AlternatingCurrent Circuits


 Dennis Lane
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1 chpter 33 AlterntingCurrent Circuits 33.1 AC Sources 33.2 esistors in n AC Circuit 33.3 Inductors in n AC Circuit 33.4 Cpcitors in n AC Circuit 33.5 The LC Series Circuit 33.6 Power in n AC Circuit 33.7 esonnce in Series LC Circuit 33.8 The Trnsformer nd Power Trnsmission 33.9 ectifiers nd Filters In this chpter, we descrie lterntingcurrent (AC) circuits. Every time you turn on television set, computer, or ny of multitude of other electricl pplinces in home, you re clling on lternting currents to provide the power to operte them. We egin our study y investigting the chrcteristics of simple series circuits tht contin resistors, inductors, nd These lrge trnsformers re used to increse the voltge t power plnt for cpcitors nd tht re driven y sinusoidl voltge. The primry im of this chnged reltively esily ecuse power is distriuted y lternting current rther distriution of energy y electricl trnsmission to the power grid. Voltges cn e thn direct current. ( Lester Lefkowitz/Getty Imges) chpter cn e summrized s follows: if n AC source pplies n lternting voltge to series circuit contining resistors, inductors, nd cpcitors, we wnt to know the mplitude nd time chrcteristics of the lternting current. We conclude this chpter with two sections concerning trnsformers, power trnsmission, nd electricl filters. 953
2 954 CHAPTE 33 AlterntingCurrent Circuits V mx v Figure 33.1 The voltge supplied y n AC source is sinusoidl with period T. T t 33.1 AC Sources An AC circuit consists of circuit elements nd power source tht provides n lternting voltge Dv. This timevrying voltge from the source is descried y Dv 5 DV mx sin vt where DV mx is the mximum output voltge of the source, or the voltge mplitude. There re vrious possiilities for AC sources, including genertors s discussed in Section 31.5 nd electricl oscilltors. In home, ech electricl outlet serves s n AC source. Becuse the output voltge of n AC source vries sinusoidlly with time, the voltge is positive during one hlf of the cycle nd negtive during the other hlf s in Figure Likewise, the current in ny circuit driven y n AC source is n lternting current tht lso vries sinusoidlly with time. From Eqution 15.12, the ngulr frequency of the AC voltge is v52pf5 2p T where f is the frequency of the source nd T is the period. The source determines the frequency of the current in ny circuit connected to it. Commercil electricpower plnts in the United Sttes use frequency of 60.0 Hz, which corresponds to n ngulr frequency of 377 rd/s. v v V mx sin vt ACTIVE FIGUE 33.2 A circuit consisting of resistor of resistnce connected to n AC source, designted y the symol esistors in n AC Circuit Consider simple AC circuit consisting of resistor nd n AC source s shown in Active Figure At ny instnt, the lgeric sum of the voltges round closed loop in circuit must e zero (Kirchhoff s loop rule). Therefore, Dv 1 Dv 5 0 or, using Eqution 27.7 for the voltge cross the resistor, Dv 2 i 5 0 If we rerrnge this expression nd sustitute DV mx sin vt for Dv, the instntneous current in the resistor is i 5 Dv 5 DV mx sin vt 5 I mx sin vt (33.1) where I mx is the mximum current: Mximum current in resistor I mx 5 DV mx Eqution 33.1 shows tht the instntneous voltge cross the resistor is (33.2) Voltge cross resistor Pitfll Prevention 33.1 TimeVrying Vlues We use lowercse symols Dv nd i to indicte the instntneous vlues of timevrying voltges nd currents. Cpitl letters represent fixed vlues of voltge nd current such s DV mx nd I mx. Dv 5 i 5 I mx sin vt (33.3) A plot of voltge nd current versus time for this circuit is shown in Active Figure At point, the current hs mximum vlue in one direction, ritrrily clled the positive direction. Between points nd, the current is decresing in mgnitude ut is still in the positive direction. At point, the current is momentrily zero; it then egins to increse in the negtive direction etween points nd c. At point c, the current hs reched its mximum vlue in the negtive direction. The current nd voltge re in step with ech other ecuse they vry identiclly with time. Becuse i nd Dv oth vry s sin vt nd rech their mximum vlues t the sme time s shown in Active Figure 33.3, they re sid to e in phse, similr to the wy two wves cn e in phse s discussed in our study of wve motion in Chpter 18. Therefore, for sinusoidl pplied voltge, the current in resistor is lwys in phse with the voltge cross the resistor. For resistors in AC circuits, there
3 33.2 esistors in n AC Circuit 955 The current nd the voltge re in phse: they simultneously rech their mximum vlues, their minimum vlues, nd their zero vlues. i, v I mx i The current nd the voltge phsors re in the sme direction ecuse the current is in phse with the voltge. i, v ACTIVE FIGUE 33.3 () Plots of the instntneous current i nd instntneous voltge Dv cross resistor s functions of time. At time t 5 T, one cycle of the timevrying voltge nd current hs een completed. () Phsor digrm for the resistive circuit showing tht the current is in phse with the voltge. V mx c T v t re no new concepts to lern. esistors ehve essentilly the sme wy in oth DC nd AC circuits. Tht, however, is not the cse for cpcitors nd inductors. To simplify our nlysis of circuits contining two or more elements, we use grphicl representtion clled phsor digrm. A phsor is vector whose length is proportionl to the mximum vlue of the vrile it represents (DV mx for voltge nd I mx for current in this discussion). The phsor rottes counterclockwise t n ngulr speed equl to the ngulr frequency ssocited with the vrile. The projection of the phsor onto the verticl xis represents the instntneous vlue of the quntity it represents. Active Figure 33.3 shows voltge nd current phsors for the circuit of Active Figure 33.2 t some instnt of time. The projections of the phsor rrows onto the verticl xis re determined y sine function of the ngle of the phsor with respect to the horizontl xis. For exmple, the projection of the current phsor in Active Figure 33.3 is I mx sin vt. Notice tht this expression is the sme s Eqution Therefore, the projections of phsors represent current vlues tht vry sinusoidlly in time. We cn do the sme with timevrying voltges. The dvntge of this pproch is tht the phse reltionships mong currents nd voltges cn e represented s vector dditions of phsors using the vector ddition techniques discussed in Chpter 3. In the cse of the singleloop resistive circuit of Active Figure 33.2, the current nd voltge phsors re in the sme direction in Active Figure 33.3 ecuse i nd Dv re in phse. The current nd voltge in circuits contining cpcitors nd inductors hve different phse reltionships. i v V mx vt I mx Pitfll Prevention 33.2 A Phsor Is Like Grph An lternting voltge cn e presented in different representtions. One grphicl representtion is shown in Figure 33.1 in which the voltge is drwn in rectngulr coordintes, with voltge on the verticl xis nd time on the horizontl xis. Active Figure 33.3 shows nother grphicl representtion. The phse spce in which the phsor is drwn is similr to polr coordinte grph pper. The rdil coordinte represents the mplitude of the voltge. The ngulr coordinte is the phse ngle. The verticlxis coordinte of the tip of the phsor represents the instntneous vlue of the voltge. The horizontl coordinte represents nothing t ll. As shown in Active Figure 33.3, lternting currents cn lso e represented y phsors. To help with this discussion of phsors, review Section 15.4, where we represented the simple hrmonic motion of rel oject y the projection of n imginry oject s uniform circulr motion onto coordinte xis. A phsor is direct nlog to this representtion. Quick Quiz 33.1 Consider the voltge phsor in Figure 33.4, shown t three instnts of time. (i) Choose the prt of the figure, (), (), or (c), tht represents the instnt of time t which the instntneous vlue of the voltge hs the lrgest mgnitude. (ii) Choose the prt of the figure tht represents the instnt of time t which the instntneous vlue of the voltge hs the smllest mgnitude. For the simple resistive circuit in Active Figure 33.2, notice tht the verge vlue of the current over one cycle is zero. Tht is, the current is mintined in the positive direction for the sme mount of time nd t the sme mgnitude s it is mintined in the negtive direction. The direction of the current, however, hs no effect on the ehvior of the resistor. We cn understnd this concept y relizing tht collisions etween electrons nd the fixed toms of the resistor result in n c Figure 33.4 (Quick Quiz 33.1) A voltge phsor is shown t three instnts of time, (), (), nd (c).
4 956 CHAPTE 33 AlterntingCurrent Circuits Figure 33.5 () Grph of the current in resistor s function of time. () Grph of the current squred in resistor s function of time, showing tht the red dshed line is the verge of I 2 mx sin2 vt. In generl, the verge vlue of sin 2 vt or cos 2 vt over one cycle is 1 2. I mx 0 i i 2 t I 2 mx 1 (i 2 ) 2 I 2 vg mx 0 t The gry shded regions under the curve nd ove the red dshed line hve the sme re s the gry shded regions ove the curve nd elow the red dshed line. increse in the resistor s temperture. Although this temperture increse depends on the mgnitude of the current, it is independent of the current s direction. We cn mke this discussion quntittive y reclling tht the rte t which energy is delivered to resistor is the power P 5 i 2, where i is the instntneous current in the resistor. Becuse this rte is proportionl to the squre of the current, it mkes no difference whether the current is direct or lternting, tht is, whether the sign ssocited with the current is positive or negtive. The temperture increse produced y n lternting current hving mximum vlue I mx, however, is not the sme s tht produced y direct current equl to I mx ecuse the lternting current hs this mximum vlue for only n instnt during ech cycle (Fig. 33.5). Wht is of importnce in n AC circuit is n verge vlue of current, referred to s the rms current. As we lerned in Section 21.1, the nottion rms stnds for rootmensqure, which in this cse mens the squre root of the men (verge) vlue of the squre of the current: I rms 5!1i 2 2 vg. Becuse i 2 vries s sin 2 vt nd ecuse the verge vlue of i 2 is 1 2I 2 mx (see Fig. 33.5), the rms current is rms current Averge power delivered to resistor rms voltge I rms 5 I mx " I mx (33.4) This eqution sttes tht n lternting current whose mximum vlue is 2.00 A delivers to resistor the sme power s direct current tht hs vlue of (0.707) (2.00 A) A. The verge power delivered to resistor tht crries n lternting current is P vg 5 I 2 rms Alternting voltge is lso est discussed in terms of rms voltge, nd the reltionship is identicl to tht for current: DV rms 5 DV mx " DV mx (33.5) When we spek of mesuring 120V lternting voltge from n electricl outlet, we re referring to n rms voltge of 120 V. A clcultion using Eqution 33.5 shows tht such n lternting voltge hs mximum vlue of out 170 V. One reson rms vlues re often used when discussing lternting currents nd voltges is tht AC mmeters nd voltmeters re designed to red rms vlues. Furthermore, with rms vlues, mny of the equtions we use hve the sme form s their directcurrent counterprts.
5 33.3 Inductors in n AC Circuit 957 Exmple 33.1 Wht Is the rms Current? The voltge output of n AC source is given y the expression Dv sin vt, where Dv is in volts. Find the rms current in the circuit when this source is connected to 100V resistor. SOLUTION Conceptulize Active Figure 33.2 shows the physicl sitution for this prolem. Ctegorize We evlute the current with n eqution developed in this section, so we ctegorize this exmple s sustitution prolem. Compring this expression for voltge output with the generl form Dv 5 DV mx sin vt shows tht DV mx V. Clculte the rms voltge from Eqution 33.5: Find the rms current: DV rms 5 DV mx " V " V I rms 5 DV rms V 100 V A 33.3 Inductors in n AC Circuit Now consider n AC circuit consisting only of n inductor connected to the terminls of n AC source s shown in Active Figure Becuse Dv L 5 2L(di L /dt) is the selfinduced instntneous voltge cross the inductor (see Eq. 32.1), Kirchhoff s loop rule pplied to this circuit gives Dv 1 Dv L 5 0, or Dv 2 L di L dt 5 0 Sustituting DV mx sin vt for Dv nd rerrnging gives Dv 5 L di L dt 5DV mx sin vt (33.6) Solving this eqution for di L gives v L L v V mx sin vt ACTIVE FIGUE 33.6 A circuit consisting of n inductor of inductnce L connected to n AC source. di L 5 DV mx sin vt dt L Integrting this expression 1 gives the instntneous current i L in the inductor s function of time: i L 5 DV mx L 3 sin vt dt 5 2 DV mx vl cos vt (33.7) Using the trigonometric identity cos vt 5 2sin(vt 2 p/2), we cn express Eqution 33.7 s i L 5 DV mx vl sin vt 2 p 2 (33.8) Compring this result with Eqution 33.6 shows tht the instntneous current i L in the inductor nd the instntneous voltge Dv L cross the inductor re out of phse y p/2 rd A plot of voltge nd current versus time is shown in Active Figure 33.7 (pge 958). When the current i L in the inductor is mximum (point in Active Fig. 33.7), it is momentrily not chnging, so the voltge cross the inductor is zero (point d). At points such s nd e, the current is zero nd the rte of chnge of Current in n inductor 1 We neglect the constnt of integrtion here ecuse it depends on the initil conditions, which re not importnt for this sitution.
6 958 CHAPTE 33 AlterntingCurrent Circuits ACTIVE FIGUE 33.7 () Plots of the instntneous current i L nd instntneous voltge Dv L cross n inductor s functions of time. () Phsor digrm for the inductive circuit. I mx V mx The current lgs the voltge y onefourth of cycle. v L, i L c d f e v L T i L t The current nd voltge phsors re t 90 to ech other. v L, i L v L V mx vt i L I mx current is t mximum. Therefore, the voltge cross the inductor is lso t mximum (points c nd f ). Notice tht the voltge reches its mximum vlue onequrter of period efore the current reches its mximum vlue. Therefore, for sinusoidl pplied voltge, the current in n inductor lwys lgs ehind the voltge cross the inductor y 90 (onequrter cycle in time). As with the reltionship etween current nd voltge for resistor, we cn represent this reltionship for n inductor with phsor digrm s in Active Figure The phsors re t 90 to ech other, representing the 90 phse difference etween current nd voltge. Eqution 33.7 shows tht the current in n inductive circuit reches its mximum vlue when cos vt 5 61: Mximum current in n inductor Inductive rectnce I mx 5 DV mx (33.9) vl This expression is similr to the reltionship etween current, voltge, nd resistnce in DC circuit, I 5 DV/ (Eq. 27.7). Becuse I mx hs units of mperes nd DV mx hs units of volts, vl must hve units of ohms. Therefore, vl hs the sme units s resistnce nd is relted to current nd voltge in the sme wy s resistnce. It must ehve in mnner similr to resistnce in the sense tht it represents opposition to the flow of chrge. Becuse vl depends on the pplied frequency v, the inductor rects differently, in terms of offering opposition to current, for different frequencies. For this reson, we define vl s the inductive rectnce X L : Therefore, we cn write Eqution 33.9 s X L ; vl (33.10) I mx 5 DV mx (33.11) X L The expression for the rms current in n inductor is similr to Eqution 33.11, with I mx replced y I rms nd DV mx replced y DV rms. Eqution indictes tht, for given pplied voltge, the inductive rectnce increses s the frequency increses. This conclusion is consistent with Frdy s lw: the greter the rte of chnge of current in the inductor, the lrger the ck emf. The lrger ck emf trnsltes to n increse in the rectnce nd decrese in the current. Using Equtions 33.6 nd 33.11, we find tht the instntneous voltge cross the inductor is Voltge cross n inductor Dv L 52L di L dt 5 2DV mx sin vt 52I mx X L sin vt (33.12)
7 33.4 Cpcitors in n AC Circuit 959 Quick Quiz 33.2 Consider the AC circuit in Figure The frequency of the AC source is djusted while its voltge mplitude is held constnt. When does the lightul glow the rightest? () It glows rightest t high frequencies. () It glows rightest t low frequencies. (c) The rightness is the sme t ll frequencies. L Figure 33.8 (Quick Quiz 33.2) At wht frequencies does the lightul glow the rightest? Exmple 33.2 A Purely Inductive AC Circuit In purely inductive AC circuit, L mh nd the rms voltge is 150 V. Clculte the inductive rectnce nd rms current in the circuit if the frequency is 60.0 Hz. SOLUTION Conceptulize Active Figure 33.6 shows the physicl sitution for this prolem. Keep in mind tht inductive rectnce increses with incresing frequency of the pplied voltge. Ctegorize We determine the rectnce nd the current from equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution to find the inductive rectnce: X L 5 vl 5 2pfL 5 2p(60.0 Hz)( H) V Use n rms version of Eqution to find the rms current: I rms 5 DV rms X L V 9.42 V A WHAT IF? If the frequency increses to 6.00 khz, wht hppens to the rms current in the circuit? Answer If the frequency increses, the inductive rectnce lso increses ecuse the current is chnging t higher rte. The increse in inductive rectnce results in lower current. Let s clculte the new inductive rectnce nd the new rms current: X L 5 2p( Hz)( H) V I rms V 942 V A 33.4 Cpcitors in n AC Circuit Active Figure 33.9 shows n AC circuit consisting of cpcitor connected cross the terminls of n AC source. Kirchhoff s loop rule pplied to this circuit gives Dv 1 Dv C 5 0, or Sustituting DV mx sin vt for Dv nd rerrnging gives Dv 2 q C 5 0 (33.13) q 5 C DV mx sin vt (33.14) where q is the instntneous chrge on the cpcitor. Differentiting Eqution with respect to time gives the instntneous current in the circuit: i C 5 dq dt 5vC DV mx cos vt (33.15) v C C v V mx sin vt ACTIVE FIGUE 33.9 A circuit consisting of cpcitor of cpcitnce C connected to n AC source.
8 960 CHAPTE 33 AlterntingCurrent Circuits ACTIVE FIGUE () Plots of the instntneous current i C nd instntneous voltge Dv C cross cpcitor s functions of time. () Phsor digrm for the cpcitive circuit. I mx V mx c The current leds the voltge y onefourth of cycle. v C, i C d f i C T v C t The current nd voltge phsors re t 90 to ech other. I mx v C, i C i C v C Vmx vt e Using the trigonometric identity cos vt 5 sin vt 1 p 2 we cn express Eqution in the lterntive form Current in cpcitor i C 5vC DV mx sin vt 1 p 2 (33.16) Compring this expression with Dv 5 DV mx sin vt shows tht the current is p/2 rd 5 90 out of phse with the voltge cross the cpcitor. A plot of current nd voltge versus time (Active Fig ) shows tht the current reches its mximum vlue onequrter of cycle sooner thn the voltge reches its mximum vlue. Consider point such s where the current is zero t this instnt. Tht occurs when the cpcitor reches its mximum chrge so tht the voltge cross the cpcitor is mximum (point d). At points such s nd e, the current is mximum, which occurs t those instnts when the chrge on the cpcitor reches zero nd the cpcitor egins to rechrge with the opposite polrity. When the chrge is zero, the voltge cross the cpcitor is zero (points c nd f ). As with inductors, we cn represent the current nd voltge for cpcitor on phsor digrm. The phsor digrm in Active Figure shows tht for sinusoidlly pplied voltge, the current lwys leds the voltge cross cpcitor y 90. Eqution shows tht the current in the circuit reches its mximum vlue when cos vt 5 61: Cpcitive rectnce Mximum current in cpcitor I mx 5vC DV mx 5 DV mx (33.17) 11/vC2 As in the cse with inductors, this looks like Eqution 27.7, so the denomintor plys the role of resistnce, with units of ohms. We give the comintion 1/vC the symol X C, nd ecuse this function vries with frequency, we define it s the cpcitive rectnce: X C ; 1 (33.18) vc We cn now write Eqution s I mx 5 DV mx X C (33.19) The rms current is given y n expression similr to Eqution 33.19, with I mx replced y I rms nd DV mx replced y DV rms.
9 33.4 Cpcitors in n AC Circuit 961 Using Eqution 33.19, we cn express the instntneous voltge cross the cpcitor s Dv C 5DV mx sin vt 5 I mx X C sin vt (33.20) Equtions nd indicte tht s the frequency of the voltge source increses, the cpcitive rectnce decreses nd the mximum current therefore increses. The frequency of the current is determined y the frequency of the voltge source driving the circuit. As the frequency pproches zero, the cpcitive rectnce pproches infinity nd the current therefore pproches zero. This conclusion mkes sense ecuse the circuit pproches direct current conditions s v pproches zero nd the cpcitor represents n open circuit. Voltge cross cpcitor C Quick Quiz 33.3 Consider the AC circuit in Figure The frequency of the AC source is djusted while its voltge mplitude is held constnt. When does the lightul glow the rightest? () It glows rightest t high frequencies. () It glows rightest t low frequencies. (c) The rightness is the sme t ll frequencies. Figure (Quick Quiz 33.3) L Quick Quiz 33.4 Consider the AC circuit in Figure The frequency of the AC source is djusted while its voltge mplitude is held constnt. When does the lightul glow the rightest? () It glows rightest t high frequencies. () It glows rightest t low frequencies. (c) The rightness is the sme t ll frequencies. C Figure (Quick Quiz 33.4) Exmple 33.3 A Purely Cpcitive AC Circuit An 8.00mF cpcitor is connected to the terminls of 60.0Hz AC source whose rms voltge is 150 V. Find the cpcitive rectnce nd the rms current in the circuit. SOLUTION Conceptulize Active Figure 33.9 shows the physicl sitution for this prolem. Keep in mind tht cpcitive rectnce decreses with incresing frequency of the pplied voltge. Ctegorize We determine the rectnce nd the current from equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution to find the cpcitive rectnce: X C 5 1 vc 5 1 2pfC 5 1 2p160.0 Hz F V Use n rms version of Eqution to find the rms current: I rms 5 DV rms X C V 332 V A WHAT IF? Wht if the frequency is douled? Wht hppens to the rms current in the circuit? Answer If the frequency increses, the cpcitive rectnce decreses, which is just the opposite from the cse of n inductor. The decrese in cpcitive rectnce results in n increse in the current. Let s clculte the new cpcitive rectnce nd the new rms current: X C 5 1 vc 5 1 2p1120 Hz F V I rms V 166 V A
10 962 CHAPTE 33 AlterntingCurrent Circuits i v v L v C v v L v C L C ACTIVE FIGUE () A series circuit consisting of resistor, n inductor, nd cpcitor connected to n AC source. () Phse reltionships etween the current nd the voltges in the series LC circuit. t t t t 33.5 The LC Series Circuit In the previous sections, we considered individul circuit elements connected to n AC source. Active Figure shows circuit tht contins comintion of circuit elements: resistor, n inductor, nd cpcitor connected in series cross n lterntingvoltge source. If the pplied voltge vries sinusoidlly with time, the instntneous pplied voltge is while the current vries s Dv 5DV mx sin vt i 5 I mx sin 1vt 2f2 where f is some phse ngle etween the current nd the pplied voltge. Bsed on our discussions of phse in Sections 33.3 nd 33.4, we expect tht the current will generlly not e in phse with the voltge in n LC circuit. Our im is to determine f nd I mx. Active Figure shows the voltge versus time cross ech element in the circuit nd their phse reltionships to the current. First, ecuse the elements re in series, the current everywhere in the circuit must e the sme t ny instnt. Tht is, the current t ll points in series AC circuit hs the sme mplitude nd phse. Bsed on the preceding sections, we know tht the voltge cross ech element hs different mplitude nd phse. In prticulr, the voltge cross the resistor is in phse with the current, the voltge cross the inductor leds the current y 90, nd the voltge cross the cpcitor lgs ehind the current y 90. Using these phse reltionships, we cn express the instntneous voltges cross the three circuit elements s Dv 5 I mx sin vt 5DV sin vt (33.21) Dv L 5 I mx X L sin vt 1 p 2 5DV L cos vt (33.22) Dv C 5 I mx X C sin vt 2 p 2 5 2DV C cos vt (33.23) The sum of these three voltges must equl the voltge from the AC source, ut it is importnt to recognize tht ecuse the three voltges hve different phse reltionships with the current, they cnnot e dded directly. Figure represents the phsors t n instnt t which the current in ll three elements is momentrily zero. The zero current is represented y the current phsor long the horizontl xis in ech prt of the figure. Next the voltge phsor is drwn t the pproprite phse ngle to the current for ech element. Becuse phsors re rotting vectors, the voltge phsors in Figure cn e comined using vector ddition s in Active Figure In Active Figure 33.15, the voltge phsors in Figure re comined on the sme coordinte xes. Active Figure shows the vector ddition of the voltge phsors. The voltge phsors DV L nd DV C re in opposite directions long the sme line, so we cn con esistor Inductor Cpcitor V L v v v I mx Figure Phse reltionships etween the voltge nd current phsors for () resistor, () n inductor, nd (c) cpcitor connected in series. I mx V I mx c V C
11 33.5 The LC Series Circuit 963 The phsors of Figure re comined on single set of xes. V L The totl voltge V mx mkes n ngle f with I mx. V L V C f V mx ACTIVE FIGUE () Phsor digrm for the series LC circuit shown in Active Figure () The inductnce nd cpcitnce phsors re dded together nd then dded vectorilly to the resistnce phsor. v V C I mx V v I mx V struct the difference phsor DV L 2 DV C, which is perpendiculr to the phsor DV. This digrm shows tht the vector sum of the voltge mplitudes DV, DV L, nd DV C equls phsor whose length is the mximum pplied voltge DV mx nd which mkes n ngle f with the current phsor I mx. From the right tringle in Active Figure 33.15, we see tht DV mx 5 "DV 2 1 1DV L 2DV C "1I mx I mx X L 2 I mx X C 2 2 DV mx 5 I mx " 2 1 1X L 2 X C 2 2 Therefore, we cn express the mximum current s DV mx I mx 5 (33.24) " 2 1 1X L 2 X C 2 2 Once gin, this expression hs the sme mthemticl form s Eqution The denomintor of the frction plys the role of resistnce nd is clled the impednce Z of the circuit: Z ; " 2 1 1X L 2 X C 2 2 (33.25) where impednce lso hs units of ohms. Therefore, Eqution cn e written in the form Mximum current in n LC circuit Impednce I mx 5 DV mx Z (33.26) Eqution is the AC equivlent of Eqution Note tht the impednce nd therefore the current in n AC circuit depend on the resistnce, the inductnce, the cpcitnce, nd the frequency (ecuse the rectnces re frequency dependent). From the right tringle in the phsor digrm in Active Figure 33.15, the phse ngle f etween the current nd the voltge is found s follows: f5tn 21 DV L 2DV C DV 5 tn 21 I mxx L 2 I mx X C I mx f5tn 21 X L 2 X C (33.27) When X L. X C (which occurs t high frequencies), the phse ngle is positive, signifying tht the current lgs the pplied voltge s in Active Figure We descrie this sitution y sying tht the circuit is more inductive thn cpcitive. When X L, X C, the phse ngle is negtive, signifying tht the current leds the pplied voltge, nd the circuit is more cpcitive thn inductive. When X L 5 X C, the phse ngle is zero nd the circuit is purely resistive. Phse ngle
12 964 CHAPTE 33 AlterntingCurrent Circuits Quick Quiz 33.5 Lel ech prt of Figure 33.16, (), (), nd (c), s representing X L. X C, X L 5 X C, or X L, X C. Figure (Quick Quiz 33.5) Mtch the phsor digrms to the reltionships etween the rectnces. Imx V mx I mx V mx c V mx I mx Exmple 33.4 Anlyzing Series LC Circuit A series LC circuit hs V, L H, nd C mf. It is connected to n AC source with f Hz nd DV mx V. (A) Determine the inductive rectnce, the cpcitive rectnce, nd the impednce of the circuit. SOLUTION Conceptulize The circuit of interest in this exmple is shown in Active Figure The current in the comintion of the resistor, inductor, nd cpcitor oscilltes t prticulr phse ngle with respect to the pplied voltge. Ctegorize The circuit is simple series LC circuit, so we cn use the pproch discussed in this section. Anlyze Find the ngulr frequency: v 5 2pf 5 2p(60.0 Hz) s 21 Use Eqution to find the inductive rectnce: X L 5 vl 5 (377 s 21 )(1.25 H) V Use Eqution to find the cpcitive rectnce: X C 5 1 vc s F V Use Eqution to find the impednce: Z 5 " 2 1 1X L 2 X C 2 2 (B) Find the mximum current in the circuit. 5 "1425 V V 2758 V V SOLUTION Use Eqution to find the mximum current: I mx 5 DV mx Z V 513 V A (C) Find the phse ngle etween the current nd voltge. SOLUTION Use Eqution to clculte the phse ngle: f5tn 21 X L 2 X C 471 V2758 V 5 tn V (D) Find the mximum voltge cross ech element. SOLUTION Use Equtions 33.2, 33.11, nd to clculte the mximum voltges: DV 5 I mx A21425 V V DV L 5 I mx X L A21471 V V DV C 5 I mx X C A21758 V V (E) Wht replcement vlue of L should n engineer nlyzing the circuit choose such tht the current leds the pplied voltge y 30.0 rther thn 34.0? All other vlues in the circuit sty the sme.
13 33.6 Power in n AC Circuit cont. SOLUTION Solve Eqution for the inductive rectnce: Sustitute Equtions nd into this expression: Solve for L: L 5 1 v 1 1 tn f vc Sustitute the given vlues: L 5 X L 5 X C 1 tn f vl 5 1 vc 1 tn f s 21 2 c V 2 tn d 1377 s F2 L H Finlize Becuse the cpcitive rectnce is lrger thn the inductive rectnce, the circuit is more cpcitive thn inductive. In this cse, the phse ngle f is negtive, so the current leds the pplied voltge. Using Equtions 33.21, 33.22, nd 33.23, the instntneous voltges cross the three elements re Dv 5 (124 V) sin 377t Dv L 5 (138 V) cos 377t Dv C 5 (2222 V) cos 377t WHAT IF? Wht if you dded up the mximum voltges cross the three circuit elements? Is tht physiclly meningful quntity? Answer The sum of the mximum voltges cross the elements is DV 1 DV L 1 DV C V. This sum is much greter thn the mximum voltge of the source, 150 V. The sum of the mximum voltges is meningless quntity ecuse when sinusoidlly vrying quntities re dded, oth their mplitudes nd their phses must e tken into ccount. The mximum voltges cross the vrious elements occur t different times. Therefore, the voltges must e dded in wy tht tkes ccount of the different phses s shown in Active Figure Power in n AC Circuit Now let s tke n energy pproch to nlyzing AC circuits nd consider the trnsfer of energy from the AC source to the circuit. The power delivered y ttery to n externl DC circuit is equl to the product of the current nd the terminl voltge of the ttery. Likewise, the instntneous power delivered y n AC source to circuit is the product of the current nd the pplied voltge. For the LC circuit shown in Active Figure 33.13, we cn express the instntneous power P s P 5 i Dv 5 I mx sin (vt 2 f) DV mx sin vt P 5 I mx DV mx sin vt sin (vt 2 f) (33.28) This result is complicted function of time nd is therefore not very useful from prcticl viewpoint. Wht is generlly of interest is the verge power over one or more cycles. Such n verge cn e computed y first using the trigonometric identity sin (vt 2 f) 5 sin vt cos f 2 cos vt sin f. Sustituting this identity into Eqution gives P 5 I mx DV mx sin 2 vt cos f 2 I mx DV mx sin vt cos vt sin f (33.29) Let s now tke the time verge of P over one or more cycles, noting tht I mx, DV mx, f, nd v re ll constnts. The time verge of the first term on the right of the equl sign in Eqution involves the verge vlue of sin 2 vt, which
14 966 CHAPTE 33 AlterntingCurrent Circuits is 1 2. The time verge of the second term on the right of the equl sign is identiclly zero ecuse sin vt cos vt sin 2vt, nd the verge vlue of sin 2vt is zero. Therefore, we cn express the verge power P vg s P vg 5 1 2I mx DV mx cos f (33.30) It is convenient to express the verge power in terms of the rms current nd rms voltge defined y Equtions 33.4 nd 33.5: Averge power delivered to n LC circuit P vg 5 I rms DV rms cos f (33.31) where the quntity cos f is clled the power fctor. Active Figure shows tht the mximum voltge cross the resistor is given y DV 5 DV mx cos f 5 I mx. Therefore, cos f 5 I mx /DV mx 5 /Z, nd we cn express P vg s P vg 5 I rms DV rms cos f5i rms DV rms Z 5 I rms DV rms Z ecognizing tht DV rms /Z 5 I rms gives P vg 5 I 2 rms (33.32) The verge power delivered y the source is converted to internl energy in the resistor, just s in the cse of DC circuit. When the lod is purely resistive, f 5 0, cos f 5 1, nd, from Eqution 33.31, we see tht P vg 5 I rms DV rms Note tht no power losses re ssocited with pure cpcitors nd pure inductors in n AC circuit. To see why tht is true, let s first nlyze the power in n AC circuit contining only source nd cpcitor. When the current egins to increse in one direction in n AC circuit, chrge egins to ccumulte on the cpcitor nd voltge ppers cross it. When this voltge reches its mximum vlue, the energy stored in the cpcitor s electric potentil energy is 1 2C(DV mx ) 2. This energy storge, however, is only momentry. The cpcitor is chrged nd dischrged twice during ech cycle: chrge is delivered to the cpcitor during two qurters of the cycle nd is returned to the voltge source during the remining two qurters. Therefore, the verge power supplied y the source is zero. In other words, no power losses occur in cpcitor in n AC circuit. Now consider the cse of n inductor. When the current in n inductor reches its mximum vlue, the energy stored in the inductor is mximum nd is given y 1 2LI 2 mx. When the current egins to decrese in the circuit, this stored energy in the inductor returns to the source s the inductor ttempts to mintin the current in the circuit. Eqution shows tht the power delivered y n AC source to ny circuit depends on the phse, result tht hs mny interesting pplictions. For exmple, fctory tht uses lrge motors in mchines, genertors, or trnsformers hs lrge inductive lod (ecuse of ll the windings). To deliver greter power to such devices in the fctory without using excessively high voltges, technicins introduce cpcitnce in the circuits to shift the phse. Quick Quiz 33.6 An AC source drives n LC circuit with fixed voltge mplitude. If the driving frequency is v 1, the circuit is more cpcitive thn inductive nd the phse ngle is 210. If the driving frequency is v 2, the circuit is more inductive thn cpcitive nd the phse ngle is 110. At wht frequency is the lrgest mount of power delivered to the circuit? () It is lrgest t v 1. () It is lrgest t v 2. (c) The sme mount of power is delivered t oth frequencies.
15 33.7 esonnce in Series LC Circuit 967 Exmple 33.5 Averge Power in n LC Series Circuit Clculte the verge power delivered to the series LC circuit descried in Exmple SOLUTION Conceptulize Consider the circuit in Active Figure nd imgine energy eing delivered to the circuit y the AC source. eview Exmple 33.4 for other detils out this circuit. Ctegorize We find the result y using equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution 33.5 nd the mximum voltge from Exmple 33.4 to find the rms voltge from the source: Similrly, find the rms current in the circuit: Use Eqution to find the power delivered y the source: DV rms 5 DV mx " V " V I rms 5 I mx " A A "2 P vg 5 I rms V rms cos f 5 (0.207 A)(106 V) cos (234.0 ) W 33.7 esonnce in Series LC Circuit We investigted resonnce in mechnicl oscillting systems in Chpter 15. A series LC circuit is n electricl oscillting system. Such circuit is sid to e in resonnce when the driving frequency is such tht the rms current hs its mximum vlue. In generl, the rms current cn e written I rms 5 DV rms (33.33) Z where Z is the impednce. Sustituting the expression for Z from Eqution into Eqution gives DV rms I rms 5 (33.34) " 2 1 1X L 2 X C 2 2 Becuse the impednce depends on the frequency of the source, the current in the LC circuit lso depends on the frequency. The ngulr frequency v 0 t which X L 2 X C 5 0 is clled the resonnce frequency of the circuit. To find v 0, we set X L 5 X C, which gives v 0 L 5 1/v 0 C, or v (33.35) "LC This frequency lso corresponds to the nturl frequency of oscilltion of n LC circuit (see Section 32.5). Therefore, the rms current in series LC circuit hs its mximum vlue when the frequency of the pplied voltge mtches the nturl oscilltor frequency, which depends only on L nd C. Furthermore, t the resonnce frequency, the current is in phse with the pplied voltge. esonnce frequency Quick Quiz 33.7 Wht is the impednce of series LC circuit t resonnce? () lrger thn () less thn (c) equl to (d) impossile to determine A plot of rms current versus ngulr frequency for series LC circuit is shown in Active Figure on pge 968. The dt ssume constnt DV rms mv, L mh, nd C nf. The three curves correspond to three vlues of. In
16 968 CHAPTE 33 AlterntingCurrent Circuits ACTIVE FIGUE () The rms current versus frequency for series LC circuit for three vlues of. () Averge power delivered to the circuit versus frequency for the series LC circuit for three vlues of. I rms (ma) The current reches its mximum vlue t the resonnce frequency v P vg (mw) As the resistnce increses, v 0 t the hlfpower point increses v v v (Mrd/s) 12 1 v v (Mrd/s) 12 ech cse, the rms current hs its mximum vlue t the resonnce frequency v 0. Furthermore, the curves ecome nrrower nd tller s the resistnce decreses. Eqution shows tht when 5 0, the current ecomes infinite t resonnce. el circuits, however, lwys hve some resistnce, which limits the vlue of the current to some finite vlue. We cn lso clculte the verge power s function of frequency for series LC circuit. Using Equtions 33.32, 33.33, nd gives Averge power s function of frequency in n LC circuit Qulity fctor P vg 5 I 2 rms 5 1DV rms 2 2 Z 2 5 1DV rms X L 2 X C 2 2 (33.36) Becuse X L 5 vl, X C 5 1/vC, nd v /LC, the term (X L 2 X C ) 2 cn e expressed s 1X L 2 X C vl 2 1 vc 2 5 L2 v 2 1v2 2v Using this result in Eqution gives 1DV rms 2 2 v 2 P vg 5 2 v 2 1 L 2 1v 2 2v 02 2 (33.37) 2 Eqution shows tht t resonnce, when v 5 v 0, the verge power is mximum nd hs the vlue (DV rms ) 2 /. Active Figure is plot of verge power versus frequency for three vlues of in series LC circuit. As the resistnce is mde smller, the curve ecomes shrper in the vicinity of the resonnce frequency. This curve shrpness is usully descried y dimensionless prmeter known s the qulity fctor, 2 denoted y Q: Q 5 v 0 Dv where Dv is the width of the curve mesured etween the two vlues of v for which P vg hs onehlf its mximum vlue, clled the hlfpower points (see Active Fig ) It is left s prolem (Prolem 76) to show tht the width t the hlfpower points hs the vlue Dv 5 /L so tht 2 The qulity fctor is lso defined s the rtio 2pE/DE, where E is the energy stored in the oscillting system nd DE is the energy decrese per cycle of oscilltion due to the resistnce.
17 33.8 The Trnsformer nd Power Trnsmission 969 Q 5 v 0L (33.38) A rdio s receiving circuit is n importnt ppliction of resonnt circuit. The rdio is tuned to prticulr sttion (which trnsmits n electromgnetic wve or signl of specific frequency) y vrying cpcitor, which chnges the receiving circuit s resonnce frequency. When the circuit is driven y the electromgnetic oscilltions rdio signl produces in n ntenn, the tuner circuit responds with lrge mplitude of electricl oscilltion only for the sttion frequency tht mtches the resonnce frequency. Therefore, only the signl from one rdio sttion is pssed on to the mplifier nd loudspekers even though signls from ll sttions re driving the circuit t the sme time. Becuse mny signls re often present over rnge of frequencies, it is importnt to design highq circuit to eliminte unwnted signls. In this mnner, sttions whose frequencies re ner ut not equl to the resonnce frequency give signls t the receiver tht re negligily smll reltive to the signl tht mtches the resonnce frequency. Exmple 33.6 A esonting Series LC Circuit Consider series LC circuit for which V, L mh, DV rms V, nd v s 21. Determine the vlue of the cpcitnce for which the current is mximum. SOLUTION Conceptulize Consider the circuit in Active Figure nd imgine vrying the frequency of the AC source. The current in the circuit hs its mximum vlue t the resonnce frequency v 0. Ctegorize We find the result y using equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution to solve for the required cpcitnce in terms of the resonnce frequency: v "LC S C 5 1 v 0 2 L Sustitute numericl vlues: C mf s H The Trnsformer nd Power Trnsmission As discussed in Section 27.6, it is economicl to use high voltge nd low current to minimize the I 2 loss in trnsmission lines when electric power is trnsmitted over gret distnces. Consequently, 350kV lines re common, nd in mny res, even highervoltge (765kV) lines re used. At the receiving end of such lines, the consumer requires power t low voltge (for sfety nd for efficiency in design). In prctice, the voltge is decresed to pproximtely V t distriuting sttion, then to V for delivery to residentil res, nd finlly to 120 V nd 240 V t the customer s site. Therefore, device is needed tht cn chnge the lternting voltge nd current without cusing pprecile chnges in the power delivered. The AC trnsformer is tht device. In its simplest form, the AC trnsformer consists of two coils of wire wound round core of iron s illustrted in Figure (Compre this rrngement to Frdy s experiment in Active Figure 31.2.) The coil on the left, which is connected to the input lterntingvoltge source nd hs N 1 turns, is clled the primry winding (or the primry). The coil on the right, consisting of N 2 turns nd connected to lod resistor L, is clled the secondry winding (or the secondry). The purposes of the iron core re to increse the mgnetic flux through the coil nd to provide An lternting voltge v 1 is pplied to the primry coil, nd the output voltge v 2 is cross the resistor of resistnce L. v 1 Primry (input) Soft iron N 1 N 2 v 2 Secondry (output) Figure An idel trnsformer consists of two coils wound on the sme iron core. L
18 970 CHAPTE 33 AlterntingCurrent Circuits medium in which nerly ll the mgnetic field lines through one coil pss through the other coil. Eddycurrent losses re reduced y using lminted core. Trnsformtion of energy to internl energy in the finite resistnce of the coil wires is usully quite smll. Typicl trnsformers hve power efficiencies from 90% to 99%. In the discussion tht follows, let s ssume we re working with n idel trnsformer, one in which the energy losses in the windings nd core re zero. Frdy s lw sttes tht the voltge Dv 1 cross the primry is Dv 1 52N 1 df B dt (33.39) where F B is the mgnetic flux through ech turn. If we ssume ll mgnetic field lines remin within the iron core, the flux through ech turn of the primry equls the flux through ech turn of the secondry. Hence, the voltge cross the secondry is Dv 2 52N df B 2 (33.40) dt Solving Eqution for df B /dt nd sustituting the result into Eqution gives Dv 2 5 N 2 N 1 Dv 1 (33.41) I 1 I 2 v 1 L v 2 N 1 N 2 Figure Circuit digrm for trnsformer. Bettmnn/COBIS Nikol Tesl Americn Physicist ( ) Tesl ws orn in Croti, ut he spent most of his professionl life s n inventor in the United Sttes. He ws key figure in the development of lterntingcurrent electricity, highvoltge trnsformers, nd the trnsport of electricl power using AC trnsmission lines. Tesl s viewpoint ws t odds with the ides of Thoms Edison, who committed himself to the use of direct current in power trnsmission. Tesl s AC pproch won out. When N 2. N 1, the output voltge Dv 2 exceeds the input voltge Dv 1. This configurtion is referred to s stepup trnsformer. When N 2, N 1, the output voltge is less thn the input voltge, nd we hve stepdown trnsformer. A circuit digrm for trnsformer connected to lod resistnce is shown in Figure When current I 1 exists in the primry circuit, current I 2 is induced in the secondry. (In this discussion, uppercse I nd DV refer to rms vlues.) If the lod in the secondry circuit is pure resistnce, the induced current is in phse with the induced voltge. The power supplied to the secondry circuit must e provided y the AC source connected to the primry circuit. In n idel trnsformer where there re no losses, the power I 1 DV 1 supplied y the source is equl to the power I 2 DV 2 in the secondry circuit. Tht is, I 1 DV 1 5 I 2 DV 2 (33.42) The vlue of the lod resistnce L determines the vlue of the secondry current ecuse I 2 5 DV 2 / L. Furthermore, the current in the primry is I 1 5 DV 1 / eq, where eq 5 N 2 1 L (33.43) N 2 is the equivlent resistnce of the lod resistnce when viewed from the primry side. We see from this nlysis tht trnsformer my e used to mtch resistnces etween the primry circuit nd the lod. In this mnner, mximum power trnsfer cn e chieved etween given power source nd the lod resistnce. For exmple, trnsformer connected etween the 1kV output of n udio mplifier nd n 8V speker ensures tht s much of the udio signl s possile is trnsferred into the speker. In stereo terminology, this process is clled impednce mtching. To operte properly, mny common household electronic devices require low voltges. A smll trnsformer tht plugs directly into the wll like the one illustrted in Figure cn provide the proper voltge. The photogrph shows the two windings wrpped round common iron core tht is found inside ll these little lck oxes. This prticulr trnsformer converts the 120V AC in the wll socket to 12.5V AC. (Cn you determine the rtio of the numers of turns in the two coils?) Some lck oxes lso mke use of diodes to convert the lternting current to direct current. (See Section 33.9.)
19 33.8 The Trnsformer nd Power Trnsmission 971 The primry winding in this trnsformer is ttched to the prongs of the plug, wheres the secondry winding is connected to the power cord on the right. Figure Electronic devices re often powered y AC dptors contining trnsformers such s this one. These dptors lter the AC voltge. In mny pplictions, the dptors lso convert lternting current to direct current.. Cengge Lerning/George Semple This trnsformer is smller thn the one in the opening photogrph of this chpter. In ddition, it is stepdown trnsformer. It drops the voltge from V to 240 V for delivery to group of residences.. Cengge Lerning/George Semple Exmple 33.7 The Economics of AC Power An electricitygenerting sttion needs to deliver energy t rte of 20 MW to city 1.0 km wy. A common voltge for commercil power genertors is 22 kv, ut stepup trnsformer is used to oost the voltge to 230 kv efore trnsmission. (A) If the resistnce of the wires is 2.0 V nd the energy costs re out 11./kWh, estimte the cost of the energy converted to internl energy in the wires during one dy. SOLUTION Conceptulize The resistnce of the wires is in series with the resistnce representing the lod (homes nd usinesses). Therefore, there is voltge drop in the wires, which mens tht some of the trnsmitted energy is converted to internl energy in the wires nd never reches the lod. Ctegorize This prolem involves finding the power delivered to resistive lod in n AC circuit. Let s ignore ny cpcitive or inductive chrcteristics of the lod nd set the power fctor equl to 1. Anlyze Clculte I rms in the wires from Eqution 33.31: I rms 5 P vg DV rms W V 5 87 A Determine the rte t which energy is delivered to the resistnce in the wires from Eqution 33.32: P wires 5 I 2 rms A V kw Clculte the energy T ET delivered to the wires over the course of dy: T ET 5 P wires Dt kw2124 h kwh Find the cost of this energy t rte of 11./kWh: Cost 5 (363 kwh)($0.11/kwh) 5 $40 (B) epet the clcultion for the sitution in which the power plnt delivers the energy t its originl voltge of 22 kv. SOLUTION Clculte I rms in the wires from Eqution 33.31: I rms 5 P vg W DV rms V A From Eqution 33.32, determine the rte t which energy is delivered to the resistnce in the wires: P wires 5 I 2 rms A V kw continued
20 972 CHAPTE 33 AlterntingCurrent Circuits 33.7 cont. Clculte the energy delivered to the wires over the course of dy: T ET 5 P wires Dt kw2124 h kwh Find the cost of this energy t rte of 11./kWh: Cost 5 ( kwh)($0.11/kwh) 5 $ Finlize Notice the tremendous svings tht re possile through the use of trnsformers nd highvoltge trnsmission lines. Such svings in comintion with the efficiency of using lternting current to operte motors led to the universl doption of lternting current insted of direct current for commercil power grids ectifiers nd Filters Portle electronic devices such s rdios nd lptop computers re often powered y direct current supplied y tteries. Mny devices come with AC DC converters such s tht shown in Figure Such converter contins trnsformer tht steps the voltge down from 120 V to, typiclly, 6 V or 9 V nd circuit tht converts lternting current to direct current. The AC DC converting process is clled rectifiction, nd the converting device is clled rectifier. The most importnt element in rectifier circuit is diode, circuit element tht conducts current in one direction ut not the other. Most diodes used in modern electronics re semiconductor devices. The circuit symol for diode is, where the rrow indictes the direction of the current in the diode. A diode hs low resistnce to current in one direction (the direction of the rrow) nd high resistnce to current in the opposite direction. To understnd how diode rectifies current, consider Figure 33.21, which shows diode nd resistor connected to the secondry of trnsformer. The trnsformer reduces the voltge from 120V AC to the lower voltge tht is needed for the device hving resistnce (the lod resistnce). Becuse the diode conducts current in only one direction, the lternting current in the lod resistor is reduced to the form shown y the solid curve in Figure The diode conducts current only when the side of the symol contining the rrowhed hs positive potentil reltive to the other side. In this sitution, the diode cts s hlfwve rectifier ecuse current is present in the circuit only during hlf of ech cycle. Diode C Primry (input) i The solid curve represents the current in the resistor with no filter cpcitor, nd the dshed curve is the current when the circuit includes the cpcitor. Figure () A hlfwve rectifier with n optionl filter cpcitor. () Current versus time in the resistor. t
21 33.9 ectifiers nd Filters 973 C The output voltge of the filter ecomes very close to the input voltge s the frequency ecomes lrge. V out / V in 1 ACTIVE FIGUE () A simple C highpss filter. () tio of output voltge to input voltge for n C highpss filter s function of the ngulr frequency of the AC source. v in v out log v When cpcitor is dded to the circuit s shown y the dshed lines nd the cpcitor symol in Figure 33.21, the circuit is simple DC power supply. The time vrition of the current in the lod resistor (the dshed curve in Fig ) is close to eing zero, s determined y the C time constnt of the circuit. As the current in the circuit egins to rise t t 5 0 in Figure 33.21, the cpcitor chrges up. When the current egins to fll, however, the cpcitor dischrges through the resistor, so the current in the resistor does not fll s quickly s the current from the trnsformer. The C circuit in Figure is one exmple of filter circuit, which is used to smooth out or eliminte timevrying signl. For exmple, rdios re usully powered y 60Hz lternting voltge. After rectifiction, the voltge still contins smll AC component t 60 Hz (sometimes clled ripple), which must e filtered. By filtered, we men tht the 60Hz ripple must e reduced to vlue much less thn tht of the udio signl to e mplified ecuse without filtering, the resulting udio signl includes n nnoying hum t 60 Hz. We cn lso design filters tht respond differently to different frequencies. Consider the simple series C circuit shown in Active Figure The input voltge is cross the series comintion of the two elements. The output is the voltge cross the resistor. A plot of the rtio of the output voltge to the input voltge s function of the logrithm of ngulr frequency (see Active Fig ) shows tht t low frequencies, DV out is much smller thn DV in, wheres t high frequencies, the two voltges re equl. Becuse the circuit preferentilly psses signls of higher frequency while locking lowfrequency signls, the circuit is clled n C highpss filter. (See Prolem 54 for n nlysis of this filter.) Physiclly, highpss filter works ecuse cpcitor locks out direct current nd AC current t low frequencies. At low frequencies, the cpcitive rectnce is lrge nd much of the pplied voltge ppers cross the cpcitor rther thn cross the output resistor. As the frequency increses, the cpcitive rectnce drops nd more of the pplied voltge ppers cross the resistor. Now consider the circuit shown in Active Figure 33.23, where we hve interchnged the resistor nd cpcitor nd where the output voltge is tken cross the cpcitor. At low frequencies, the rectnce of the cpcitor nd the voltge cross the cpcitor is high. As the frequency increses, the voltge cross the cpcitor drops. Therefore, this filter is n C lowpss filter. The rtio of output voltge to input voltge (see Prolem 56), plotted s function of the logrithm of v in Active Figure 33.23, shows this ehvior. You my e fmilir with crossover networks, which re n importnt prt of the speker systems for highqulity udio systems. These networks use lowpss filters to direct low frequencies to specil type of speker, the woofer, which is designed to reproduce the low notes ccurtely. The high frequencies re sent to the tweeter speker. v in The output voltge of the filter ecomes very close to the input voltge s the frequency ecomes smll. V out / V in 1 ACTIVE FIGUE C v out log v () A simple C lowpss filter. () tio of output voltge to input voltge for n C lowpss filter s function of the ngulr frequency of the AC source.
22 974 CHAPTE 33 AlterntingCurrent Circuits Definitions Summry In AC circuits tht contin inductors nd cpcitors, it is useful to define the inductive rectnce X L nd the cpcitive rectnce X C s X L ; vl (33.10) X C ; 1 (33.18) vc where v is the ngulr frequency of the AC source. The SI unit of rectnce is the ohm. The impednce Z of n LC series AC circuit is Z ; " 2 1 1X L 2 X C 2 2 (33.25) This expression illustrtes tht we cnnot simply dd the resistnce nd rectnces in circuit. We must ccount for the pplied voltge nd current eing out of phse, with the phse ngle f etween the current nd voltge eing f5tn 21 X L 2 X C (33.27) The sign of f cn e positive or negtive, depending on whether X L is greter or less thn X C. The phse ngle is zero when X L 5 X C. Concepts nd Principles The rms current nd rms voltge in n AC circuit in which the voltges nd current vry sinusoidlly re given y I rms 5 I mx " I mx (33.4) DV rms 5 DV mx " DV mx (33.5) where I mx nd DV mx re the mximum vlues. If n AC circuit consists of source nd resistor, the current is in phse with the voltge. Tht is, the current nd voltge rech their mximum vlues t the sme time. If n AC circuit consists of source nd n inductor, the current lgs the voltge y 90. Tht is, the voltge reches its mximum vlue onequrter of period efore the current reches its mximum vlue. If n AC circuit consists of source nd cpcitor, the current leds the voltge y 90. Tht is, the current reches its mximum vlue onequrter of period efore the voltge reches its mximum vlue. The verge power delivered y the source in n LC circuit is P vg 5 I rms DV rms cos f (33.31) An equivlent expression for the verge power is The rms current in series LC circuit is I rms 5 DV rms " 2 1 1X L 2 X C 2 2 (33.34) P vg 5 I 2 rms (33.32) The verge power delivered y the source results in incresing internl energy in the resistor. No power loss occurs in n idel inductor or cpcitor. A series LC circuit is in resonnce when the inductive rectnce equls the cpcitive rectnce. When this condition is met, the rms current given y Eqution hs its mximum vlue. The resonnce frequency v 0 of the circuit is v (33.35) "LC The rms current in series LC circuit hs its mximum vlue when the frequency of the source equls v 0, tht is, when the driving frequency mtches the resonnce frequency. AC trnsformers llow for esy chnges in lternting voltge ccording to Dv 2 5 N 2 Dv 1 (33.41) N 1 where N 1 nd N 2 re the numers of windings on the primry nd secondry coils, respectively, nd Dv 1 nd Dv 2 re the voltges on these coils.
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