Direct Current Circuits. Chapter Outline Electromotive Force 28.2 Resistors in Series and in Parallel 28.3 Kirchhoff s Rules 28.


 Marsha Carpenter
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1 P U Z Z L E R If ll these pplinces were operting t one time, circuit reker would proly e tripped, preventing potentilly dngerous sitution. Wht cuses circuit reker to trip when too mny electricl devices re plugged into one circuit? (George Semple) c h p t e r Direct Current Circuits Chpter Outline 28.1 Electromotive Force 28.2 Resistors in Series nd in Prllel 28.3 Kirchhoff s Rules 28.4 RC Circuits 28.5 (Optionl) Electricl Instruments 28.6 (Optionl) Household Wiring nd Electricl Sfety 868
2 T his chpter is concerned with the nlysis of some simple electric circuits tht contin tteries, resistors, nd cpcitors in vrious comintions. The nlysis of these circuits is simplified y the use of two rules known s Kirchhoff s rules, which follow from the lws of conservtion of energy nd conservtion of electric chrge. Most of the circuits nlyzed re ssumed to e in stedy stte, which mens tht the currents re constnt in mgnitude nd direction. In Section 28.4 we discuss circuits in which the current vries with time. Finlly, we descrie vriety of common electricl devices nd techniques for mesuring current, potentil difference, resistnce, nd emf Electromotive Force ELECTROMOTIVE FORCE In Section 27.6 we found tht constnt current cn e mintined in closed circuit through the use of source of emf, which is device (such s ttery or genertor) tht produces n electric field nd thus my cuse chrges to move round circuit. One cn think of source of emf s chrge pump. When n electric potentil difference exists etween two points, the source moves chrges uphill from the lower potentil to the higher. The emf descries the work done per unit chrge, nd hence the SI unit of emf is the volt. Consider the circuit shown in Figure 28.1, consisting of ttery connected to resistor. We ssume tht the connecting wires hve no resistnce. The positive terminl of the ttery is t higher potentil thn the negtive terminl. If we neglect the internl resistnce of the ttery, the potentil difference cross it (clled the terminl voltge) equls its emf. However, ecuse rel ttery lwys hs some internl resistnce r, the terminl voltge is not equl to the emf for ttery in circuit in which there is current. To understnd why this is so, consider the circuit digrm in Figure 28.2, where the ttery of Figure 28.1 is represented y the dshed rectngle contining n emf in series with n internl resistnce r. Now imgine moving through the ttery clockwise from to nd mesuring the electric potentil t vrious loctions. As we pss from the negtive terminl to the positive terminl, the potentil increses y n mount. However, s we move through the resistnce r, the potentil decreses y n mount Ir, where I is the current in the circuit. Thus, the terminl voltge of the ttery is 1 V V V Bttery + Resistor Figure 28.1 A circuit consisting of resistor connected to the terminls of ttery. 1 The terminl voltge in this cse is less thn the emf y n mount Ir. In some situtions, the terminl voltge my exceed the emf y n mount Ir. This hppens when the direction of the current is opposite tht of the emf, s in the cse of chrging ttery with nother source of emf.
3 870 CHAPTER 28 Direct Current Circuits I IR V Figure d r R () () () Circuit digrm of source of emf (in this cse, ttery), of internl resistnce r, connected to n externl resistor of resistnce R. () Grphicl representtion showing how the electric potentil chnges s the circuit in prt () is trversed clockwise. r c c R Ir d I V Ir (28.1) From this expression, note tht is equivlent to the opencircuit voltge tht is, the terminl voltge when the current is zero. The emf is the voltge leled on ttery for exmple, the emf of D cell is 1.5 V. The ctul potentil difference etween the terminls of the ttery depends on the current through the ttery, s descried y Eqution Figure 28.2 is grphicl representtion of the chnges in electric potentil s the circuit is trversed in the clockwise direction. By inspecting Figure 28.2, we see tht the terminl voltge V must equl the potentil difference cross the externl resistnce R, often clled the lod resistnce. The lod resistor might e simple resistive circuit element, s in Figure 28.1, or it could e the resistnce of some electricl device (such s toster, n electric heter, or lightul) connected to the ttery (or, in the cse of household devices, to the wll outlet). The resistor represents lod on the ttery ecuse the ttery must supply energy to operte the device. The potentil difference cross the lod resistnce is V IR. Comining this expression with Eqution 28.1, we see tht IR Ir (28.2) Solving for the current gives I (28.3) R r This eqution shows tht the current in this simple circuit depends on oth the lod resistnce R externl to the ttery nd the internl resistnce r. If R is much greter thn r, s it is in mny relworld circuits, we cn neglect r. If we multiply Eqution 28.2 y the current I, we otin I I 2 R I 2 r (28.4) This eqution indictes tht, ecuse power I V (see Eq ), the totl power output I of the ttery is delivered to the externl lod resistnce in the mount I 2 R nd to the internl resistnce in the mount I 2 r. Agin, if r V R, then most of the power delivered y the ttery is trnsferred to the lod resistnce. EXAMPLE 28.1 Terminl Voltge of Bttery A ttery hs n emf of 12.0 V nd n internl resistnce of Its terminls re connected to lod resistnce of () Find the current in the circuit nd the terminl voltge of the ttery. Solution Using first Eqution 28.3 nd then Eqution 28.1, we otin 12.0 V I 3.93 A R r 3.05 V Ir 12.0 V (3.93 A)(0.05 ) To check this result, we cn clculte the voltge cross the lod resistnce R : V IR (3.93 A)(3.00 ) 11.8 V 11.8 V () Clculte the power delivered to the lod resistor, the power delivered to the internl resistnce of the ttery, nd the power delivered y the ttery. Solution The power delivered to the lod resistor is R I 2 R (3.93 A) 2 (3.00 ) The power delivered to the internl resistnce is r I 2 r (3.93 A) 2 (0.05 ) 46.3 W W Hence, the power delivered y the ttery is the sum of these quntities, or 47.1 W. You should check this result, using the expression I.
4 28.2 Resistors in Series nd in Prllel 871 EXAMPLE 28.2 Mtching the Lod Show tht the mximum power delivered to the lod resistnce R in Figure 28.2 occurs when the lod resistnce mtches the internl resistnce tht is, when R r. mx Solution The power delivered to the lod resistnce is equl to I 2 R, where I is given y Eqution 28.3: I 2 R 2 R When is plotted versus R s in Figure 28.3, we find tht reches mximum vlue of 2 /4r t R r. We cn lso prove this y differentiting with respect to R, setting the result equl to zero, nd solving for R. The detils re left s prolem for you to solve (Prolem 57). (R r) 2 r 2r 3r Figure 28.3 Grph of the power delivered y ttery to lod resistor of resistnce R s function of R. The power delivered to the resistor is mximum when the lod resistnce equls the internl resistnce of the ttery. R 28.2 RESISTORS IN SERIES AND IN PARALLEL Suppose tht you nd your friends re t crowded sketll gme in sports ren nd decide to leve erly. You hve two choices: (1) your whole group cn exit through single door nd wlk down long hllwy contining severl concession stnds, ech surrounded y lrge crowd of people witing to uy food or souvenirs; or () ech memer of your group cn exit through seprte door in the min hll of the ren, where ech will hve to push his or her wy through single group of people stnding y the door. In which scenrio will less time e required for your group to leve the ren? It should e cler tht your group will e le to leve fster through the seprte doors thn down the hllwy where ech of you hs to push through severl groups of people. We could descrie the groups of people in the hllwy s cting in series, ecuse ech of you must push your wy through ll of the groups. The groups of people round the doors in the ren cn e descried s cting in prllel. Ech memer of your group must push through only one group of people, nd ech memer pushes through different group of people. This simple nlogy will help us understnd the ehvior of currents in electric circuits contining more thn one resistor. When two or more resistors re connected together s re the lightuls in Figure 28.4, they re sid to e in series. Figure 28.4 is the circuit digrm for the lightuls, which re shown s resistors, nd the ttery. In series connection, ll the chrges moving through one resistor must lso pss through the second resistor. (This is nlogous to ll memers of your group pushing through the crowds in the single hllwy of the sports ren.) Otherwise, chrge would ccumulte etween the resistors. Thus, for series comintion of resistors, the currents in the two resistors re the sme ecuse ny chrge tht psses through R 1 must lso pss through R 2. The potentil difference pplied cross the series comintion of resistors will divide etween the resistors. In Figure 28.4, ecuse the voltge drop 2 from to 2 The term voltge drop is synonymous with decrese in electric potentil cross resistor nd is used often y individuls working with electric circuits.
5 872 CHAPTER 28 Direct Current Circuits R 1 R 2 R 1 R 2 c R eq c + I I I Bttery V + V + () Figure 28.4 () () A series connection of two resistors R 1 nd R 2. The current in R 1 is the sme s tht in R 2. () Circuit digrm for the tworesistor circuit. (c) The resistors replced with single resistor hving n equivlent resistnce R eq R 1 R 2. (c) equls IR 1 nd the voltge drop from to c equls IR 2, the voltge drop from to c is V IR 1 IR 2 I(R 1 R 2 ) Therefore, we cn replce the two resistors in series with single resistor hving n equivlent resistnce R eq, where R eq R 1 R 2 (28.5) The resistnce R eq is equivlent to the series comintion R 1 R 2 in the sense tht the circuit current is unchnged when R eq replces R 1 R 2. The equivlent resistnce of three or more resistors connected in series is R eq R 1 R 2 R 3 (28.6) This reltionship indictes tht the equivlent resistnce of series connection of resistors is lwys greter thn ny individul resistnce. Quick Quiz 28.1 If piece of wire is used to connect points nd c in Figure 28.4, does the rightness of ul R 1 increse, decrese, or sty the sme? Wht hppens to the rightness of ul R 2? A series connection of three lightuls, ll rted t 120 V ut hving power rtings of 60 W, 75 W, nd 200 W. Why re the intensities of the uls different? Which ul hs the gretest resistnce? How would their reltive intensities differ if they were connected in prllel? Now consider two resistors connected in prllel, s shown in Figure When the current I reches point in Figure 28.5, clled junction, it splits into two prts, with I 1 going through R 1 nd I 2 going through R 2. A junction is ny point in circuit where current cn split ( just s your group might split up nd leve the ren through severl doors, s descried erlier.) This split results in less current in ech individul resistor thn the current leving the ttery. Becuse chrge must e conserved, the current I tht enters point must equl the totl current leving tht point: I I 1 I 2
6 28.2 Resistors in Series nd in Prllel 873 R 1 R 2 + () Figure 28.5 Bttery I 1 I I 2 R 1 R 2 V + () I R eq V + () A prllel connection of two resistors R 1 nd R 2. The potentil difference cross R 1 is the sme s tht cross R 2. () Circuit digrm for the tworesistor circuit. (c) The resistors replced with single resistor hving n equivlent resistnce R eq (R 1 1 R 1 2 ) 1. (c) QuickL Tpe one pir of drinking strws end to end, nd tpe second pir side y side. Which pir is esier to low through? Wht would hppen if you were compring three strws tped end to end with three tped side y side? As cn e seen from Figure 28.5, oth resistors re connected directly cross the terminls of the ttery. Thus, when resistors re connected in prllel, the potentil differences cross them re the sme. Strws in series Strws in prllel Becuse the potentil differences cross the resistors re the sme, the expression V IR gives From this result, we see tht the equivlent resistnce of two resistors in prllel is given y or I I 1 I 2 V R 1 V V R V R 1 R 2 R eq 1 R eq 1 R 1 1 R 2 R eq 1 1 R 1 1 R 2 An extension of this nlysis to three or more resistors in prllel gives (28.7) 1 R eq 1 R 1 1 R 2 1 R 3 (28.8) The equivlent resistnce of severl resistors in prllel
7 874 CHAPTER 28 Direct Current Circuits We cn see from this expression tht the equivlent resistnce of two or more resistors connected in prllel is lwys less thn the lest resistnce in the group. Household circuits re lwys wired such tht the pplinces re connected in prllel. Ech device opertes independently of the others so tht if one is switched off, the others remin on. In ddition, the devices operte on the sme voltge. Three lightuls hving power rtings of 25 W, 75 W, nd 150 W, connected in prllel to voltge source of out 100 V. All uls re rted t the sme voltge. Why do the intensities differ? Which ul drws the most current? Which hs the lest resistnce? Quick Quiz 28.2 Assume tht the ttery of Figure 28.1 hs zero internl resistnce. If we dd second resistor in series with the first, does the current in the ttery increse, decrese, or sty the sme? How out the potentil difference cross the ttery terminls? Would your nswers chnge if the second resistor were connected in prllel to the first one? Quick Quiz 28.3 Are utomoile hedlights wired in series or in prllel? How cn you tell? EXAMPLE 28.3 Find the Equivlent Resistnce Four resistors re connected s shown in Figure () Find the equivlent resistnce etween points nd c. Solution The comintion of resistors cn e reduced in steps, s shown in Figure The 8.0 nd 4.0 resistors re in series; thus, the equivlent resistnce etween nd is 12 (see Eq. 28.5). The 6.0 nd 3.0 resistors re in prllel, so from Eqution 28.7 we find tht the equivlent resistnce from to c is 2.0. Hence, the equivlent resistnce from to c is 14. () Wht is the current in ech resistor if potentil difference of 42V is mintined etween nd c? Solution The currents in the 8.0 nd 4.0 resistors re the sme ecuse they re in series. In ddition, this is the sme s the current tht would exist in the 14 equivlent resistor suject to the 42V potentil difference. Therefore, using Eqution 27.8 (R V/I ) nd the results from prt (), we otin I V c 42 V 3.0 A R eq 14 This is the current in the 8.0 nd 4.0 resistors. When this 3.0A current enters the junction t, however, it splits, with prt pssing through the 6.0 resistor (I 1 ) nd prt through the 3.0 resistor (I 2 ). Becuse the potentil difference is V c cross ech of these resistors (since they re in prllel), we see tht (6.0 ) I 1 (3.0 )I 2, or I 2 2I 1. Using this result nd the fct tht I 1 I A, we find tht I A nd I A. We could hve guessed this t the strt y noting tht the current through the 3.0 resistor hs to e twice tht through the 6.0 resistor, in view of their reltive resistnces nd the fct tht the sme voltge is pplied to ech of them. As finl check of our results, note tht V c (6.0 )I 1 (3.0 )I V nd V (12 )I 36 V; therefore, V c V V c 42 V, s it must. () () (c) 8.0 Ω I 12 Ω I Ω I 2 14 Ω Figure Ω 3.0 Ω 2.0 Ω c c c
8 28.2 Resistors in Series nd in Prllel 875 EXAMPLE 28.4 Three Resistors in Prllel Three resistors re connected in prllel s shown in Figure A potentil difference of 18 V is mintined etween points nd. () Find the current in ech resistor. Solution The resistors re in prllel, nd so the potentil difference cross ech must e 18 V. Applying the reltionship V IR to ech resistor gives I 1 V R 1 I 2 V R 2 I 3 V R 3 () Clculte the power delivered to ech resistor nd the totl power delivered to the comintion of resistors. Solution We pply the reltionship ( V ) 2 /R to ech resistor nd otin 1 V 2 R 1 2 V 2 R 2 3 V 2 R 3 18 V V V 9.0 (18 V)2 3.0 (18 V)2 6.0 (18 V) A 3.0 A 2.0 A 110 W 54 W 36 W This shows tht the smllest resistor receives the most power. Summing the three quntities gives totl power of 200 W. (c) Clculte the equivlent resistnce of the circuit. Solution We cn use Eqution 28.8 to find R eq : Exercise R eq Use R eq to clculte the totl power delivered y the ttery. Answer 200 W. Figure R eq V I Three resistors connected in prllel. The voltge cross ech resistor is 18 V. I 1 I 2 I Ω 6.0 Ω 9.0 Ω EXAMPLE 28.5 Finding R eq y Symmetry Arguments Consider five resistors connected s shown in Figure Find the equivlent resistnce etween points nd. Solution In this type of prolem, it is convenient to ssume current entering junction nd then pply symmetry 5 Ω c 1 Ω 1 Ω 1 Ω 1 Ω 1/2 Ω 1/2 Ω 1 Ω 5 Ω c,d c,d 1 Ω 1 Ω 1 Ω 1 Ω d () () (c) (d) Figure 28.8 Becuse of the symmetry in this circuit, the 5 resistor does not contriute to the resistnce etween points nd nd therefore cn e disregrded when we clculte the equivlent resistnce.
9 876 CHAPTER 28 Direct Current Circuits rguments. Becuse of the symmetry in the circuit (ll 1 resistors in the outside loop), the currents in rnches c nd d must e equl; hence, the electric potentils t points c nd d must e equl. This mens tht V cd 0 nd, s result, points c nd d my e connected together without ffecting the circuit, s in Figure Thus, the 5 resistor my e removed from the circuit nd the remining circuit then reduced s in Figures 28.8c nd d. From this reduction we see tht the equivlent resistnce of the comintion is 1. Note tht the result is 1 regrdless of the vlue of the resistor connected etween c nd d. CONCEPTUAL EXAMPLE 28.6 Figure 28.9 illustrtes how threewy lightul is constructed to provide three levels of light intensity. The socket of the lmp is equipped with threewy switch for selecting different light intensities. The ul contins two filments. When the lmp is connected to 120V source, one filment receives 100 W of power, nd the other receives 75 W. Explin how the two filments re used to provide three different light intensities. Solution The three light intensities re mde possile y pplying the 120 V to one filment lone, to the other filment lone, or to the two filments in prllel. When switch S 1 is closed nd switch S 2 is opened, current psses only through the 75W filment. When switch S 1 is open nd switch S 2 is closed, current psses only through the 100W filment. When oth switches re closed, current psses through oth filments, nd the totl power is 175 W. If the filments were connected in series nd one of them were to rek, no current could pss through the ul, nd the ul would give no illumintion, regrdless of the switch position. However, with the filments connected in prllel, if one of them (for exmple, the 75W filment) reks, the ul will still operte in two of the switch positions s current psses through the other (100W) filment. Opertion of ThreeWy Lightul Exercise Determine the resistnces of the two filments nd their prllel equivlent resistnce. Answer 144, 192, Figure W filment 75W filment S 1 S V A threewy lightul. APPLICATION Strings of Lights Strings of lights re used for mny ornmentl purposes, such s decorting Christms trees. Over the yers, oth prllel nd series connections hve een used for multilight strings powered y 120 V. 3 Serieswired uls re sfer thn prllelwired uls for indoor Christmstree use ecuse serieswired uls operte with less light per ul nd t lower temperture. However, if the filment of single ul fils (or if the ul is removed from its socket), ll the lights on the string re extinguished. The populrity of serieswired light strings diminished ecuse trouleshooting filed ul ws tedious, timeconsuming chore tht involved trilnderror sustitution of good ul in ech socket long the string until the defective ul ws found. In prllelwired string, ech ul opertes t 120 V. By design, the uls re righter nd hotter thn those on serieswired string. As result, these uls re inherently more dngerous (more likely to strt fire, for instnce), ut if one ul in prllelwired string fils or is removed, the rest of the uls continue to glow. (A 25ul string of 4W uls results in power of 100 W; the totl power ecomes sustntil when severl strings re used.) A new design ws developed for soclled miniture lights wired in series, to prevent the filure of one ul from extinguishing the entire string. The solution is to crete connection (clled jumper) cross the filment fter it fils. (If n lternte connection existed cross the filment efore 3 These nd other household devices, such s the threewy lightul in Conceptul Exmple 28.6 nd the kitchen pplinces shown in this chpter s Puzzler, ctully operte on lternting current (c), to e introduced in Chpter 33.
10 28.3 Kirchhoff s Rules 877 it filed, ech ul would represent prllel circuit; in this circuit, the current would flow through the lternte connection, forming short circuit, nd the ul would not glow.) When the filment reks in one of these miniture lightuls, 120 V ppers cross the ul ecuse no current is present in the ul nd therefore no drop in potentil occurs cross the other uls. Inside the lightul, smll loop covered y n insulting mteril is wrpped round the filment leds. An rc urns the insultion nd connects the filment leds when 120 V ppers cross the ul tht is, when the filment fils. This short now completes the circuit through the ul even though the filment is no longer ctive (Fig ). Suppose tht ll the uls in 50ul miniturelight string re operting. A 2.4V potentil drop occurs cross ech ul ecuse the uls re in series. The power input to this style of ul is 0.34 W, so the totl power supplied to the string is only 17 W. We clculte the filment resistnce t the operting temperture to e (2.4 V) 2 /(0.34 W) 17. When the ul fils, the resistnce cross its terminls is reduced to zero ecuse of the lternte jumper connection mentioned in the preceding prgrph. All the other uls not only sty on ut glow more rightly ecuse the totl resistnce of the string is reduced nd consequently the current in ech ul increses. Let us ssume tht the operting resistnce of ul remins t 17 even though its temperture rises s result of the incresed current. If one ul fils, the potentil drop cross ech of the remining uls increses to 2.45 V, the current increses from 0.142A to A, nd the power increses to W. As more lights fil, the current keeps rising, the filment of ech ul opertes t higher temperture, nd the lifetime of the ul is reduced. It is therefore good ide to check for filed (nonglowing) uls in such serieswired string nd replce them s soon s possile, in order to mximize the lifetimes of ll the uls. Filment Jumper Glss insultor () () Figure () Schemtic digrm of modern miniture holidy lightul, with jumper connection to provide current pth if the filment reks. () A Christmstree lightul KIRCHHOFF S RULES As we sw in the preceding section, we cn nlyze simple circuits using the expression V IR nd the rules for series nd prllel comintions of resistors. Very often, however, it is not possile to reduce circuit to single loop. The procedure for nlyzing more complex circuits is gretly simplified if we use two principles clled Kirchhoff s rules: 1. The sum of the currents entering ny junction in circuit must equl the sum of the currents leving tht junction: I in I out (28.9)
11 878 CHAPTER 28 Direct Current Circuits 2. The sum of the potentil differences cross ll elements round ny closed circuit loop must e zero: V 0 closed loop (28.10) Kirchhoff s first rule is sttement of conservtion of electric chrge. All current tht enters given point in circuit must leve tht point ecuse chrge cnnot uild up t point. If we pply this rule to the junction shown in Figure 28.11, we otin Gustv Kirchhoff ( ) Kirchhoff, professor t Heidelerg, Germny, nd Roert Bunsen invented the spectroscope nd founded the science of spectroscopy, which we shll study in Chpter 40. They discovered the elements cesium nd ruidium nd invented stronomicl spectroscopy. Kirchhoff formulted nother Kirchhoff s rule, nmely, cool sustnce will sor light of the sme wvelengths tht it emits when hot. (AIP ESVA/W. F. Meggers Collection) QuickL Drw n ritrrily shped closed loop tht does not cross over itself. Lel five points on the loop,, c, d, nd e, nd ssign rndom numer to ech point. Now strt t nd work your wy round the loop, clculting the difference etween ech pir of djcent numers. Some of these differences will e positive, nd some will e negtive. Add the differences together, mking sure you ccurtely keep trck of the lgeric signs. Wht is the sum of the differences ll the wy round the loop? I 1 I 2 I 3 Figure represents mechnicl nlog of this sitution, in which wter flows through rnched pipe hving no leks. The flow rte into the pipe equls the totl flow rte out of the two rnches on the right. Kirchhoff s second rule follows from the lw of conservtion of energy. Let us imgine moving chrge round the loop. When the chrge returns to the strting point, the chrge circuit system must hve the sme energy s when the chrge strted from it. The sum of the increses in energy in some circuit elements must equl the sum of the decreses in energy in other elements. The potentil energy decreses whenever the chrge moves through potentil drop IR cross resistor or whenever it moves in the reverse direction through source of emf. The potentil energy increses whenever the chrge psses through ttery from the negtive terminl to the positive terminl. Kirchhoff s second rule pplies only for circuits in which n electric potentil is defined t ech point; this criterion my not e stisfied if chnging electromgnetic fields re present, s we shll see in Chpter 31. In justifying our clim tht Kirchhoff s second rule is sttement of conservtion of energy, we imgined crrying chrge round loop. When pplying this rule, we imgine trveling round the loop nd consider chnges in electric potentil, rther thn the chnges in potentil energy descried in the previous prgrph. You should note the following sign conventions when using the second rule: Becuse chrges move from the highpotentil end of resistor to the lowpotentil end, if resistor is trversed in the direction of the current, the chnge in potentil V cross the resistor is IR (Fig ). If resistor is trversed in the direction opposite the current, the chnge in potentil V cross the resistor is IR (Fig ). If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction of the emf (from to ), the chnge in potentil V is (Fig c). The emf of the ttery increses the electric potentil s we move through it in this direction. If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction opposite the emf (from to ), the chnge in potentil V is (Fig d). In this cse the emf of the ttery reduces the electric potentil s we move through it. Limittions exist on the numers of times you cn usefully pply Kirchhoff s rules in nlyzing given circuit. You cn use the junction rule s often s you need, so long s ech time you write n eqution you include in it current tht hs not een used in preceding junctionrule eqution. In generl, the numer of times you cn use the junction rule is one fewer thn the numer of junction
12 28.3 Kirchhoff s Rules 879 I 1 I 2 Flow in Figure I 3 () () Flow out () Kirchhoff s junction rule. Conservtion of chrge requires tht ll current entering junction must leve tht junction. Therefore, I 1 I 2 I 3. () A mechnicl nlog of the junction rule: the mount of wter flowing out of the rnches on the right must equl the mount flowing into the single rnch on the left. () () (c) I V = IR I V = +IR + V = + + (d) V = Figure Rules for determining the potentil chnges cross resistor nd ttery. (The ttery is ssumed to hve no internl resistnce.) Ech circuit element is trversed from left to right. points in the circuit. You cn pply the loop rule s often s needed, so long s new circuit element (resistor or ttery) or new current ppers in ech new eqution. In generl, in order to solve prticulr circuit prolem, the numer of independent equtions you need to otin from the two rules equls the numer of unknown currents. Complex networks contining mny loops nd junctions generte gret numers of independent liner equtions nd correspondingly gret numer of unknowns. Such situtions cn e hndled formlly through the use of mtrix lger. Computer progrms cn lso e written to solve for the unknowns. The following exmples illustrte how to use Kirchhoff s rules. In ll cses, it is ssumed tht the circuits hve reched stedystte conditions tht is, the currents in the vrious rnches re constnt. Any cpcitor cts s n open circuit; tht is, the current in the rnch contining the cpcitor is zero under stedystte conditions. ProlemSolving Hints Kirchhoff s Rules Drw circuit digrm, nd lel ll the known nd unknown quntities. You must ssign direction to the current in ech rnch of the circuit. Do not e lrmed if you guess the direction of current incorrectly; your result will e negtive, ut its mgnitude will e correct. Although the ssignment of current directions is ritrry, you must dhere rigorously to the ssigned directions when pplying Kirchhoff s rules. Apply the junction rule to ny junctions in the circuit tht provide new reltionships mong the vrious currents.
13 880 CHAPTER 28 Direct Current Circuits Apply the loop rule to s mny loops in the circuit s re needed to solve for the unknowns. To pply this rule, you must correctly identify the chnge in potentil s you imgine crossing ech element in trversing the closed loop (either clockwise or counterclockwise). Wtch out for errors in sign! Solve the equtions simultneously for the unknown quntities. EXAMPLE 28.7 A SingleLoop Circuit A singleloop circuit contins two resistors nd two tteries, s shown in Figure (Neglect the internl resistnces of the tteries.) () Find the current in the circuit. Solution We do not need Kirchhoff s rules to nlyze this simple circuit, ut let us use them nywy just to see how they re pplied. There re no junctions in this singleloop circuit; thus, the current is the sme in ll elements. Let us ssume tht the current is clockwise, s shown in Figure Trversing the circuit in the clockwise direction, strting t, we see tht : represents potentil chnge of 1, : c represents potentil chnge of IR 1, c : d represents potentil chnge of 2, nd d : represents potentil chnge of IR 2. Applying Kirchhoff s loop rule gives Figure d 1 = 6.0 V + R 2 = 10 Ω I R 1 = 8.0 Ω + 2 = 12 V A series circuit contining two tteries nd two resistors, where the polrities of the tteries re in opposition. c Solving for I nd using the vlues given in Figure 28.13, we otin The negtive sign for I indictes tht the direction of the current is opposite the ssumed direction. () Wht power is delivered to ech resistor? Wht power is delivered y the 12V ttery? Solution I 1 2 R 1 R 2 V 0 1 IR 1 2 IR V 12 V I 2 R 1 (0.33 A) 2 (8.0 ) 2 I 2 R 2 (0.33 A) 2 (10 ) 0.33 A 0.87 W 1.1 W Hence, the totl power delivered to the resistors is W. The 12V ttery delivers power I W. Hlf of this power is delivered to the two resistors, s we just clculted. The other hlf is delivered to the 6V ttery, which is eing chrged y the 12V ttery. If we hd included the internl resistnces of the tteries in our nlysis, some of the power would pper s internl energy in the tteries; s result, we would hve found tht less power ws eing delivered to the 6V ttery. EXAMPLE 28.8 Applying Kirchhoff s Rules Find the currents I 1, I 2, nd I 3 in the circuit shown in Figure Solution Notice tht we cnnot reduce this circuit to simpler form y mens of the rules of dding resistnces in series nd in prllel. We must use Kirchhoff s rules to nlyze this circuit. We ritrrily choose the directions of the currents s leled in Figure Applying Kirchhoff s junction rule to junction c gives (1) I 1 I 2 I 3 We now hve one eqution with three unknowns I 1, I 2, nd I 3. There re three loops in the circuit cd, efc, nd efd. We therefore need only two loop equtions to determine the unknown currents. (The third loop eqution would give no new informtion.) Applying Kirchhoff s loop rule to loops cd nd efc nd trversing these loops clockwise, we otin the expressions (2) cd 10 V (6 )I 1 (2 )I 3 0 (3) efc 14 V (6 )I 1 10 V (4 )I 2 0
14 28.3 Kirchhoff s Rules 881 Note tht in loop efc we otin positive vlue when trversing the 6 resistor ecuse our direction of trvel is opposite the ssumed direction of I 1. Expressions (1), (2), nd (3) represent three independent equtions with three unknowns. Sustituting Eqution (1) into Eqution (2) gives 10 V (6 )I 1 (2 ) (I 1 I 2 ) 0 (4) 10 V (8 )I 1 (2 )I 2 Dividing ech term in Eqution (3) y 2 nd rerrnging gives (5) 12 V (3 )I 1 (2 )I 2 Sutrcting Eqution (5) from Eqution (4) elimintes I 2, giving Using this vlue of I 1 in Eqution (5) gives vlue for I 2 : I 2 3 A 22 V (11 )I 1 I 1 2 A (2 )I 2 (3 )I 1 12 V (3 ) (2 A) 12 V 6 V e 14 V + f Finlly, I 3 I 1 I 2 1 A 4 Ω + 10 V 6 Ω I 1 c I 2 I 3 The fct tht I 2 nd I 3 re oth negtive indictes only tht the currents re opposite the direction we chose for them. However, the numericl vlues re correct. Wht would hve hppened hd we left the current directions s leled in Figure ut trversed the loops in the opposite direction? Figure d 2 Ω A circuit contining three loops. Exercise nd c. Answer 2 V. Find the potentil difference etween points EXAMPLE 28.9 A Multiloop Circuit () Under stedystte conditions, find the unknown currents I 1, I 2, nd I 3 in the multiloop circuit shown in Figure Solution First note tht ecuse the cpcitor represents n open circuit, there is no current etween g nd long pth gh under stedystte conditions. Therefore, when the chrges ssocited with I 1 rech point g, they ll go through the 8.00V ttery to point ; hence, I g I 1. Leling the currents s shown in Figure nd pplying Eqution 28.9 to junction c, we otin (1) I 1 I 2 I 3 Eqution pplied to loops defcd nd cfgc, trversed clockwise, gives (2) defcd 4.00 V (3.00 )I 2 (5.00 )I 3 0 (3) cfgc (3.00 )I 2 (5.00 )I V 0 From Eqution (1) we see tht I 1 I 3 I 2, which, when sustituted into Eqution (3), gives (4) (8.00 )I 2 (5.00 )I V 0 Sutrcting Eqution (4) from Eqution (2), we eliminte I 3 nd find tht I V A Becuse our vlue for I 2 is negtive, we conclude tht the direction of I 2 is from c to f through the resistor. Despite Figure I 3 I 1 d c 5.00 Ω I V 4.00 V + e 3.00 Ω V 5.00 Ω + I = µ F A multiloop circuit. Kirchhoff s loop rule cn e pplied to ny closed loop, including the one contining the cpcitor. f g h I 3 I 1
15 882 CHAPTER 28 Direct Current Circuits this interprettion of the direction, however, we must continue to use this negtive vlue for I 2 in susequent clcultions ecuse our equtions were estlished with our originl choice of direction. Using I A in Equtions (3) nd (1) gives () Wht is the chrge on the cpcitor? Solution I A I A We cn pply Kirchhoff s loop rule to loop gh (or ny other loop tht contins the cpcitor) to find the potentil difference V cp cross the cpcitor. We enter this potentil difference in the eqution without reference to sign convention ecuse the chrge on the cpcitor depends only on the mgnitude of the potentil difference. Moving clockwise round this loop, we otin 8.00 V V cp 3.00 V 0 V cp 11.0 V Becuse Q C V cp (see Eq. 26.1), the chrge on the cpcitor is Q (6.00 F)(11.0 V) 66.0 C Why is the left side of the cpcitor positively chrged? Exercise Find the voltge cross the cpcitor y trversing ny other loop. Answer 11.0 V. Exercise Reverse the direction of the 3.00V ttery nd nswer prts () nd () gin. Answer () () 30 C. I A, I A, I A; 28.4 RC CIRCUITS So fr we hve een nlyzing stedystte circuits, in which the current is constnt. In circuits contining cpcitors, the current my vry in time. A circuit contining series comintion of resistor nd cpcitor is clled n RC circuit. Chrging Cpcitor Let us ssume tht the cpcitor in Figure is initilly unchrged. There is no current while switch S is open (Fig ). If the switch is closed t t 0, however, chrge egins to flow, setting up current in the circuit, nd the cpcitor egins to chrge. 4 Note tht during chrging, chrges do not jump cross the cpcitor pltes ecuse the gp etween the pltes represents n open circuit. Insted, chrge is trnsferred etween ech plte nd its connecting wire due to the electric field estlished in the wires y the ttery, until the cpcitor is fully chrged. As the pltes ecome chrged, the potentil difference cross the cpcitor increses. The vlue of the mximum chrge depends on the voltge of the ttery. Once the mximum chrge is reched, the current in the circuit is zero ecuse the potentil difference cross the cpcitor mtches tht supplied y the ttery. To nlyze this circuit quntittively, let us pply Kirchhoff s loop rule to the circuit fter the switch is closed. Trversing the loop clockwise gives q C IR 0 (28.11) where q/c is the potentil difference cross the cpcitor nd IR is the potentil 4 In previous discussions of cpcitors, we ssumed stedystte sitution, in which no current ws present in ny rnch of the circuit contining cpcitor. Now we re considering the cse efore the stedystte condition is relized; in this sitution, chrges re moving nd current exists in the wires connected to the cpcitor.
16 28.4 RC Circuits 883 Resistor Cpcitor R R + Switch C q + q I Figure Bttery () S () t < 0 (c) t > 0 () A cpcitor in series with resistor, switch, nd ttery. () Circuit digrm representing this system t time t 0, efore the switch is closed. (c) Circuit digrm t time t 0, fter the switch hs een closed. S difference cross the resistor. We hve used the sign conventions discussed erlier for the signs on nd IR. For the cpcitor, notice tht we re trveling in the direction from the positive plte to the negtive plte; this represents decrese in potentil. Thus, we use negtive sign for this voltge in Eqution Note tht q nd I re instntneous vlues tht depend on time (s opposed to stedystte vlues) s the cpcitor is eing chrged. We cn use Eqution to find the initil current in the circuit nd the mximum chrge on the cpcitor. At the instnt the switch is closed (t 0), the chrge on the cpcitor is zero, nd from Eqution we find tht the initil current in the circuit I 0 is mximum nd is equl to I 0 R (current t t 0) (28.12) Mximum current At this time, the potentil difference from the ttery terminls ppers entirely cross the resistor. Lter, when the cpcitor is chrged to its mximum vlue Q, chrges cese to flow, the current in the circuit is zero, nd the potentil difference from the ttery terminls ppers entirely cross the cpcitor. Sustituting I 0 into Eqution gives the chrge on the cpcitor t this time: Q C (mximum chrge) (28.13) To determine nlyticl expressions for the time dependence of the chrge nd current, we must solve Eqution single eqution contining two vriles, q nd I. The current in ll prts of the series circuit must e the sme. Thus, the current in the resistnce R must e the sme s the current flowing out of nd into the cpcitor pltes. This current is equl to the time rte of chnge of the chrge on the cpcitor pltes. Thus, we sustitute I dq /dt into Eqution nd rerrnge the eqution: dq dt R To find n expression for q, we first comine the terms on the righthnd side: dq C dt RC q C q RC RC q RC Mximum chrge on the cpcitor
17 884 CHAPTER 28 Direct Current Circuits Now we multiply y dt nd divide y q C to otin Integrting this expression, using the fct tht q 0 t t 0, we otin q 0 dq q C 1 RC dt dq q C 1 t RC ln q C C t RC From the definition of the nturl logrithm, we cn write this expression s 0 dt Chrge versus time for cpcitor eing chrged q(t ) C (1 e t/rc ) Q(1 e t /RC ) (28.14) where e is the se of the nturl logrithm nd we hve mde the sustitution C Q from Eqution We cn find n expression for the chrging current y differentiting Eqution with respect to time. Using I dq /dt, we find tht Current versus time for chrging cpcitor I(t ) R e t /RC (28.15) Plots of cpcitor chrge nd circuit current versus time re shown in Figure Note tht the chrge is zero t t 0 nd pproches the mximum vlue C s t :. The current hs its mximum vlue I 0 /R t t 0 nd decys exponentilly to zero s t :. The quntity RC, which ppers in the exponents of Equtions nd 28.15, is clled the time constnt of the circuit. It represents the time it tkes the current to decrese to 1/e of its initil vlue; tht is, in time, I e 1 I In time 2, I e I 0. I I 0, nd so forth. Likewise, in time, the chrge increses from zero to C (1 e 1 ) 0.632C. The following dimensionl nlysis shows tht hs the units of time: [ ] [RC] V Q I V Q [ t] T Q / t q I C C τ =RC I 0 I 0 = R 0.368I 0 τ t τ t Figure () () () Plot of cpcitor chrge versus time for the circuit shown in Figure After time intervl equl to one time constnt hs pssed, the chrge is 63.2% of the mximum vlue C. The chrge pproches its mximum vlue s t pproches infinity. () Plot of current versus time for the circuit shown in Figure The current hs its mximum vlue I 0 /R t t 0 nd decys to zero exponentilly s t pproches infinity. After time intervl equl to one time constnt hs pssed, the current is 36.8% of its initil vlue.
18 28.4 RC Circuits 885 RC Becuse hs units of time, the comintion t /RC is dimensionless, s it must e in order to e n exponent of e in Equtions nd Q The energy output of the ttery s the cpcitor is fully chrged is After the cpcitor is fully chrged, the energy stored in the cpcitor 1 is Q C C 2, which is just hlf the energy output of the ttery. It is left s prolem (Prolem 60) to show tht the remining hlf of the energy supplied y the ttery ppers s internl energy in the resistor. Dischrging Cpcitor Now let us consider the circuit shown in Figure 28.18, which consists of cpcitor crrying n initil chrge Q, resistor, nd switch. The initil chrge Q is not the sme s the mximum chrge Q in the previous discussion, unless the dischrge occurs fter the cpcitor is fully chrged (s descried erlier). When the switch is open, potentil difference Q /C exists cross the cpcitor nd there is zero potentil difference cross the resistor ecuse I 0. If the switch is closed t t 0, the cpcitor egins to dischrge through the resistor. At some time t during the dischrge, the current in the circuit is I nd the chrge on the cpcitor is q (Fig ). The circuit in Figure is the sme s the circuit in Figure except for the sence of the ttery. Thus, we eliminte the emf from Eqution to otin the pproprite loop eqution for the circuit in Figure 28.18: q C IR 0 (28.16) C C Q +Q q +q S t < 0 () R R I When we sustitute I dq /dt into this expression, it ecomes R dq dt Integrting this expression, using the fct tht q Q t t 0, gives q Q q C dq q 1 RC dt dq q 1 RC t ln q Q t RC q(t ) Qe t /RC 0 dt (28.17) Figure S t > 0 () () A chrged cpcitor connected to resistor nd switch, which is open t t 0. () After the switch is closed, current tht decreses in mgnitude with time is set up in the direction shown, nd the chrge on the cpcitor decreses exponentilly with time. Chrge versus time for dischrging cpcitor Differentiting this expression with respect to time gives the instntneous current s function of time: I(t) dq dt d dt (Qe t /RC ) Q RC e t /RC (28.18) Current versus time for dischrging cpcitor where Q /RC I 0 is the initil current. The negtive sign indictes tht the current direction now tht the cpcitor is dischrging is opposite the current direction when the cpcitor ws eing chrged. (Compre the current directions in Figs c nd ) We see tht oth the chrge on the cpcitor nd the current decy exponentilly t rte chrcterized y the time constnt RC.
19 886 CHAPTER 28 Direct Current Circuits CONCEPTUAL EXAMPLE Mny utomoiles re equipped with windshield wipers tht cn operte intermittently during light rinfll. How does the opertion of such wipers depend on the chrging nd dischrging of cpcitor? Solution The wipers re prt of n RC circuit whose time constnt cn e vried y selecting different vlues of R Intermittent Windshield Wipers through multiposition switch. As it increses with time, the voltge cross the cpcitor reches point t which it triggers the wipers nd dischrges, redy to egin nother chrging cycle. The time intervl etween the individul sweeps of the wipers is determined y the vlue of the time constnt. EXAMPLE Chrging Cpcitor in n RC Circuit An unchrged cpcitor nd resistor re connected in series to ttery, s shown in Figure If 12.0 V, C 5.00 F, nd R , find the time constnt of the circuit, the mximum chrge on the cpcitor, the mximum current in the circuit, nd the chrge nd current s functions of time. Solution The time constnt of the circuit is ( )( F) 4.00 s. The mximum chrge on the cpcitor is Q C (5.00 F) (12.0 V) 60.0 C. The mximum current in the circuit is I 0 /R (12.0 V)/( ) 15.0 A. Using these vlues nd Equtions nd 28.15, we find tht q(t) I(t) (60.0 C)(1 e t/4.00 s ) (15.0 A) e t/4.00 s Grphs of these functions re provided in Figure RC Exercise Clculte the chrge on the cpcitor nd the current in the circuit fter one time constnt hs elpsed. Answer 37.9 C, 5.52 A. q(µc) µ Q = 60.0 µc µ I(µA) µ () t = τ I 0 = 15.0 µa µ t(s) R 5 t = τ C + S Figure The switch of this series RC circuit, open for times t 0, is closed t t 0. Figure () t(s) Plots of () chrge versus time nd () current versus time for the RC circuit shown in Figure 28.19, with 12.0 V, R , nd C 5.00 F. EXAMPLE Dischrging Cpcitor in n RC Circuit Consider cpcitor of cpcitnce C tht is eing dischrged through resistor of resistnce R, s shown in Figure () After how mny time constnts is the chrge on the cpcitor onefourth its initil vlue? Solution The chrge on the cpcitor vries with time ccording to Eqution 28.17, q(t) Qe t /RC. To find the time it tkes q to drop to onefourth its initil vlue, we sustitute q(t) Q /4 into this expression nd solve for t:
20 28.5 Electricl Instruments 887 Tking logrithms of oth sides, we find ln 4 Q 4 t RC Qe t /RC 1 4 e t /RC t RC(ln 4) 1.39RC 1.39 () The energy stored in the cpcitor decreses with time s the cpcitor dischrges. After how mny time constnts is this stored energy onefourth its initil vlue? Solution Using Equtions (U Q 2 /2C) nd 28.17, we cn express the energy stored in the cpcitor t ny time t s where U 0 Q 2 /2C is the initil energy stored in the cpcitor. As in prt (), we now set U U 0 /4 nd solve for t: Agin, tking logrithms of oth sides nd solving for t gives Exercise After how mny time constnts is the current in the circuit onehlf its initil vlue? Answer U q 2 2C (Qe t /RC ) 2 2C t 1 2RC(ln 4) 0.693RC 0.693RC Q 2 2C e 2t /RC U 0 e 2t /RC U 0 4 U 2t /RC 0e 1 4 e 2t /RC EXAMPLE Energy Delivered to Resistor A F cpcitor is chrged to potentil difference of 800 V nd then dischrged through 25.0k resistor. How much energy is delivered to the resistor in the time it tkes to fully dischrge the cpcitor? Solution We shll solve this prolem in two wys. The first wy is to note tht the initil energy in the circuit equls the energy stored in the cpcitor, C 2 /2 (see Eq ). Once the cpcitor is fully dischrged, the energy stored in it is zero. Becuse energy is conserved, the initil energy stored in the cpcitor is trnsformed into internl energy in the resistor. Using the given vlues of C nd, we find Energy 1 2 C ( F)(800 V) J The second wy, which is more difficult ut perhps more instructive, is to note tht s the cpcitor dischrges through the resistor, the rte t which energy is delivered to the resistor is given y I 2 R, where I is the instntneous current given y Eqution Becuse power is defined s the time rte of chnge of energy, we conclude tht the energy delivered to the resistor must equl the time integrl of I 2 R dt: To evlute this integrl, we note tht the initil current is equl to /R nd tht ll prmeters except t re constnt. Thus, we find (1) Energy 2 e 2t/RC dt R 0 This integrl hs vlue of RC/2; hence, we find which grees with the result we otined using the simpler pproch, s it must. Note tht we cn use this second pproch to find the totl energy delivered to the resistor t ny time fter the switch is closed y simply replcing the upper limit in the integrl with tht specific vlue of t. Exercise Energy I 2 R dt (I 0 e t /RC ) 2 R dt 0 Energy 1 2 C 2 Show tht the integrl in Eqution (1) hs the vlue RC/2. 0 I 0 Optionl Section 28.5 The Ammeter ELECTRICAL INSTRUMENTS A device tht mesures current is clled n mmeter. The current to e mesured must pss directly through the mmeter, so the mmeter must e connected in se
21 888 CHAPTER 28 Direct Current Circuits + A Figure R 1 R 2 Current cn e mesured with n mmeter connected in series with the resistor nd ttery of circuit. An idel mmeter hs zero resistnce. R 1 Figure V R 2 The potentil difference cross resistor cn e mesured with voltmeter connected in prllel with the resistor. An idel voltmeter hs infinite resistnce. ries with other elements in the circuit, s shown in Figure When using n mmeter to mesure direct currents, you must e sure to connect it so tht current enters the instrument t the positive terminl nd exits t the negtive terminl. Idelly, n mmeter should hve zero resistnce so tht the current eing mesured is not ltered. In the circuit shown in Figure 28.21, this condition requires tht the resistnce of the mmeter e much less thn R 1 R 2. Becuse ny mmeter lwys hs some internl resistnce, the presence of the mmeter in the circuit slightly reduces the current from the vlue it would hve in the meter s sence. The Voltmeter A device tht mesures potentil difference is clled voltmeter. The potentil difference etween ny two points in circuit cn e mesured y ttching the terminls of the voltmeter etween these points without reking the circuit, s shown in Figure The potentil difference cross resistor R 2 is mesured y connecting the voltmeter in prllel with R 2. Agin, it is necessry to oserve the polrity of the instrument. The positive terminl of the voltmeter must e connected to the end of the resistor tht is t the higher potentil, nd the negtive terminl to the end of the resistor t the lower potentil. An idel voltmeter hs infinite resistnce so tht no current psses through it. In Figure 28.22, this condition requires tht the voltmeter hve resistnce much greter thn R 2. In prctice, if this condition is not met, corrections should e mde for the known resistnce of the voltmeter. The Glvnometer The glvnometer is the min component in nlog mmeters nd voltmeters. Figure illustrtes the essentil fetures of common type clled the D Arsonvl glvnometer. It consists of coil of wire mounted so tht it is free to rotte on pivot in mgnetic field provided y permnent mgnet. The sic op Scle N S Figure Spring () Coil () The principl components of D Arsonvl glvnometer. When the coil situted in mgnetic field crries current, the mgnetic torque cuses the coil to twist. The ngle through which the coil rottes is proportionl to the current in the coil ecuse of the countercting torque of the spring. () A lrgescle model of glvnometer movement. Why does the coil rotte out the verticl xis fter the switch is closed? ()
22 28.5 Electricl Instruments 889 Glvnometer 60 Ω Glvnometer R s 60 Ω R p Figure () () When glvnometer is to e used s n mmeter, shunt resistor R p is connected in prllel with the glvnometer. () When the glvnometer is used s voltmeter, resistor R s is connected in series with the glvnometer. () ertion of the glvnometer mkes use of the fct tht torque cts on current loop in the presence of mgnetic field (Chpter 29). The torque experienced y the coil is proportionl to the current through it: the lrger the current, the greter the torque nd the more the coil rottes efore the spring tightens enough to stop the rottion. Hence, the deflection of needle ttched to the coil is proportionl to the current. Once the instrument is properly clirted, it cn e used in conjunction with other circuit elements to mesure either currents or potentil differences. A typicl offtheshelf glvnometer is often not suitle for use s n mmeter, primrily ecuse it hs resistnce of out 60. An mmeter resistnce this gret considerly lters the current in circuit. You cn understnd this y considering the following exmple: The current in simple series circuit contining 3V ttery nd 3 resistor is 1 A. If you insert 60 glvnometer in this circuit to mesure the current, the totl resistnce ecomes 63 nd the current is reduced to A! A second fctor tht limits the use of glvnometer s n mmeter is the fct tht typicl glvnometer gives fullscle deflection for currents of the order of 1 ma or less. Consequently, such glvnometer cnnot e used directly to mesure currents greter thn this vlue. However, it cn e converted to useful mmeter y plcing shunt resistor R p in prllel with the glvnometer, s shown in Figure The vlue of R p must e much less thn the glvnometer resistnce so tht most of the current to e mesured psses through the shunt resistor. A glvnometer cn lso e used s voltmeter y dding n externl resistor R s in series with it, s shown in Figure In this cse, the externl resistor must hve vlue much greter thn the resistnce of the glvnometer to ensure tht the glvnometer does not significntly lter the voltge eing mesured. The Whetstone Bridge An unknown resistnce vlue cn e ccurtely mesured using circuit known s Whetstone ridge (Fig ). This circuit consists of the unknown resistnce R x, three known resistnces R 1, R 2, nd R 3 (where R 1 is clirted vrile resistor), glvnometer, nd ttery. The known resistor R 1 is vried until the glvnometer reding is zero tht is, until there is no current from to. Under this condition the ridge is sid to e lnced. Becuse the electric potentil t + Figure I 1 I 2 R 1 R 2 R 3 Circuit digrm for Whetstone ridge, n instrument used to mesure n unknown resistnce R x in terms of known resistnces R 1, R 2, nd R 3. When the ridge is lnced, no current is present in the glvnometer. The rrow superimposed on the circuit symol for resistor R 1 indictes tht the vlue of this resistor cn e vried y the person operting the ridge. G R x
23 890 CHAPTER 28 Direct Current Circuits The strin guge, device used for experimentl stress nlysis, consists of thin coiled wire onded to flexile plstic cking. The guge mesures stresses y detecting chnges in the resistnce of the coil s the strip ends. Resistnce mesurements re mde with this device s one element of Whetstone ridge. Strin guges re commonly used in modern electronic lnces to mesure the msses of ojects. Figure Voltges, currents, nd resistnces re frequently mesured with digitl multimeters like this one. point must equl the potentil t point when the ridge is lnced, the potentil difference cross R 1 must equl the potentil difference cross R 2. Likewise, the potentil difference cross R 3 must equl the potentil difference cross R x. From these considertions we see tht (1) I 1 R 1 I 2 R 2 (2) I 1 R 3 I 2 R x Dividing Eqution (1) y Eqution (2) elimintes the currents, nd solving for R x, we find tht R x R 2R 3 (28.19) R 1 A numer of similr devices lso operte on the principle of null mesurement (tht is, djustment of one circuit element to mke the glvnometer red zero). One exmple is the cpcitnce ridge used to mesure unknown cpcitnces. These devices do not require clirted meters nd cn e used with ny voltge source. Whetstone ridges re not useful for resistnces ove 10 5, ut modern electronic instruments cn mesure resistnces s high s Such instruments hve n extremely high resistnce etween their input terminls. For exmple, input resistnces of re common in most digitl multimeters, which re devices tht re used to mesure voltge, current, nd resistnce (Fig ). The Potentiometer A potentiometer is circuit tht is used to mesure n unknown emf x y comprison with known emf. In Figure 28.27, point d represents sliding contct tht is used to vry the resistnce (nd hence the potentil difference) etween points nd d. The other required components re glvnometer, ttery of known emf 0, nd ttery of unknown emf x. With the currents in the directions shown in Figure 28.27, we see from Kirchhoff s junction rule tht the current in the resistor R x is I I x, where I is the current in the left rnch (through the ttery of emf 0 ) nd I x is the current in the right rnch. Kirchhoff s loop rule pplied to loop cd trversed clockwise gives x (I I x )R x 0 Becuse current I x psses through it, the glvnometer displys nonzero reding. The sliding contct t d is now djusted until the glvnometer reds zero (indicting lnced circuit nd tht the potentiometer is nother nullmesurement device). Under this condition, the current in the glvnometer is zero, nd the potentil difference etween nd d must equl the unknown emf x : x IR x Next, the ttery of unknown emf is replced y stndrd ttery of known emf s, nd the procedure is repeted. If R s is the resistnce etween nd d when lnce is chieved this time, then s IR s where it is ssumed tht I remins the sme. Comining this expression with the preceding one, we see tht x R x (28.20) R s s
24 28.6 Household Wiring nd Electricl Sfety 891 If the resistor is wire of resistivity, its resistnce cn e vried y using the sliding contct to vry the length L, indicting how much of the wire is prt of the circuit. With the sustitutions R s L s /A nd R x L x /A, Eqution ecomes x L x L s (28.21) where L x is the resistor length when the ttery of unknown emf x is in the circuit nd L s is the resistor length when the stndrd ttery is in the circuit. The slidingwire circuit of Figure without the unknown emf nd the glvnometer is sometimes clled voltge divider. This circuit mkes it possile to tp into ny desired smller portion of the emf 0 y djusting the length of the resistor. s 0 I I I x Figure d R x Circuit digrm for potentiometer. The circuit is used to mesure n unknown emf x. I x G c x Optionl Section 28.6 HOUSEHOLD WIRING AND ELECTRICAL SAFETY Household circuits represent prcticl ppliction of some of the ides presented in this chpter. In our world of electricl pplinces, it is useful to understnd the power requirements nd limittions of conventionl electricl systems nd the sfety mesures tht prevent ccidents. In conventionl instlltion, the utility compny distriutes electric power to individul homes y mens of pir of wires, with ech home connected in prllel to these wires. One wire is clled the live wire, 5 s illustrted in Figure 28.28, nd the other is clled the neutrl wire. The potentil difference etween these two wires is out 120 V. This voltge lterntes in time, with the neutrl wire connected to ground nd the potentil of the live wire oscillting reltive to ground. Much of wht we hve lerned so fr for the constntemf sitution (direct current) cn lso e pplied to the lternting current tht power compnies supply to usinesses nd households. (Alternting voltge nd current re discussed in Chpter 33.) A meter is connected in series with the live wire entering the house to record the household s usge of electricity. After the meter, the wire splits so tht there re severl seprte circuits in prllel distriuted throughout the house. Ech circuit contins circuit reker (or, in older instlltions, fuse). The wire nd circuit reker for ech circuit re crefully selected to meet the current demnds for tht circuit. If circuit is to crry currents s lrge s 30 A, hevy wire nd n pproprite circuit reker must e selected to hndle this current. A circuit used to power only lmps nd smll pplinces often requires only 15 A. Ech circuit hs its own circuit reker to ccommodte vrious lod conditions. As n exmple, consider circuit in which toster oven, microwve oven, nd coffee mker re connected (corresponding to R 1, R 2, nd R 3 in Figure nd s shown in the chpteropening photogrph). We cn clculte the current drwn y ech pplince y using the expression I V. The toster oven, rted t W, drws current of W/120 V 8.33 A. The microwve oven, rted t W, drws 10.8 A, nd the coffee mker, rted t 800 W, drws 6.67 A. If the three pplinces re operted simultneously, they drw totl cur 5 Live wire is common expression for conductor whose electric potentil is ove or elow ground potentil. Live Neutrl R 1 Figure Meter Circuit reker R 2 R 3 Wiring digrm for household circuit. The resistnces represent pplinces or other electricl devices tht operte with n pplied voltge of 120 V. 120 V 0 V
25 892 CHAPTER 28 Direct Current Circuits Figure A power connection for 240V pplince. rent of 25.8 A. Therefore, the circuit should e wired to hndle t lest this much current. If the rting of the circuit reker protecting the circuit is too smll sy, 20 A the reker will e tripped when the third pplince is turned on, preventing ll three pplinces from operting. To void this sitution, the toster oven nd coffee mker cn e operted on one 20A circuit nd the microwve oven on seprte 20A circuit. Mny hevyduty pplinces, such s electric rnges nd clothes dryers, require 240 V for their opertion (Fig ). The power compny supplies this voltge y providing third wire tht is 120 V elow ground potentil. The potentil difference etween this live wire nd the other live wire (which is 120 V ove ground potentil) is 240 V. An pplince tht opertes from 240V line requires hlf the current of one operting from 120V line; therefore, smller wires cn e used in the highervoltge circuit without overheting. Figure A threepronged power cord for 120V pplince. Electricl Sfety When the live wire of n electricl outlet is connected directly to ground, the circuit is completed nd shortcircuit condition exists. A short circuit occurs when lmost zero resistnce exists etween two points t different potentils; this results in very lrge current. When this hppens ccidentlly, properly operting circuit reker opens the circuit nd no dmge is done. However, person in contct with ground cn e electrocuted y touching the live wire of fryed cord or other exposed conductor. An exceptionlly good (lthough very dngerous) ground contct is mde when the person either touches wter pipe (normlly t ground potentil) or stnds on the ground with wet feet. The ltter sitution represents good ground ecuse norml, nondistilled wter is conductor ecuse it contins lrge numer of ions ssocited with impurities. This sitution should e voided t ll cost. Electric shock cn result in ftl urns, or it cn cuse the muscles of vitl orgns, such s the hert, to mlfunction. The degree of dmge to the ody depends on the mgnitude of the current, the length of time it cts, the prt of the ody touched y the live wire, nd the prt of the ody through which the current psses. Currents of 5 ma or less cuse senstion of shock ut ordinrily do little or no dmge. If the current is lrger thn out 10 ma, the muscles contrct nd the person my e unle to relese the live wire. If current of out 100 ma psses through the ody for only few seconds, the result cn e ftl. Such lrge current prlyzes the respirtory muscles nd prevents rething. In some cses, currents of out 1 A through the ody cn produce serious (nd sometimes ftl) urns. In prctice, no contct with live wires is regrded s sfe whenever the voltge is greter thn 24 V. Mny 120V outlets re designed to ccept threepronged power cord such s the one shown in Figure (This feture is required in ll new electricl instlltions.) One of these prongs is the live wire t nominl potentil of 120 V. The second, clled the neutrl, is nominlly t 0 V nd crries current to ground. The third, round prong is sfety ground wire tht normlly crries no current ut is oth grounded nd connected directly to the csing of the pplince. If the live wire is ccidentlly shorted to the csing (which cn occur if the wire insultion wers off), most of the current tkes the lowresistnce pth through the pplince to ground. In contrst, if the csing of the pplince is not properly grounded nd short occurs, nyone in contct with the pplince experiences n electric shock ecuse the ody provides lowresistnce pth to ground.
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