electrc feld see emnders magnetc feld see emnders threads see emnders stubs see emnders When you fnsh ths chapter, you wll be able to: descrbe explan explan. Introducton Intro

2 . An Inductor n a D rcut A col of wre placed n a crcut s called an nductor. A number of nductors are llustrated n Fg... Whle nductors can take on a varety of forms, the smplest nductor s just a solenod. Fgure.. A varety of nductors. et s consder an nductor n the smple seres D crcut llustrated n Fg... Fgure.. A D crcut wth an nductor. We know that the current passng through the nductor wll create a magnetc feld and that the nductor wll have a small resstance, but other than that, the nductor s smply a length of wre. If we gnore ts resstance, there wll be a current gven by Ohm s aw: /. Now let s take a magnet and brng near the nductor, as shown n Fg..3. N Fgure.3. A D crcut wth an nductor and a magnet.

3 There wll be feld lnes from the magnet comng down nto the nductor. Snce the feld s ncreasng, the nductor wll create an nduced magnetc feld that wll pont upward. Ths wll cause a current to low n the crcut. (Dependng on your pont of vew, you may see the cols wound one way or the other. But, f we assume we re lookng somewhat downward on the cols, current wll flow from the top of the nductor toward the postve termnal of the battery.) If the south pole of the magnet s brought near the nductor n the same fashon, current wll flow n the opposte drecton. If the magnet s statonary, t doesn t affect the crcut at all. In other words, an nductor wll change a crcut f there s a changng magnetc feld passng through t, so that an nduced current wll be produced. Thngs to remember: In a D crcut, an nductor normally behaves just as a long segment of wre. If there s a changng magnetc feld n an nductor, current wll be nduced n the crcut... An Inductor n an A rcut Now let s attach the nductor and resstor to an A power supply, as shown n Fg..4. Because the power supply s changng snusodally, the magnetc feld produced by the nductor also vares snusodally. And snce the magnetc feld n the nductor vares n tme, an nduced EMF s produced n the col. Ths process s called self nducton. ε Fgure.4. An A crcut wth an nductor. The easest way of descrbng self nducton s to fnd the EMF produced by the nductor. We wll gnore any resstance n the nductor s cols, so the voltage across the nductor s just the nduced EMF. Knowng the EMF wll allow us to calculate how the current n the crcut behaves n tme. Now let s use faraday s aw to see just what n nductor does n a crcut. et s assume we have a solenodal nductor wth a cross sectonal area A and N turns of wre over a length l. We defne n N/l to be the number of turns per unt length as we dd n hapter 8. Puttng ths all together, we get: 3

4 B µ n Φ B NΦ BA µ na B dφ ε N Nµ na µ n la B d µ n la As we can see, the EMF equals some quanttes that depend only on the sze and shape of the solenod and on the rate at whch current changes n crcut. Ths sn t too surprsng, as the rate flux changes n the solenod must be ted to the rate current changes n the crcut. For convenence, we lump all of the geometrcal factors nto one term called the nductance of the nductor. The nductance of a solenodal nductor s, then (. Inductance of a solenod) n la. In general, nductance can be defned from the relatonshp (.) where: µ Inductance d s the voltage across the nductor measured n volts (). s the nductance of the nductor measured n henres (H). I s the current through the nductor measured n amperes (A). As wth many equatons, the sgn of the nductance equaton can be confusng. The rule s that voltage s postve f t tends to drve current n the drecton current s already flowng. Ths s just a consequence of enz s aw. If current n the crcut s decreasng, the voltage of the nductor pushes () charge n the drecton of the current to ncrease the current to oppose ts decrease. If the current s ncreasng, the nductor pushes charge aganst the current n order to reduce the current and oppose ts ncrease. Note the smlarty between the equatons for voltage across a resstor, a capactor, and an nductor: dq q d d q 4

5 It may not take a great deal of magnaton to beleve that nductance adds lke resstance rather than capactance n seres and parallel combnatons. To be more rgorous, we can see that the voltages of two nductors n parallel must be the same and that d/ as well as must the same for two nductors n seres. In seres: In parallel: d d d d d d Thngs to remember: d Inductance s defned by the equaton: Inductance depends on the geometry of the nductor, not on the current, etc. Inductance adds lke resstance n seres and parallel combnatons..3 Energy n Inductors and Magnetc Felds et s take a very smple crcut consstng only of an nductor and an A power supply, as llustrated n Fg..5. 5

6 ε Fgure.5. An A crcut wth an nductor. Not worryng too much about sgns, we know that the voltage across the power supply must equal the voltage across the nductor: d ε. and the power provded by the power supply must be d P ε d Snce power s the rate of change of energy, and the only energy s the potental energy of the nductor, we must conclude:. (.3 Energy stored n an nductor) U. If we take the specal case of a solenodal nductor, we can wrte the energy as: U µ n Al U µ n Al 6

7 l Snce the magnetc feld s B µ n and the volume of the solenod s vol A, we can wrte the energy densty n the nductor as : (.4 Energy densty n a magnetc feld) u U vol B µ. Note that these equaton bear strong resemblance to the equatons for energy stored n a capactor and the energy densty of the electrc feld: U U vol, u ε E. Thngs to remember: The energy stored n an nductor s U. U The energy densty of a magnetc feld s u B. vol µ.4. rcuts et s return agan to a smple crcut contanng a battery, a resstor, and an nductor all connected n seres; however, now let s add a swtch to the crcut. Fgure.6. A seres crcut. Intally there s no current and no magnetc feld n the nductor. As soon as the swtch s closed, current starts flowng from the battery and magnetc starts beng produced n the nductor. The nductor tres to oppose change n the system. That s, t produces an nduced current that opposes the current from the battery and opposes the creaton of a magnetc feld n the nductor. et s see f we can fnd the current as a functon of tme n the crcut. To do ths, we apply Krchoff s oop aw to the crcut. One thng we need to be careful of s to get the sgns 7

8 8 correct n the loop equaton. Just after the swtch s closed, we know the EMF of the nductor opposes that of the battery, so we can put and sgns on the crcut elements as n Fg..7. Fgure.7. An crcut wth and sgns of the voltages added. Now, we can wrte out the loop equaton: d We need to remove the absolute value removed. To do ths, we have to ask whether the current s ncreasng n tme or decreasng n tme. It may not be obvous, but t turns out that the ntal current s zero and t rses to a fnal value of /, the current that would flow f there were no nductor. Hence: > d d We mght guess (and snce I know the answer already, the guess s correct) that the soluton to ths equaton would be smlar to the equaton of a chargng capactor. So let s try: ( ) ( ) e e e e e e e t t t t t t t t τ τ τ τ τ τ τ τ τ τ τ ) ( / / / / / / /

9 We see then that the current obeys an exponental equaton, much as s the charge n a chargng capactor; however the tme constant s now τ /. If the tme constant s large, t takes a long tme for the current to reach ts maxmum value. It makes sense that a large nductor would be better able to oppose the battery s current and that t would take a relatvely large tme for the current to ncrease to ts fnal value. The dependence of the tme constant on resstance may be a lttle harder to understand ntutvely. However, we thnk of the nductor as creatng an nduced current that contnues to flow opposte the battery. The smaller the resstance, the longer tme t takes for the nduced current to de out leavng only the current of the battery. There s one more type of crcut we can consder, that shown n Fg..8 below. Fgure.8. A varaton of the crcut. In ths crcut, as swtch s ntally n poston and a steady-state current / s flowng through the nductor. Then at tme t, the swtch s moved to poston, removng the battery from the crcut. The nductor tres to keep current flowng through the crcut as long as t can. The nductor then acts lke a battery pushng current around the crcut n the orgnal drecton, as shown n Fg..9. Fgure.9. The crcut of Fg..8 wth the swtch flpped to poston. The loop equaton for the crcut wth the swtch n poston s d. 9

10 The current s now decreasng, so to remove the absolute values we need to do the followng: d < d Ths tme we mght expect the current to be an exponentally decreasng functon of tme. ( t) e e τ τ t / τ t / τ e t / τ Just as wth crcuts, the same tme constant governs both Thngs to remember: In crcuts, current changes n tme ether by The tme constant s τ..5. rcuts and Phases t /τ e t /τ e /τ or by e t. and /τ e t equatons. The next thng we can do s consder what happens when we charge a capactor to a voltage and connect t n seres wth an nductor, as shown n Fg..9. Fgure.9. An crcut. At frst glance, t mght not seem that ths crcut would be much dfferent than a crcut wth a battery; however, the current changes as the capactor dscharges, so an nduced s produced on the nductor. We can qualtatvely guess what should happen wth ths crcut ether by consder the current and charge or by consderng energy. Snce both are nstructve n dfferent ways, we ll look at each n turn.

11 Frst, let s thnk about charge and current. We begn by connectng a charged capactor to the nductor. As tme progresses, the crcut goes through the followng stages: ) The capactor begns to dscharge, but the nductor opposes the flow of current. ) The current from the capactor ncreases as the capactor dscharges. 3) The capactor s fully dscharged but current s flowng through the crcut. The nductor keeps current n the same drecton. 4) The capactor begns to charge n the opposte drecton. 5) The capactor has charge Q n the opposte drecton and current ceases to flow. Fgure.. Successve stages of the capactor dschargng and chargng agan n the opposte drecton. Note that snce there s no resstance n ths crcut, the charge oscllates back and forth ndefntely. Now let s thnk about the same process n terms of energy. We ll follow the same stages as before: ) All the energy s n the electrc feld of the capactor. ) As current begns to flow, some of the capactor s energy s transferred to the magnetc feld of the nductor. 3) The capactor s fully dscharged and all the energy n the system s n the nductor. Ths mples that the current reaches a maxmum at ths pont. 4) The capactor begns rechargng and some of the energy s transferred back to the capactor. 5) All the energy goes back to the electrc feld of the nductor. At ths pont we could guess that the soluton to the problem must be somethng lke q( t) snωt, but we don t know whatω s. et s see f we can apply Krchoff s oop aw as we dd before. The trcky part s to get the sgns rght. To do that, all we have to do s fnd any tme where the sgns are all consstent and wrte down the equaton at that tme. The sgns at other tmes wll be consstent wth that tme. At ths pont, we need to establsh a sgn conventon for voltages. The reason we need to do ths s that sgns can quckly become confusng when the drecton of the current s constantly changng. Our bass for the sgn conventon s that we want to use Ohm s aw the same way n A crcuts as n D

12 crcuts. Note that the voltage across a resstor s taken to be postve when the voltage opposes current flow. We then use ths same conventon for capactors and nductors: Sgn conventon for voltages n A crcuts. Defne a postve sense for current. oltage across a resstor, capactor, or nductor s postve f t pushes current n the negatve drecton and negatve f t pushes current n the postve drecton. There are some very mportant consequences to ths sgn conventon, so we should take a lttle whle to go over these. et s consder the case of resstors, capactors, and nductors ndvdually. For resstors, the sgn conventon s qute smple. When the current s postve, the resstor pushes current n the negatve drecton, so the voltage s postve. When the current s negatve, t pushes current n the postve drecton, so the voltage s negatve. We can wrte. (t) (t) t Fgure.. urrent through and the voltage across a resstor n an A crcut. We say that the voltage across a resstor s n phase wth the current through the resstor. That s, the peaks and valleys of the two functons occur at the same tme. For nductors, we know the nduced voltage wll oppose the change n current. When the current s postve and ncreasng, the nduced EMF wll oppose the ncrease, pushng charge n the negatve drecton. et s consder the sgns for the voltage n each of the possble combnatons of postve and negatve current and ncreasng and decreasng current. The one trcky part of the table s to remember that f the current s negatve and d/ s postve, the current s gettng more postve meanng that the magntude of the current s droppng. Smlarly f s negatve and d/ s also negatve, the current s gettng smaller (more negatve) so ts magntude s gettng larger. Be sure you thnk about that a bt before you go on.

13 urrent d urrent Magntude Induced urrent Drecton Inductor oltage postve postve ncreasng negatve postve postve negatve decreasng postve negatve negatve postve decreasng negatve postve negatve negatve ncreasng postve negatve The mportant thng to note about ths table s that the nductor voltage s postve when the slope of the current s postve and the nductor voltage s negatve when the slope of the current s negatve. Because of ths, we can wrte: d (.5 For the A crcut sgn conventon). Ths equaton s qute confusng because the sgn seems to be reversed from our earler result, but t the consequence of our conventon that postve voltage causes current to flow n the negatve drecton. (t) 9 (t) t Fgure.. urrent through and the voltage across an nductor n an A crcut. Thnk About It ook at Fg.. and convnce yourself that the voltage across the nductor s postve whenever the slope of the current s postve. 3

14 In ths case we say that the voltage across an nductor leads the current by 9, or that the phase angle s 9. Note that the phase angle s the angular dfference between the maxmum voltage and the maxmum current, as shown by the arrow n Fg... apactors are just a lttle harder. Wth D crcuts, we always thought of the charge on a capactor as postve. Wth A crcuts we can no longer do that. By our sgn conventon, we must take the charge on a capactor to be postve when t tends to drve charge aganst the current. But what we really want to know s what that means n terms of current. If the charge on a capactor s postve, the capactor voltage s postve. If current s decreasng, then current must be flowng n the negatve drecton. et s make a table of all such results: apactor harge apactor oltage urrent Drecton dq d or postve postve negatve negatve postve postve postve postve negatve negatve negatve negatve negatve negatve postve postve dq Ths table tells us that. But snce the charge and voltage have the same sgn, we see that whenever the voltage has a negatve slope, the current must be negatve. Smlarly, whenever the voltage has a postve slope, the current must be postve. These results lead to the graph shown n Fg..3. (t) (t) 9 t Fgure.3. urrent through and the voltage across a capactor n an A crcut. In ths case we say that the voltage across a capactor lags the current by 9, or that the phase angle s 9. Agan, the phase angle s the angular dfference between the maxmum voltage and the maxmum current. 4

15 Now let s return to our smple crcut. For smplcty, we take a tme shortly after current begns to flow from the capactor, as shown n Fg... et s take postve current to be n the clockwse drecton. The charge on the capactor tends to drve current n the postve drecton, so the capactor charge and voltage wll both be negatve ntally (by our confusng sgn conventon). Snce the current s gettng larger n magntude, the EMF on the nductor wll drve the current n the negatve drecton, and hence be postve. goes ths way. Fgure.4. An crcut shortly after the capactor begns dschargng. In ths case, we can wrte the loop equaton as: c q d We need to get rd of the absolute value notaton. In Fg.. the charge on the capactor s negatve and ncreasng (gettng less negatve) and the current s postve and ncreasng. Thus, we have: q < To keep all the sgns consstent, we must have: dq > > d > dq d d q > q d q 5

16 Ths then leads to the dfferental equaton for crcuts: d q (.6) q As you may easly verfy, the soluton to ths equaton must be a combnaton of snes and cosnes. Snce we want charge to be a mnmum at tme t, we choose q( t) cosωt. Then: ω cosωt cosωt ω Ths tells us the frequency at whch the charge on the capactor, the current n the crcut, the energy n the crcut, the voltage on the capactor, the voltage on the nductor everythng n the crcut oscllates. Once we know the charge on the capactor, we can fnd anythng we want. ω q( t) q( t) dq ω snωt d f π cosωt cosωt snωt ω cosωt cosωt In Fg..5, we plot the current and the voltages across the capactor and the nductor as a functon of tme. 6

17 (t) (t) (t) t Fgure.5. urrent and voltages n an rcut. Note how the capactor s voltage peaks after the current, but the nductor s voltage peaks before the current, just as we had suggested above. A convenent way to remember these phase relatonshp s to use the mnemonc devce below. EI the IE man In an nductor, the EMF ( ) leads the current by 9. In a capactor, the current leads the EMF ( ) by 9. Thngs to remember: An crcut oscllates at an angular frequency ω. Energy s transferred back and forth between the electrc feld of the capactor and the magnetc feld of the nductor as the crcut oscllates. EI the IE man and ts meanng. Know how to derve Krchoff s loop equaton, Eq.6. You may be a bt cavaler about sgns. 7

18 .6. Phasors The word phasor s short for phase vector. It s a way to represent a sne or cosne functon graphcally. If you have taken Physcs 3, you may have used phasors to analyze the nterference of lght through slts. In ths course, phasors are very helpful n vsualzng and analyzng A crcuts. In A crcuts, currents and voltages are all snusodal functons. The general mathematcal form of such a functon s: where A t) A sn t ( ( ω φ ) A (t) s the value of A (generally a current or voltage) at tme t. A s the maxmum value of A. ω s the angular frequency n rad/s. φ s the phase angle. A phasor s a vector whch has length A and s drected at an angle θ ωt φ to the x axs, as shown n Fg..6. A sn( ω t φ ) A θ ω t φ Fgure.7. A phasor representng the functon A t) A sn( ω t φ ) (. As any other vector the phasor A r can be expressed n terms of components: r A A cos y ( ω t φ) xˆ A cos( ωt φ) ˆ. From ths we can see the relaton between the phasor and the functon s that the functon s just the y component of the phasor. The angle of such a phasor changes n tme, so as tme progresses, the phasor rotates about the orgn at an angular velocty ω. Ths s llustrated n Fg..8 or n an anmated verson on the course webste at 8

19 Fgure.8. A phasor rotatng as a functon of tme. et s assume that we have an A crcut wth a current gven by the equaton ( t) sn( ωt). Assume that both and ω are known. We wsh to then construct phasors for the voltages across resstors, nductors, and capactors. A. esstors In order to construct a phasor for the voltage across any crcut element, we need to know the magntude and the angle of the phasor. Ths s easy for a resstor, as we only need Ohm s aw and the knowledge that the phase angle for a resstor s. We then have; where ( r) sn( ω t) s the maxmum voltage across the resstor n volts (). s maxmum current through the resstor n amperes (A). s the resstance n ohms ( Ω ). We can draw the phasor when the current s at any angle. For smplcty, let s draw t for tme t. r r Fgure.9. The voltage and current phasors for a resstor. An anmated verson of ths can be found at 9

20 on the course webste at B. Inductors Now we can go through the same process for nductors. We know the angle of the voltage phasor as t s 9 ahead of the current. The magntude of the phasor comes from the relatonshp: d ω ω cos( ω t ) Note that ths equaton looks a lot lke Ohm s aw. Even though an nductor has no resstance and no energy loss, the nductance offers an effectve resstance to lmt the flow of current through a crcut. We call ths effectve resstance the nductve reactance and wrte t as: (.5 Inductve eactance) X ω X It s reasonable that the effectve resstance for an nductor s ω snce hgher frequences and larger nductance both lead to larger nduced currents. Then we can draw the phasors for an nductor as follows: r r Fgure.. The voltage and current phasors for an nductor. An anmated verson of ths can be found at on the course webste at

21 . apactors Fnally, we come to capactors. The angle of the voltage phasor as t s 9 behnd the current. Snce dq/, the magntude of the phasor comes from the relatonshp: q ω ω cos( ω t) As wth an nductor, a capactor has no resstance and no energy loss, but t does produce an effectve resstance n a crcut. We call ths effectve resstance the caapctve reactance and wrte t as: (.6 apactve eactance) X ω X To understand ths relatonshp, we should remember that a capactor offers resstance n a crcut when t charges and opposes current flows. The larger the charge the capactor develops, the larger ts effectve resstance n a crcut. If frequency s very hgh, a capactor has lttle chance to charge before the current reverses drecton, so t offers lttle resstance to current. Smlarly, f the capactance s large, a large amount of charge can collect on a capactor plates wthout ncreasng the voltage much. Then we can draw the phasors for an nductor as follows: r r Fgure.. The voltage and current phasors for an nductor. An anmated verson of ths can be found at on the course webste at

22 Thngs to remember: A sne wave can be represented by the projecton of a phasor onto the y axs. The length of a phasor s the ampltude of the sne wave. The angle of a phasor wth respect to the x axs s the anglular argument (the phase angle) of the sne functon. Phasors of waves can be added as vectors to produce the sum of two sne functons. For A crcuts, the phase angle s ω t, so phasors rotate counterclockwse at an angular speed of ω. We usually wsh to construct current and voltage phasors for each crcut element. For resstors, the current and voltage phasors are n phase. For nductors, the voltage phasor s at an angle of 9 from the current phasor. For capactors, the voltage phasor s at an angle of 9 from the current phasor..7. ules for A rcuts We can use smlar rules for A crcuts as we had for D crcuts, but wth small modfcatons to take nto account the snusodal varaton n voltages and currents. ules for A rcuts. If two crcut elements are n seres, they have the same current phasor and ther voltage phasors add as vectors.. If two crcut elements are n parallel, they have the same voltage phasor and ther current phasors add as vectors. 3. The sum of current phasors nto a juncton equals the sum of current phasors out of a juncton. 4. The sum of voltage phasors from crcut elements around a loop s the sum of voltage phasors from EMFs around a loop. (Note that they don t sum to zero because of our standard defnton of postve voltage. That s, voltage phasors for crcut elements are voltage drops, whereas voltages for EMFs are voltage gans.) In order to see how to apply these rules, let s take a specfc example. Example.. An A crcut wth seres and parallel elements.

23 ε 3 Fgure.. An A crcut wth both seres and parallel components. Frst, we want to gve numercal values for a number of the quanttes n the problem: ω Ω.Ω 3.A Hz H F We wsh to fnd the voltage of the power supply and all the other voltages and currents n the crcuts whle we re at t. When we worked wth D crcuts usng Krchoff s aws, the frst thng we dd was assgn a drecton for the current. Wth A crcuts, we need to defne the drecton we take to be postve. Wth a sngle EMF, we should thnk of the power supply as a battery and draw the currents so they are consstent wth flow of current from a battery. Ths s done n Fg... Frst, let s draw current and voltage phasors for the nductor. For smplcty, we can draw the current phasor along the x axs. We know the voltage phasor wll be along the y axs and that ts magntude wll be: X ω 4.Ω. 3

24 Ths gves us: r r Next we add a phasor for the resstor,. Snce ths resstor s n seres wth the nductor, the share the same current phasor. The length of the voltage phasor for the resstor s: 3.. The voltage of the resstor s n phase wth the current. r r r r Next, we add the phasors for the nductor and resstor voltage to gve us the voltage phasor for the combnaton of the two crcut elements. 4

25 r r φ r We would lke to fnd the magntude of r the angle that r and the components of r. Ths can be accomplshed by smple geometry: r makes wth respect to the x axs, tanφ r 4, 3 cosφ xˆ.37 φ snφ yˆ xˆ yˆ 3. xˆ. yˆ Before we leave the resstor and nductor, there s one more thng we can fnd, the mpedance of the combnaton. Z or Z Z X 4.3Ω 4.3Ω X Now that we ve found the voltage, current, and mpedance for the branch of the crcut wth, we turn our attenton to the capactor. Snce the capactor s n parallel wth the nductor-resstor combnaton, we know the capactor s voltage phasor wll be the same as r. In the capactor (IE), the current leads the voltage, so we know the drecton of the current phasor, r 3. We can fnd the magntude of the current phasor by consderng the voltage phasors: 5

26 X 3 4.Ω ω 3 X 3.9A Now, let s draw the phasor dagram for the capactor..37 r r r 3 r For the next step, we note that the currents n the two parallel branches, r and r 3, add as phasors to gve the total current, r. r r r r 3 r 6

27 et s algebracally work out the components of r and r 3. We know that the angle between r 3 and the negatve x axs s 9 φ 4. 4, and the length of the phasor s 3.9A, so: r 3 r (3.9A)cos(4.4 ) xˆ (3.9A)sn(4.4 ) yˆ 3.Axˆ.75Ayˆ r r.75ayˆ 3 Note that we add phasors exactly the same way we add any other vectors. Now that we have the current through, we can easly fnd the voltage phasor for ths resstor: r r.5 yˆ. We can solve for the EMF snce the voltage phasors around any closed loop s zero. Keepng n mnd the sgn conventon for voltages across crcut elements s opposte that for the power supply, we have: r r ε r r ε r r.5 yˆ 3. xˆ. yˆ 3. xˆ 3.5 yˆ ε ( 3. ) ( 3.5 ) 3.83 φ ε r r r r r Fnally, we would lke to know the phase angle between the current r and the EMF, εr. In ths case, t s easy to fnd the angle wth smple geometry, snce the current s n the y drecton. 7

28 However, I ll use a lttle trck that s handy when we want to fnd the angle between arbtrary vectors. For two general vectors: Applyng ths rule to our vectors: r r A B AB cosφ r r A B cosφ AB r r ε cosφ ε φ ±.53 Snce r leads εr, the phase angle s negatve (more lke a capactor) and we conclude that φ. 53 Thngs to remember: Know the rules outlned n the box at the begnnng of ths secton..8 Power n A rcuts We know that n a D crcut P. In an A crcut, ths same result must hold; however, power wll be a functon of tme. Frst, let s consder a resstor. Assumng a snusodal current, we know the voltage across the resstor wll be n phase wth the current. Hence: ( ω t) P( t) ( t) ( t) sn. Although ths s true, we often fnd that t s more useful to know the average power dsspated by the resstor over one full cycle (or many full cycles). To fnd the average power, we can frst thnk of takng N samples over one full cycle. P N N j P j N N j sn ( ω t ) Ths, of course, s just an approxmaton to the average, snce N would have to be nfnte for the average to be exact. We actually can take the sum over an nfnte number of terms by turnng the sum nto an ntegral. To do ths, let s dvde one cycle nto N tme ntervals each of length t, so that t T / N where T s one full perod. Then P N t N j sn ( ω t ) t sn ( ω t) T T 8

29 The ntegral s a standard one for calculus courses, but we don t actually do t here. It s suffcent to know that the result s just T/. Ths then gves us: (.7 A Power, esstor) P. Ths equaton brngs up a practcal queston. When we say an A power supply provdes a gven voltage, what number should we use for the voltage? We could use the average voltage, but that s just zero. Another good dea would be to average the absolute value of the voltage over one cycle. The drawback to ths s that absolute values are rather messy mathematcal functons. The soluton physcsts chose was based on average power. They suggested that a good defnton for effectve voltage s the voltage that would yeld the same power as a D voltage. That s, we let: eff P eff Mathematcally, the way we obtan ths effectve voltage s essentally what we dd to fnd the average power: we square the voltage, average the squared voltage over one complete cycle, and then take the square root. Because of ths, the effectve voltage s called the root-mean-square voltage, or just the rms voltage. So, when we say that our outlets provde 5 A, what we really mean s that the rms voltage s 5. Thnk About It What s the peak voltage n your A power outlet? Of course we can defne rms currents n a smlar fashon to rms voltages, and the power n terms of these rms quanttes. P rms rms rms rms So far, we have only consdered the power dsspated by a resstor. We also want to be able to calculate the power provded by a power supply. We can follow the same method we used above, but we do have to take nto consderaton that there s a phase dfference between the 9

30 voltage and the current. Denotng the phase angle from the current phasor to the voltage phasor as φ (φ 9 for an nductor, for example), the nstantaneous power s: Ths leads to an average power: P( t) ( t) ε ( t) sn ε sn P ε P T ( ωt) ε sn( ωt φ) ( ωt) [ sn( ωt) cosφ cos( ωt) snφ] sn ε cosφ T T P ε cosφ ( ωt) [ sn( ωt) cosφ cos( ωt) snφ] T sn T ε snφ T ( ωt) sn( ωt) cos( ωt) The power provded by an A power supply, then depends on the phase angle between the current and EMF phasors. The quantty cos φ s called the power factor for the crcut. When the current s n phase n wth the EMF, the power s just what t was for D crcuts, but the power provded by the power supply s less when the phase angle gets larger. Thnk About It What s the power factor for a resstor? a capactor? and nductor? a capactor? What power s dsspated n an nductor? a capactor? The power provded by a power supply (or really any crcut element) s gven by: (.8 A Power, general) P cosφ rms cosφ. r r rms Example.. Power n the crcut of Example.. What power s dsspated n each of the resstors of Fg..? What power s provded by the power supply? 3

31 esstor has a resstance of. Ω and a current of power s P (.75 A) (.Ω). 565W. esstor has a resstance of. Ω and a current of power s P ( 3.A) (.Ω) 4. 5W.. 75A passes through t, so the 3. A passes through t, so the The power provded by the power supply s (wth no ntermedate roundng): P ε cosφ 5. 65W Note that the power provded by the power supply equals the total power dsspated by the resstors. Thngs to remember: The rms values of voltage or current are the peak values dvded by. The average power dsspated by a resstor n an A crcut s P The average power provded by an A power source s P cosφ. cos φ s called the power factor. rmsrms..9. The Seres rcut The smplest and most mportant A crcut we can analyze s the seres crcut, llustrated n Fg... ε Fgure.. The seres crcut. 3

32 The analyss of ths crcut s qute easy snce all the crcut elements share the same current. We can draw a phasor dagram for the current and voltages across the nductor, capactor, and resstor. X r r ε Z X εr Fgure.3. Phasors for the seres crcut. Snce the elements are n seres, we recognze that the voltages phasors of the resstor, capactor, r r r r and nductor add together to get the total EMF, ε. In Fg..3, we have added r r frst and then added that sum to r to get. At ths pont, I want to ntroduce a lttle trck to smplfy ths partcular problem. We note that every voltage n the system, ncludng the EMF, has a common factor of. If we dvde ths out, we can obtan a dagram for the resstance, reactances, and mpedance that s qute useful. We ll call ths the mpedance dagram of the crcut. It s shown n fgure.4. 3

33 X X X X Z φ Fgure.4 The mpedance dagram of seres crcut. From ths dagram, we can deduce some smple equatons for the total mpedance and the phase angle n seres crcuts: (.9 For seres crcuts) Z tanφ cosφ ( X X ) X Z X Thngs to remember: Be able to reproduce the mpedance dagram for seres crcuts, Fg..4. Understand the phase relatons that appear n the dagram. X X Know that ( ) Know that.. esonance Z X and be able to deduce ths from the mpedance dagram. X tan φ and be able to deduce ths from the mpedance dagram. If we consder the mpedance equaton along wth the equatons for the nductve and capactve reactance, we see that mpedance has a rather complcated dependence on the frequency of the oscllator. 33

34 Z X X ( X X ) ω ω When the frequency s very small, the capactve reactance s large and Z X. When the frequency s very large, the nductve reactance s large and Z X. Z s a mnmum when X X, and Z s a mnmum, the current n the crcut s a maxmum. When ths happens, the resstance provdes the only mpedance n the crcut, Z. Ths condton s called resonance and s electrcal analog to resonance n harmonc oscllators such as a swngng pendulum or a mass on the end of a sprng. Frst, let us fnd the frequency at whch resonance s acheved: (Eq.., resonance frequency) X X ω ω ω Note that ths s just the frequency at whch the capactor and nductor would oscllate f there were power supply or resstance n the crcut. Note that the condton for maxmzng the current n an crcut s to drve the crcut at the frequency t wants to naturally oscllate. Ths s smlar to a swng the maxmum ampltude s obtaned when we push the swng at ts natural frequency of oscllaton. X X X Z X Fgure.5. The mpedance dagram at resonance. 34

35 Thngs to remember: esonance s when the mpedance of a crcut s mnmzed so that the current s maxmzed. esonance n a seres crcut s acheved when X X. At resonance Z. The resonant frequency s the natural oscllaton frequency, ω. 35