Chaptr 6 Currnt and Rsistanc 6.1 Elctric Currnt... 1 6.1.1 Currnt Dnsity... 1 6. Ohm s Law... 3 6.3 Elctrical Enrgy and Powr... 6 6.4 Summary... 6.5 Solvd Problms... 8 6.5.1 Rsistivity of a Cabl... 8 6.5. Charg at a Junction... 9 6.5.3 Drift Vlocity... 10 6.5.4 Rsistanc of a Truncatd Con... 11 6.5.5 Rsistanc of a Hollow Cylindr... 1 6.6 Concptual Qustions... 13 6. Additional Problms... 13 6..1 Currnt and Currnt Dnsity... 13 6.. Powr Loss and Ohm s Law... 13 6..3 Rsistanc of a Con... 14 6..4 Currnt Dnsity and Drift Spd... 14 6..5 Currnt Sht... 15 6..6 Rsistanc and Rsistivity... 15 6.. Powr, Currnt, and Voltag... 16 6..8 Charg Accumulation at th Intrfac... 16 0
Currnt and Rsistanc 6.1 Elctric Currnt Elctric currnts ar flows of lctric charg. Suppos a collction of chargs is moving prpndicular to a surfac of ara A, as shown in Figur 6.1.1. Figur 6.1.1 Chargs moving through a cross sction. Th lctric currnt is dfind to b th rat at which chargs flow across any crosssctional ara. If an amount of charg Q passs through a surfac in a tim intrval t, thn th avrag currnt I avg is givn by Q Iavg = (6.1.1) t Th SI unit of currnt is th ampr (A), with 1 A = 1 coulomb/sc. Common currnts rang from mga-amprs in lightning to nano-amprs in your nrvs. In th limit t 0, th instantanous currnt I may b dfind as I dq = (6.1.) dt Sinc flow has a dirction, w hav implicitly introducd a convntion that th dirction of currnt corrsponds to th dirction in which positiv chargs ar flowing. Th flowing chargs insid wirs ar ngativly chargd lctrons that mov in th opposit dirction of th currnt. Elctric currnts flow in conductors: solids (mtals, smiconductors), liquids (lctrolyts, ionizd) and gass (ionizd), but th flow is impdd in nonconductors or insulators. 6.1.1 Currnt Dnsity To rlat currnt, a macroscopic quantity, to th microscopic motion of th chargs, lt s xamin a conductor of cross-sctional ara A, as shown in Figur 6.1.. 1
Figur 6.1. A microscopic pictur of currnt flowing in a conductor. Lt th total currnt through a surfac b writtn as I = d A (6.1.3) J whr J is th currnt dnsity (th SI unit of currnt dnsity ar A/m ). If q is th charg of ach carrir, and n is th numbr of charg carrirs pr unit volum, th total amount of charg in this sction is thn Q = q( na x). Suppos that th charg carrirs mov with a spd vd ; thn th displacmnt in a tim intrval t will b x = vd t, which implis I avg Q = = nqvd A t (6.1.4) Th spd at which th charg carrirs ar moving is known as th drift spd. v d Physically, v d is th avrag spd of th charg carrirs insid a conductor whn an xtrnal lctric fild is applid. Actually an lctron insid th conductor dos not travl in a straight lin; instad, its path is rathr rratic, as shown in Figur 6.1.3. Figur 6.1.3 Motion of an lctron in a conductor. From th abov quations, th currnt dnsity J can b writtn as J = nqv d (6.1.5) Thus, w s that J and v d point in th sam dirction for positiv charg carrirs, in opposit dirctions for ngativ charg carrirs.
To find th drift vlocity of th lctrons, w first not that an lctron in th conductor xprincs an lctric forc F = E which givs an acclration a F m = = E m (6.1.6) Lt th vlocity of a givn lctron immdiat aftr a collision b v i. Th vlocity of th lctron immdiatly bfor th nxt collision is thn givn by whr t is th tim travld. Th avrag of f = i + t = i E v v a v t m v f ovr all tim intrvals is (6.1.) v f = i E v m t (6.1.8) which is qual to th drift vlocity v d. Sinc in th absnc of lctric fild, th vlocity of th lctron is compltly random, it follows that v 0. If τ = t is th avrag charactristic tim btwn succssiv collisions (th man fr tim), w hav Th currnt dnsity in Eq. (6.1.5) bcoms v d = = E v f m τ i = E n τ J = nvd = n τ = E m m (6.1.9) (6.1.10) Not that J and E will b in th sam dirction for ithr ngativ or positiv charg carrirs. 6. Ohm s Law In many matrials, th currnt dnsity is linarly dpndnt on th xtrnal lctric fild E. Thir rlation is usually xprssd as J = σ E (6..1) 3
whr σ is calld th conductivity of th matrial. Th abov quation is known as th (microscopic) Ohm s law. A matrial that obys this rlation is said to b ohmic; othrwis, th matrial is non-ohmic. Comparing Eq. (6..1) with Eq. (6.1.10), w s that th conductivity can b xprssd as n τ σ = (6..) m To obtain a mor usful form of Ohm s law for practical applications, considr a sgmnt of straight wir of lngth l and cross-sctional ara A, as shown in Figur 6..1. Figur 6..1 A uniform conductor of lngth l and potntial diffrnc V = V V. Suppos a potntial diffrnc V = Vb V a is applid btwn th nds of th wir, crating an lctric fild E and a currnt I. Assuming E to b uniform, w thn hav b a V = V V = d s = El b a b E a (6..3) Th currnt dnsity can thn b writtn as J V σe σ = = (6..4) l With J = I / A, th potntial diffrnc bcoms l l V = J = I = RI (6..5) σ σ A whr V l R = = (6..6) I σ A is th rsistanc of th conductor. Th quation 4
V = IR (6..) is th macroscopic vrsion of th Ohm s law. Th SI unit of R is th ohm (Ω, Grk lttr Omga), whr 1V 1 Ω (6..8) 1A Onc again, a matrial that obys th abov rlation is ohmic, and non-ohmic if th rlation is not obyd. Most mtals, with good conductivity and low rsistivity, ar ohmic. W shall focus mainly on ohmic matrials. Figur 6.. Ohmic vs. Non-ohmic bhavior. Th rsistivity ρ of a matrial is dfind as th rciprocal of conductivity, 1 m ρ = σ = n τ (6..9) From th abov quations, w s that ρ can b rlatd to th rsistanc R of an objct by or E V / l RA ρ = = = J I / A l ρl R = (6..10) A Th rsistivity of a matrial actually varis with tmpratur T. For mtals, th variation is linar ovr a larg rang of T: [ T T ] ρ = ρ0 1 + α( 0 ) (6..11) whr α is th tmpratur cofficint of rsistivity. Typical valus of ρ, σ and α (at 0 C) for diffrnt typs of matrials ar givn in th Tabl blow. 5
Matrial Elmnts Silvr Rsistivity ρ ( Ω m) 1.59 10 8 8 Coppr 1. 10 8 Aluminum.8 10 8 Tungstn 5.6 10 Conductivity σ ( Ω m) 1 Tmpratur Cofficint α (C) 1 6.9 10 0.0038 5.81 10 0.0039 3.55 10 0.0039 1.8 10 0.0045 Iron 8 10.0 10 1.0 10 0.0050 Platinum 8 10.6 10 1.0 10 0.0039 Alloys Brass 10 8 1.4 10 0.00 Manganin 8 44 10 0.3 10 1.0 10 5 Nichrom 8 100 10 0.1 10 0.0004 Smiconductors Carbon (graphit) 3.5 10 5 4.9 10 0.0005 Grmanium (pur) 0.46. 0.048 Silicon (pur) 640 3 1.6 10 0.05 Insulators Glass 10 14 10 10 14 10 10 10 Sulfur 15 10 15 10 16 Quartz (fusd) 5 10 1.33 10 18 6.3 Elctrical Enrgy and Powr Considr a circuit consisting of a battry and a rsistor with rsistanc R (Figur 6.3.1). Lt th potntial diffrnc btwn two points a and b b V = Vb Va > 0. If a charg q is movd from a through th battry, its lctric potntial nrgy is incrasd by U = q V. On th othr hand, as th charg movs across th rsistor, th potntial nrgy is dcrasd du to collisions with atoms in th rsistor. If w nglct th intrnal rsistanc of th battry and th conncting wirs, upon rturning to a th potntial nrgy of q rmains unchangd. Figur 6.3.1 A circuit consisting of a battry and a rsistor of rsistanc R. 6
Thus, th rat of nrgy loss through th rsistor is givn by U q P = = V = I V t t (6.3.1) This is prcisly th powr supplid by th battry. Using V = IR, on may rwrit th abov quation as ( V ) P= I R= (6.3.) R 6.4 Summary Th lctric currnt is dfind as: I dq = dt Th avrag currnt in a conductor is I avg = nqv A d whr n is th numbr dnsity of th charg carrirs, q is th charg ach carrir has, v is th drift spd, and A is th cross-sctional ara. d Th currnt dnsity J through th cross sctional ara of th wir is J = nqv Microscopic Ohm s law: th currnt dnsity is proportional to th lctric fild, and th constant of proportionality is calld conductivity σ : d J = σ E Th rciprocal of conductivity σ is calld rsistivity ρ : 1 ρ = σ Macroscopic Ohm s law: Th rsistanc R of a conductor is th ratio of th potntial diffrnc V btwn th two nds of th conductor and th currnt I:
V R = I Rsistanc is rlatd to rsistivity by ρl R = A whr l is th lngth and A is th cross-sctional ara of th conductor. Th drift vlocity of an lctron in th conductor is v d = E m τ whr m is th mass of an lctron, and succssiv collisions. τ is th avrag tim btwn Th rsistivity of a mtal is rlatd to τ by 1 m ρ = = σ n τ Th tmpratur variation of rsistivity of a conductor is ( ) ρ = ρ0 1+ α T T 0 whr α is th tmpratur cofficint of rsistivity. Powr, or rat at which nrgy is dlivrd to th rsistor is 6.5 Solvd Problms P= I V = I R= ( ) V R 6.5.1 Rsistivity of a Cabl A 3000-km long cabl consists of svn coppr wirs, ach of diamtr 0.3 mm, bundld togthr and surroundd by an insulating shath. Calculat th rsistanc of th 6 cabl. Us 3 10 Ω cm for th rsistivity of th coppr. 8
Solution: Th rsistanc R of a conductor is rlatd to th rsistivity ρ by R = ρl/ A, whr l and A ar th lngth of th conductor and th cross-sctional ara, rspctivly. Sinc th cabl consists of N = coppr wirs, th total cross sctional ara is d (0.03cm) A= Nπ r = N π = π 4 4 Th rsistanc thn bcoms R ρl A 6 8 ( Ω )( ) 3 10 cm 3 10 cm 3.1 10 π 0.03cm / 4 4 = = = Ω ( ) 6.5. Charg at a Junction Show that th total amount of charg at th junction of th two matrials in Figur 6.5.1 1 1 is ε0i( σ σ1 ), whr I is th currnt flowing through th junction, andσ 1 and σ ar th conductivitis for th two matrials. Solution: Figur 6.5.1 Charg at a junction. In a stady stat of currnt flow, th normal componnt of th currnt dnsity J must b th sam on both sids of th junction. Sinc J = σ E, w hav σ1e1 = σ E or E σ 1 E = σ 1 Lt th charg on th intrfac b, w hav, from th Gauss s law: q in or E S q = = ( ) in d A E E1 A ε 0 9
E E = 1 qin Aε 0 Substituting th xprssion for E from abov thn yilds Sinc th currnt is ( σ ) σ 1 1 1 qin = ε0ae1 1 = ε0aσ1 E1 σ σ σ 1 I = JA = E A, th amount of charg on th intrfac bcoms 1 1 q 1 1 = ε I σ σ 1 in 0 6.5.3 Drift Vlocity Th rsistivity of sawatr is about 5 Ω cm. Th charg carrirs ar chifly Na + and Cl ions, and of ach thr ar about 0 3 3 10 / cm. If w fill a plastic tub mtrs long with sawatr and connct a 1-volt battry to th lctrods at ach nd, what is th rsulting avrag drift vlocity of th ions, in cm/s? Solution: Th currnt in a conductor of cross sctional ara A is rlatd to th drift spd charg carrirs by I = nav d v d of th whr n is th numbr of chargs pr unit volum. W can thn rwrit th Ohm s law as ρl V = IR = ( navd) = nvdρl A which yilds v d V = nρl Substituting th valus, w hav v d 1V 5 V cm 5 cm = =.5 10 =.5 10 C Ω s 0 3 19 ( 6 10 /cm )( 1.6 10 C)( 5Ω cm)( 00cm) 10
In convrting th units w hav usd V V 1 A = = = s Ω C Ω C C 1 6.5.4 Rsistanc of a Truncatd Con Considr a matrial of rsistivity ρ in a shap of a truncatd con of altitud h, and radii a and b, for th right and th lft nds, rspctivly, as shown in th Figur 6.5.. Figur 6.5. A truncatd Con. Assuming that th currnt is distributd uniformly throughout th cross-sction of th con, what is th rsistanc btwn th two nds? Solution: Considr a thin disk of radius r at a distanc x from th lft nd. From th figur shown on th right, w hav b r b a = x h or r = ( a b) x + b h Sinc rsistanc R is rlatd to rsistivity ρ by R = ρl/ A, whr l is th lngth of th conductor and A is th cross sction, th contribution to th rsistanc from th disk having a thicknss dy is ρ dx dr = = π π ρ dx + r [ b ( a b) x/ h] 11
Straightforward intgration thn yilds h ρ dx ρh R = = 0 π[ b+ ( a b) x/ h] π ab whr w hav usd du 1 = α + β α α + β ( u ) ( u ) Not that if b= a, Eq. (6..9) is rproducd. 6.5.5 Rsistanc of a Hollow Cylindr Considr a hollow cylindr of lngth L and innr radius a and outr radius b, as shown in Figur 6.5.3. Th matrial has rsistivity ρ. Figur 6.5.3 A hollow cylindr. (a) Suppos a potntial diffrnc is applid btwn th nds of th cylindr and producs a currnt flowing paralll to th axis. What is th rsistanc masurd? (b) If instad th potntial diffrnc is applid btwn th innr and outr surfacs so that currnt flows radially outward, what is th rsistanc masurd? Solution: (a) Whn a potntial diffrnc is applid btwn th nds of th cylindr, currnt flows paralll to th axis. In this cas, th cross-sctional ara is A= π ( b a ), and th rsistanc is givn by ρl ρl R = = A π b a ( ) 1
(b) Considr a diffrntial lmnt which is mad up of a thin cylindr of innr radius r and outr radius r + dr and lngth L. Its contribution to th rsistanc of th systm is givn by ρ dl ρ dr dr = = A π rl whr A= π rl is th ara normal to th dirction of currnt flow. Th total rsistanc of th systm bcoms b ρdr ρ b R = = ln a πrl πl a 6.6 Concptual Qustions 1. Two wirs A and B of circular cross-sction ar mad of th sam mtal and hav qual lngths, but th rsistanc of wir A is four tims gratr than that of wir B. Find th ratio of thir cross-sctional aras.. From th point of viw of atomic thory, xplain why th rsistanc of a matrial incrass as its tmpratur incrass. 3. Two conductors A and B of th sam lngth and radius ar connctd across th sam potntial diffrnc. Th rsistanc of conductor A is twic that of B. To which conductor is mor powr dlivrd? 6. Additional Problms 6..1 Currnt and Currnt Dnsity 9 A sphr of radius 10 mm that carris a charg of 8 nc = 8 10 C is whirld in a circl at th nd of an insulatd string. Th rotation frquncy is 100π rad/s. (a) What is th basic dfinition of currnt in trms of charg? (b) What avrag currnt dos this rotating charg rprsnt? (c) What is th avrag currnt dnsity ovr th ara travrsd by th sphr? 6.. Powr Loss and Ohm s Law A 1500 W radiant hatr is constructd to oprat at 115 V. 13
(a) What will b th currnt in th hatr? [Ans. ~10 A] (b) What is th rsistanc of th hating coil? [Ans. ~10 Ω] (c) How many kilocaloris ar gnratd in on hour by th hatr? (1 Calori = 4.18 J) 6..3 Rsistanc of a Con A coppr rsistor of rsistivity ρ is in th shap of a cylindr of radius b and lngth L 1 appndd to a truncatd right circular con of lngth L and nd radii b and a as shown in Figur 6..1. Figur 6..1 (a) What is th rsistanc of th cylindrical portion of th rsistor? (b) What is th rsistanc of th ntir rsistor? (Hint: For th taprd portion, it is ncssary to writ down th incrmntal rsistanc dr of a small slic, dx, of th rsistor at an arbitrary position, x, and thn to sum th slics by intgration. If th tapr is small, on may assum that th currnt dnsity is uniform across any cross sction.) (c) Show that your answr rducs to th xpctd xprssion if a = b. (d) If L 1 = 100 mm, L = 50 mm, a = 0.5 mm, b = 1.0 mm, what is th rsistanc? 6..4 Currnt Dnsity and Drift Spd (a) A group of chargs, ach with charg q, movs with vlocity v. Th numbr of particls pr unit volum is n. What is th currnt dnsity J of ths chargs, in magnitud and dirction? Mak sur that your answr has units of A/m. (b) W want to calculat how long it taks an lctron to gt from a car battry to th startr motor aftr th ignition switch is turnd. Assum that th currnt flowing is115 A, and that th lctrons travl through coppr wir with cross-sctional ara 31. mm and lngth 85.5 c m. What is th currnt dnsity in th wir? Th numbr dnsity of th 8 3 conduction lctrons in coppr is 8.49 10 /m. Givn this numbr dnsity and th currnt dnsity, what is th drift spd of th lctrons? How long dos it tak for an 14
6 lctron starting at th battry to rach th startr motor? [Ans: 3.69 10 A/m, 4.1 10 m/s,5.5 min.] 6..5 Currnt Sht A currnt sht, as th nam implis, is a plan containing currnts flowing in on dirction in that plan. On way to construct a sht of currnt is by running many paralll wirs in a plan, say th yz -plan, as shown in Figur 6..(a). Each of ths wirs carris currnt I out of th pag, in th j ˆ dirction, with n wirs pr unit lngth in th z-dirction, as shown in Figur 6..(b). Thn th currnt pr unit lngth in th z dirction is ni. W will us th symbol K to signify currnt pr unit lngth, so that K = nl hr. Figur 6.. A currnt sht. Anothr way to construct a currnt sht is to tak a non-conducting sht of charg with fixd charg pr unit ara σ and mov it with som spd in th dirction you want currnt to flow. For xampl, in th sktch to th lft, w hav a sht of charg moving out of th pag with spd v. Th dirction of currnt flow is out of th pag. (a) Show that th magnitud of th currnt pr unit lngth in th z dirction, K, is givn by σ v. Chck that this quantity has th propr dimnsions of currnt pr lngth. This is in fact a vctor rlation, K (t) = σ v( t), sinc th sns of th currnt flow is in th sam dirction as th vlocity of th positiv chargs. (b) A blt transfrring charg to th high-potntial innr shll of a Van d Graaff acclrator at th rat of.83 mc/s. If th width of th blt carrying th charg is 50 cm and th blt travls at a spd of 30 m /s, what is th surfac charg dnsity on th blt? [Ans: 189 µc/m ] 6..6 Rsistanc and Rsistivity A wir with a rsistanc of 6.0 Ω is drawn out through a di so that its nw lngth is thr tims its original lngth. Find th rsistanc of th longr wir, assuming that th 15
rsistivity and dnsity of th matrial ar not changd during th drawing procss. [Ans: 54 Ω]. 6.. Powr, Currnt, and Voltag A 100-W light bulb is pluggd into a standard 10-V outlt. (a) How much dos it cost pr month (31 days) to lav th light turnd on? Assum lctricity costs 6 cnts pr kw h. (b) What is th rsistanc of th bulb? (c) What is th currnt in th bulb? [Ans: (a) $4.46; (b) 144 Ω; (c) 0.833 A]. 6..8 Charg Accumulation at th Intrfac Figur 6..3 shows a thr-layr sandwich mad of two rsistiv matrials with rsistivitis ρ 1 and ρ. From lft to right, w hav a layr of matrial with rsistivity ρ 1 of width d /3, followd by a layr of matrial with rsistivity ρ, also of width d /3, followd by anothr layr of th first matrial with rsistivity ρ 1, again of width d /3. Figur 6..3 Charg accumulation at intrfac. Th cross-sctional ara of all of ths matrials is A. Th rsistiv sandwich is boundd on ithr sid by mtallic conductors (black rgions). Using a battry (not shown), w maintain a potntial diffrnc V across th ntir sandwich, btwn th mtallic conductors. Th lft sid of th sandwich is at th highr potntial (i.., th lctric filds point from lft to right). Thr ar four intrfacs btwn th various matrials and th conductors, which w labl a through d, as indicatd on th sktch. A stady currnt I flows through this sandwich from lft to right, corrsponding to a currnt dnsity J = I / A. (a) What ar th lctric filds E 1 and E in th two diffrnt dilctric matrials? To obtain ths filds, assum that th currnt dnsity is th sam in vry layr. Why must this b tru? [Ans: All filds point to th right, E 1 = ρ 1 I / A, E = ρ I / A; th currnt dnsitis must b th sam in a stady stat, othrwis thr would b a continuous buildup of charg at th intrfacs to unlimitd valus.] 16
(b) What is th total rsistanc R of this sandwich? Show that your xprssion rducs to th xpctd rsult if ρ 1 = ρ = ρ. [Ans: R= d( ρ1+ ρ) / 3 A; if ρ 1 = ρ = ρ, thn R = d ρ / A, as xpctd.] (c) As w mov from right to lft, what ar th changs in potntial across th thr V ρ / ρ ρ,v ρ / ρ + ρ, layrs, in trms of V and th rsistivitis? [Ans: 1 ( 1+ ) ( 1 ) V ρ /( ρ + ρ ), summing to a total potntial drop of V, as rquird]. 1 1 (d) What ar th chargs pr unit ara, σ a through σ d, at th intrfacs? Us Gauss's Law and assum that th lctric fild in th conducting caps is zro. [Ans: σ a = σd = 3 ε0vρ1/ d( ρ1+ ρ), σ b = σc = 3 ε0v( ρ ρ1) / d ( ρ1+ ρ).] () Considr th limit ρ ρ 1. What do your answrs abov rduc to in this limit? 1