Electricity and Magnetism Inductance

Lana Sheridan

De Anza College

Mar 13, 2018 Last time

• relativity and ﬁelds Overview

• inductors and inductance

• resistor-inductor circuits Inductors A capacitor is a device that stores an electric ﬁeld as a component of a circuit.

inductor a device that stores a magnetic ﬁeld in a circuit

It is typically a coil of wire. 782 Chapter 26 Capacitance and Dielectrics

▸ 26.2 continued Categorize Because of the spherical symmetry of the sys- ϪQ tem, we can use results from previous studies of spherical Figure 26.5 (Example 26.2) systems to find the capacitance. A spherical capacitor consists of an inner sphere of radius a sur- ϩQ a Analyze As shown in Chapter 24, the direction of the rounded by a concentric spherical electric field outside a spherically symmetric charge shell of radius b. The electric field b distribution is radial and its magnitude is given by the between the spheres is directed expression E 5 k Q /r 2. In this case, this result applies to radially outward when the inner e sphere is positively charged. 820 the field Cbetweenhapter 27 the C spheresurrent and ( aR esistance, r , b). b S Write an expression for the potential difference between 2 52 ? S 782 Chapter 26 Capacitance and DielectricsTable 27.3 CriticalVb TemperaturesVa 3 E d s a the two conductors from Equation 25.3: for Various Superconductors

▸ 26.2 continued Material Tc (K) b b dr 1 b ApplyCategorize the result Because of Example of the spherical 24.3 forsymmetry the electric ofHgBa the sys 2fieldCa- 2Cu 3O8 2 52134 52 Ϫ5Q Vb Va 3 Er dr ke Q 3 2 ke Q Tl—Ba—Ca—Cu—O 125a a r r a outsidetem, wea sphericallycan use results symmetric from previous charge studies distribution of spherical Figure 26.5 (Example 26.2) systems to findS the capacitance. S Bi—Sr—Ca—Cu—O 105 c d and note that E is parallel to d s along a radial line: A spherical capacitor consists of 1 1 a 2 b YBa2Cu3O7 an inner sphere of radius a sur- 92 ϩQ a (1) Vb 2 Va 5 ke Q 2 5 ke Q Analyze As shown in Chapter 24, the directionNb of Ge the rounded by a concentric spherical23.2 782 Chapter 26 Capacitance and Dielectrics 3 b a ab electric field outside a spherically symmetric Nb chargeSn shell of radius b. The electric field18.05 3 a b b distribution is radial and its magnitude is givenNb by the between the spheres is directed 9.46

Courtesy of IBM Research Laboratory Q Q ▸ 26.2 continued 2 radially outward when the inner ab Substituteexpression the E 5absolute keQ /r . Invalue this case,of D Vthis into result Equation appliesPb to 26.1: 5 5 7.18 5 sphere is positivelyC charged. Ϫ (26.6) A small permanentCategorize magnet levi- Because of the spherical, symmetry, of the sys- DV V 2 QV k b 2 a the fieldtem, we between can use the results spheres from previous (a studiesr b). of spherical Hg b4.15 a e tated above a disk of the super- Figure 26.5 (Example 26.2) systems to find the capacitance. Sn A spherical capacitor consists of b 3.72 conductor YBa Cu O , which is in S ϩQ a Write2 3 an7 expression for the potential difference betweenAl an inner sphere ofV radius2 Va sur52- E ? d Ss 0 1.19 0 1 2 liquid nitrogenFinalize at 77Analyze TheK. Ascapacitance shown in Chapter depends 24, the direction on a of and the b roundedas expected. by a concentricb sphericalThea potential3 difference between the spheres in Equation the twoelectric conductors field outside from a spherically Equation symmetric 25.3: charge Zn shell of radius b. The electric field a 0.88 b (1) is negativedistribution because is radial and Q its is magnitude positive is given and by b the. a. betweenTherefore, the spheres isin directed Equation 26.6, when we take the absolute value, we change expression E 5 k Q /r 2. In this case, this result applies to radially outward when the inner b b b e sphere is positively charged. dr 1 a 2Apply b tothe bthe field2 result a.between The of the Example resultspheres ( ais ,24.3 ar ,positive bfor). the electric number field. V 2 V 52 E dr 52k Q 5 k Q Today, thousands of superconductorsb a are known,3 r and e as3 Table2 27.3e rillustrates, b a a r a outside a spherically symmetric charge distribution S WHAT WriteIF ? an expressionIf Sthe radius for the potential b of the difference outer between sphere approaches2 52 ?infinity,S what does the capacitance become? and note that E is parallelthe to criticald Ss along temperatures a radial line: V bof Vrecentlya 3 E discoveredd s superconductors are substantiallyc d the two conductors from Equation 25.3: a 1 1 a 2 b higher than initially thought possible.(1) V b Two2 Va 5 kindske Q of2 superconductors5 ke Q are recog- Answer In Equation 26.6, we let b `: b b b a ab nized. The Smore recently identified ones are dressentially1 b ceramics with high criti- Apply the result of Example 24.3 for the electric field V 2 V 52 E dr 52k Q 5 k Q b aab 3 r abe 3 r 2 a ae r b outside a spherically symmetriccal temperatures,charge distribution whereas superconductinga Q a materialsQ sucha ab as those observed by S S D C 5 lim 5 5 5 c4pd P0a 980 SubstituteChapterand note the32 that absolute EInductance is parallel value to d s of along V ainto radial Equation line: S26.1: C 5 5 5 (26.6) Kamerlingh-Onnes areb `metals.ke b 2 If a room-temperature1ke 1b ka e2 b superconductor is ever iden- (1) Vb 2 Va 5 ke Q DV2 5Vbke2Q Va ke b 2 a tified, its effect on technology could beb tremendous.a ab Notice that this expression is the same as Equation 26.2, thea capacitanceb of an isolated spherical conductor. The value of T is sensitive1Q to chemicalQ2 1 2ab composition,0 0 1 pressure,2 and molecular Substitute the absolute value of DV into Equation 26.1:c C 5 5 5 (26.6) ▸ 32.5 continuedFinalize The capacitance depends on a and b as expected.D The 2potential difference2 between the spheres in Equation structure. Copper, silver, andV gold,Vb whichVa k eareb aexcellent conductors, do not exhibit (1) is negative because Q is positive and b . a. Therefore, in Equation 26.6, when we take the absolute value, we change superconductivity. 0 0 1 2 SOLUTI O N a 2 bFinalize to b 2 The a. Thecapacitance result depends is a positive on a and number b as expected.. The potential difference between the spheres in Equation (1) is negative because Q is positiveOne and truly b . a . remarkableTherefore, in Equation feature 26.6, whenof superconductors we take the absolute value, is wethat change once a current is set up WHAT IF ? If the radius b of the outer sphere approaches infinity, what does the capacitance become? Conceptualize Bea sure 2 b to byou 2 a. Thecan result identify isin a positivethem, the number it two 26.3persists. coils without inCombinations the any situation applied andpotential understand ofdifference Capacitors that(because a changing R 5 0). currentSteady cur in -one coil induces a currentAnswerWHA T In inIF Equation? theIf the second radius 26.6, rentsb of coil.we the let outerhave b S sphere been`: approaches observed infinity, to whatpersist does the in capacitance superconducting become? loops for several years with Answer In Equation 26.6,no we letapparent b S `: decay! ab ab a TwoC 5 orlimab more ab capacitorsa5 5often5 4arepP0a combined in electric circuits. We can calculate Categorize We will determine the result using conceptsb S ` discussed in this section, so we categorize this example as a An importantC 5 lim andk 5eusefulb 2 a5 application5k4epPb0a k eof superconductivity is in the development b S ` k b 2 a k b k substitution problem. of superconductingthe equivalente magnets,e capacitancee in which the of magnitudes certain combinations of the magnetic using field methods are described in NoticeNotice that that this this expression expression is theis the same same as Equation as Equation 26.2, the1 capacitance26.2,2 the ofcapacitance an1 2isolated spherical of an conductor.isolated spherical conductor. Capacitor approximatelythis section. ten1 times2 Throughout1 2 greater than this those section, produced we assume by the the best capacitors normal elec to- be combined are Circuit component symbols NB Use Equation 30.17symbol to express the magnetictromagnets. initiallyfield Such in the superconductinguncharged. B 5m magnets i are being considered as a means of 0 , interior of the base solenoid: storing energy. In Superconductingstudying electric magnets circuits, are we currently use a simplified used in medical pictorial magnetic representation called a Combinations of Capacitors resonance26.3circuit imaging, diagram. or MRI, Suchunits, awhich diagramF produce uses high-quality circuit symbols images toof internalrepresent various circuit Battery ϩ 26.3 Combinations NofH CapacitorsBH N H BA NBNH Find the mutual inductance, notingorgans thatTwo the withoutor more magnetic capacitors the need often for are excessivecombinedM 5 in exposure electric 5circuits. of patientsWe can5 calculate mto0 x-rays orA other harm- batterysymbol V Ϫ the equivalentelements. capacitance The of certain circuit combinations symbolsi using are methods connectedi described in by , straight lines that represent the flux F through the handle’s coilful caused radiation. by the mag- BH Capacitor this Twosection.wires or Throughout more between capacitors this section, the often we circuit assume are thecombined elements. capacitors toin be Theelectric combined circuit circuits. are symbols We can forcalculate capacitors, batteries, netic fieldThe ofdirection the base of thesymbol coil is BA: initiallytheand uncharged. equivalent switches capacitance as well as of the certain color combinations codes used using for methods them in described this text in are given in Fig- effective flowCapacitor of positive Inthis studying section. electric Throughout circuits, we use this a simplified section, pictorial we assume representation the capacitors called a to be combined are circuit diagram. Such a diagram uses circuit symbols to represent various circuit Wirelesscapacitorcharge charging is clockwise.Switch Csymbol is usedBattery inOpen ϩa number ofinitially otherure 26.6.“cordless”uncharged. The devices.symbol Onefor the significant capacitor example reflects is the inductivegeometry charging of the most common symbol Ϫ 27.6elements. Electrical The circuit symbols Power are connected by straight lines that represent the used by some manufacturerssymbol of electricwires cars betweenInmodel that studying theavoids for circuit electric a direct elements.capacitor, circuits, metal-to-metal The circuita we pair use symbols ofa simplified parallel forcontact capacitors, pictorialbetweenplates. batteries, Therepresentation the positivecar and thecalledterminal charg a - of the battery I and switches as well as the color codes used for them in this text are given in Fig- ing apparatus. Battery Closedϩ In typicalcircuit iselectric at diagram. the circuits, higher Such energypotential a diagram TET and isuses transferred iscircuit represented symbols by electrical toin represent the transmission circuit various symbol circuitfrom by the longer line. Switch Open ure elements.26.6. The symbol The for circuit the capacitor symbols reflects are the connected geometry of the by most straight common lines that represent the switch Ssymbolsymbol Ϫ a sourcemodel forsuch a capacitor, as a battery a pair of parallelto some plates. device The positive such terminal as a oflightbulb the battery or a radio receiver. Figure 26.6 CircuitClosed symbolsLet’s for is determine atwires the higher between potential an expression the and circuitis represented that elements. inwill the allowcircuit The symbol us circuit to by calculate the symbols longer line. the for capacitors,rate of this batteries, energy capacitors, batteries, and switches. b Figure 26.6 Circuitc symbols fortransfer. and First,Parallel switches consider as Combination well the as simple the color circuit codes in used Figure for 27.11,them inwhere this textenergy are isgiven delivered in Fig - ϩ Notice thatcapacitors,Switch capacitors batteries,Open andare switches. in Parallelure 26.6.Combination The symbol for the capacitor reflects the geometry of the most common Notice that capacitors are in to a resistor.Two (Resistors capacitors are designatedconnected by as the shown circuit in symbol Figure 26.7a are.) Because known theas a parallel combi- ⌬V R Two capacitors connected as shown in Figure 26.7a are known as a parallel combi- resistorblue, batteriesblue, Rsymbol batteries are arein ingreen, green, and and model for a capacitor, a pair of parallel plates. The positive terminal of the battery Ϫ connectingnation of capacitors.wires also Figure have 26.7b resistance,shows a circuit somediagram energyfor this combination is delivered of to the wires and switches areswitches in are red. in red. TheClosed The closed32.5 isOscillations atnation the higher of capacitors.potential in and Figurean is represented LC 26.7b Circuit in shows the circuit a circuit symbol diagram by the longer for line.this combination of a switch can carry current,d whereassome capacitors. to the Theresistor. left plates Unless of the capacitors noted areotherwise, connected to we the shallpositive assume terminal of the resistance of the switch canthe carryopen one current, cannot. whereas the batterycapacitors. by a conducting The wire andleft are plates therefore of both the at the capacitors same electric potential are connected to the positive terminal of ϩ Figure 26.6 Circuit symbolsWhen for a capacitor is connected to an inductor as illustrated in Figure 32.10, the C the open one cannot. wires is smallthe comparedbattery by with a conducting the resistance wire of andthe circuitare therefore element bothso that at the the energy same electric potential Ϫ capacitors, batteries,L and switches. Parallel Combination inductorQ L combinationdelivered to isthe an wires LC iscircuit. negligible. If the capacitor is initially charged and the switch is max Notice that capacitors are in Two capacitors connected as shown in Figure 26.7a are known as a parallel combi- blue, batteries are in green,then and Imagine closed, followingboth the acurrent positive in quantity the circuit of charge and Qthe moving charge clockwise on the aroundcapacitor the oscil - Figure 27.11 A circuit consist- nation of capacitors. Figure 26.7b shows a circuit diagram for this combination of switches are in red. The closedlatecircuit between in Figure maximum 27.11 from positive point a and through negative the battery values. and If resistor the resistance back to point of the a. cir- ing of a resistorswitch of canresistance carry current, R whereas capacitors. The left plates of the capacitors are connected to the positive terminal of S We identify the entire circuit as our system. As the charge moves from a to b through and a batterythe having open onea potential cannot. cuit is zero,the batteryno energy by a conducting is transformed wire and to are internal therefore energy. both at the In same the electricfollowing potential analysis, difference DV across its terminals. the battery, the electric potential energy of the system increases by an amount Q DV Figure 32.10 A simple LC cir- the resistance in the circuit is neglected. We also assume an idealized situation in cuit. The capacitor has an initial which energy is not radiated away from the circuit. This radiation mechanism is charge Q max, and the switch is discussed in Chapter 34. open for t , 0 and then closed at Assume the capacitor has an initial charge Q (the maximum charge) and t 5 0. max the switch is open for t , 0 and then closed at t 5 0. Let’s investigate what happens from an energy viewpoint. When the capacitor is fully charged, the energy U in the circuit is stored in 2 the capacitor’s electric field and is equal to Q max/2C (Eq. 26.11). At this time, the current in the circuit is zero; therefore, no energy is stored in the inductor. After the switch is closed, the rate at which charges leave or enter the capacitor plates (which is also the rate at which the charge on the capacitor changes) is equal to the current in the circuit. After the switch is closed and the capacitor begins to discharge, the energy stored in its electric field decreases. The capacitor’s discharge represents a current in the circuit, and some energy is now stored in the magnetic field of the inductor. Therefore, energy is transferred from the electric field of the capacitor to the magnetic field of the inductor. When the capacitor is fully discharged, it stores no energy. At this time, the current reaches its maximum value and all the energy in the circuit is stored in the inductor. The current continues in the same direction, decreasing in magnitude, with the capacitor eventually becoming fully charged again but with the polarity of its plates now opposite the initial polarity. This process is followed by another discharge until

the circuit returns to its original state of maximum charge Q max and the plate polarity shown in Figure 32.10. The energy continues to oscillate between inductor and capacitor. The oscillations of the LC circuit are an electromagnetic analog to the mechanical oscillations of the particle in simple harmonic motion studied in Chapter 15. Much of what was discussed there is applicable to LC oscillations. For example, we investigated the effect of driving a mechanical oscillator with an external force, Inductance

Just like capacitors have a capacitance that depends on the geometry of the capacitor, inductors have an inductance that depends on their geometry.

Capacitance is deﬁned as being the constant of proportionality relating the charge on the plates to the potential diﬀerence across the plates Q = C (∆V ) .

Inductance is deﬁned in a similar way. Inductance

The magnetic ﬂux linkage is NΦB .

Inductance, L the constant of proportionality relating the magnetic ﬂux linkage (NΦB ) to the current, I: NΦ NΦ = L I ; L = B B I

ΦB is the magnetic ﬂux through the coil, and I is the current in the coil.

Units: henries, H. 1 henry = 1 H = 1 T m2 /A Inductance

Just like capacitors have a capacitance that depends on the geometry of the capacitor, inductors have an inductance that depends on their structure.

For a solenoid inductor:

2 L = µ0n A`

where n is the number of turns per unit length, A is the cross sectional area, and ` is the length of the inductor.

(Doesn’t depend on current or ﬂux, only geometry of device.) Value of µ0: New units

The magnetic permeability of free space µ0 is a constant.

−7 µ0 = 4π × 10 T m / A

It can also be written in terms of henries:

−7 µ0 = 4π × 10 H / m

(Remember, 1 H = 1 T m2 / A) Inductance of Solenoid Inductors

Suppose now that the only source of magnetic ﬂux in the solenoid is the ﬂux produced by a current in the wire.

Then the ﬁeld produced within the solenoid is:

B = µ0In

where n is the number of turns per unit length.

That means the ﬂux will be:

◦ ΦB = BA cos(0 ) = BA = µ0InA

where A is the cross sectional area of the solenoid. Inductance of Solenoid Inductors

NΦ L = B I

Replacing N = n`, ΦB = µ0InA: n`(µ In)A L = 0 I

So we conﬁrm our expression for a solenoid inductor:

2 L = µ0n A` This inductance1, L, is the self-inductance of the coil.

NΦB = L I

We are assuming the flux ΦB is entirely due to the B-field resulting from the current in the solenoid and there is no external source of magnetic field adding to the flux.

Induction from an external ﬂux vs Self-Induction

Previously, we considered the effect of a changing magnetic field from some external source causing and emf and current flow in a wire loop.

However, a changing current in the solenoid itself can be causing the changing field that affects the flow of the current.

1In this textbook, and most sources, L is the self-inductance. Induction from an external ﬂux vs Self-Induction

Previously, we considered the effect of a changing magnetic field from some external source causing and emf and current flow in a wire loop.

However, a changing current in the solenoid itself can be causing the changing field that affects the flow of the current.

This inductance1, L, is the self-inductance of the coil.

NΦB = L I

We are assuming the flux ΦB is entirely due to the B-field resulting from the current in the solenoid and there is no external source of magnetic field adding to the flux.

1In this textbook, and most sources, L is the self-inductance. Self-Induction

When the current in the solenoid circuit is changing there is a (self-) induced emf in the coil.

From Faraday’s Law, we have

d(NΦ ) E = − B dt

Since L is a constant for a particular inductor and Li = NΦB , di E = −L L dt

EL is the self-induced emf.

The emf opposes the change in current. Inductors vs. Resistors

Inductors are a bit similar to resistors.

Resistors resist the ﬂow of current.

Inductors resist any change in current.

If the current is high and lowered, the emf acts to keep the current ﬂowing. If the current is low and increased, the emf acts to resist the increase. 32.1 Self-Induction and Inductance 971

Consider a circuit consisting of a switch, a resistor, and a source of emf as shown in Figure 32.1. The circuit diagram is represented in perspective to show the orien- tations of some of the magnetic field lines due to the current in the circuit. When the switch is thrown to its closed position, the current does not immediately jump from zero to its maximum value e/R. Faraday’s law of electromagnetic induction (Eq. 31.1) can be used to describe this effect as follows. As the current increases with time, the magnetic field lines surrounding the wires pass through the loop represented by the circuit itself. This magnetic field passing through the loop causes a magnetic flux through the loop. This increasing flux creates an induced emf in the circuit. The direction of the induced emf is such that it would cause an induced current in the loop (if the loop did not already carry a current), which would establish a magnetic field opposing the change in the original magnetic Brady-Handy Collection, Library of Congress Prints and Photographs Division [LC-BH83-997] field. Therefore, the direction of the induced emf is opposite the direction of the Joseph Henry emf of the battery, which results in a gradual rather than instantaneous increase in American Physicist (1797–1878) the current to its final equilibrium value. Because of the direction of the induced Henry became the first director of emf, it is also called a back emf, similar to that in a motor as discussed in Chapter 31. the Smithsonian Institution and first president of the Academy of Natural This effect is called self-induction because the changing flux through the circuit Science. He improved the design of the and the resultant induced emf arise from the circuit itself. The emf eL set up in this electromagnet and constructed one of case is called a self-induced emf. the first motors. He also discovered the To obtain a quantitative description of self-induction, recall from Faraday’s law phenomenon of self-induction, but he that the induced emf is equal to the negative of the time rate of change of the mag- failed to publish his findings. The unit of inductance, the henry, is named in netic flux. The magnetic flux is proportional to the magnetic field, which in turn his honor. is proportional to the current in the circuit. Therefore, a self-induced emf is always proportional to the time rate of change of the current. For any loop of wire, we can write this proportionality as di e 52L (32.1) L dt where L is a proportionality constant—called the inductance of the loop—that depends on the geometry of the loop and other physical characteristics. If we consider a closely spaced coil of N turns (a toroid or an ideal solenoid) carrying a current i and containing N turns, Faraday’s law tells us that eL 5 2N dFB /dt. Com- bining this expression with Equation 32.1 gives

NFB L 5 (32.2) Inductance of an N-turn coil i where it is assumed the same magnetic flux passes through each turn and L is the After the switch is closed, the inductance of the entire coil. current produces a magnetic flux From Equation 32.1, we can also write the inductance as the ratio through the area enclosed by the loop. As the current increases e toward its equilibrium value, this L 52 L (32.3) di/dt Self-Inductionmagnetic flux changes in time and induces an emf in the loop. Recall that resistance is a measure of the opposition to current as given by Equa- S tion 27.7, R 5 DV/I ; in comparison, Equation 32.3, being of the same mathematical B form as Equation 27.7, shows us that inductance is a measure of the opposition to a S change in current. i The SI unit of inductance is the henry (H), which as we can see from Equation 32.3 is 1 volt-second per ampere: 1 H 5 1 V ? s/A. As shown in Example 32.1, the inductance of a coil depends on its geometry. This ϩ dependence is analogous to the capacitance of a capacitor depending on the geome- R Ϫ e try of its plates as we found in Equation 26.3 and the resistance of a resistor depending on the length and area of the conducting material in Equation 27.10. Inductance i calculations can be quite difficult to perform for complicated geometries, but the Figure 32.1 Self-induction in a examples below involve simple situations for which inductances are easily evaluated. simple circuit. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) halliday_c30_791-825hr.qxd 11-12-2009 12:19we cannot Page 807define an electric potential within the inductor itself, where the flux is changing. However, potentials can still be defined at points of the circuit that are not within the inductor—points where the electric fields are due to charge distributions and their associated electric potentials.Self-Induction L Moreover, we can define a self-induced potential difference VL across an L inductor (between its terminals, which we assume to be outside the region of PART 3 The changing changing flux). For an ideal inductor (its wire has negligible resistance), the mag- 807 current changes 30-9 RL CIRCUITS (a) nitude of VL is equal to the magnitude of the self-induced emf ᏱL. the flux, which If, instead, the wire in the inductor has resistance r,we mentally separate the creates an emf This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) i (decreasing) inductor into a resistance r (which we take to be outside the region of changing that opposes we cannot define an electric potential within the inductor itself, where the flux flux) and an ideal inductor of self-induced emf ᏱL.As with a real battery of emf the change. is changing. However, potentialsᏱ andcan internalstill be defined resistance at points r,the ofpotential the circuit difference that across the terminals of a real are not within the inductor—pointsinductor where then the differs electric from fields the areemf. due Unless to charge otherwise indicated, we assume here L distributions and their associatedthat electric inductors potentials. are ideal. L Moreover, we can define a self-induced potential difference VL across an L L inductor (between its terminals, which we assume to be outside the region of The changing changing flux). For an ideal inductorCHECKPOINT(its wire has negligible 5 resistance), the mag- current changes (a) (b) nitude of VL is equal to the magnitudeThe figure of the shows self-induced an emf Ᏹ emfinduced ᏱL. in a coil. Which of the flux, which L L If, instead, the wire in the inductorthe following has resistance can describe r,we the mentally current through separate the the coil: (a) creates an emf i (decreasing) Fig. 30-14 (a) The current i is increasing, inductor into a resistance r (whichconstant we take and to rightward, be outside (b) the constant region and of leftward, changing (c) in- that opposesand the self-induced emf Ᏹ appears along creasing and rightward,Ᏹ (d) decreasing and rightward, L flux) and an ideal inductor of self-induced emf L.As with a real battery of emf the change. the coil in a direction such that it opposes Ᏹ and internal resistance r,the potential(e) increasing difference and leftward, across (f) the decreasing terminals and of leftward? a real the increase.The arrow representing ᏱL can inductor then differs from the emf. Unless otherwise indicated, we assume here L be drawn along a turn of the coil or along- that inductors are ideal. side the coil. Both are shown. (b) The cur-

L rent i is decreasing, and the self-induced emf appears in a direction such that it opposes CHECKPOINT 5 the decrease. The figure shows an emf Ᏹ induced30-9 in aRL coil.Circuits Which of (b) L L the following can describe the current through the coil: (a) In Section 27-9 we saw that if we suddenly introduceFig. an emf30-14 Ᏹ into(a) Thea single-loop current i is increasing, constant and rightward, (b) constant and leftward, (c) in- circuit containing a resistor R and a capacitor C,the chargeand the self-inducedon the capacitor emf Ᏹ doesappears along creasing and rightward, (d) decreasing and rightward, L not build up immediately to its final equilibrium valuethe C coilᏱ but in a approaches direction such it that in an it opposes (e) increasing and leftward, (f) decreasing and leftward? exponential fashion: the increase.The arrow representing ᏱL can be drawn along a turn of the coil or along- q ϭ CᏱ(1 Ϫ eϪt/␶C). side the coil. Both are shown.(30-36) (b) The current i is decreasing, and the self-induced emf The rate at which the charge builds up is determinedappears by in the a direction capacitive such that time it opposes 30-9 RL Circuits constant tC,defined in Eq.27-36 as the decrease. t ϭ RC In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into aC single-loop. (30-37) circuit containing a resistor R and aIf capacitor we suddenly C,the remove charge onthe the emf capacitor from this does same circuit, the charge does not not build up immediately to its immediatelyfinal equilibrium fall tovalue zero C butᏱ but approaches approaches zero it in in an an exponential fashion: exponential fashion: Ϫt/␶C q ϭ q0e . (30-38) q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The time constant tC describes the fall of the charge as well as its rise. The rate at which the charge buildsAn upanalogous is determined slowing byof the rise capacitive (or fall) time of the current occurs if we introduce constant tC,defined in Eq.27-36an as emf Ᏹ into (or remove it from) a single-loop circuit containing a resistor R and an inductortC ϭ RC L..When the switch S in Fig.30-15(30-37) is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current If we suddenly remove the emf from this same circuit, the charge does not would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S immediately fall to zero but approaches zero in an exponential fashion: induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b + R the qcurrent,ϭ q eϪt which/␶C. means that it opposes (30-38)the battery emf Ᏹ in polarity. Thus, the 0 – current in the resistor responds to the difference between two emfs, a constant Ᏹ L The time constant t describes the fall of the charge as well as its rise. C due to the battery and a variable Ᏹ (ϭϪL di/dt) due to self-induction.As long as An analogous slowing of the rise (or fall) of the current occurs ifL we introduce an emf Ᏹ into (or remove it from)ᏱL ais single-looppresent, the circuit current containing will be less a resistor than Ᏹ /RR.and an inductor L.When the switch S inAs Fig.30-15 time goes is closed on, the on ratea,for at example,the which the current cur- increases becomes less rapid Fig. 30-15 An RL circuit.When switch rent in the resistor starts to rise.and If the the magnitude inductor were of notthe present, self-induced the current emf, which is proportional to di/dt, S is closed on a,the current rises and approaches a limiting value Ᏹ/R. would rise rapidly to a steady valuebecomes Ᏹ/R.Because smaller. Thus,of the theinductor,however,a current in the circuit self- approaches Ᏹ/R asymptotically.a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Fig. 30-15 An RL circuit.When switch and the magnitude of the self-induced emf, which is proportional to di/dt, S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically. proaches a limiting value Ᏹ/R. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

This means that when a self-induced emf is produced in the inductor of Fig. 30-13, i (increasing) we cannot define an electric potential within the inductor itself, where the flux is changing. However, potentials can still be defined at points of the circuit that are not within the inductor—points where the electric fields are due to charge distributions and their associated electric potentials. L

Moreover, we can define a self-induced potential difference VL across an L inductor (between its terminals, which we assume to be outside the region of The changing changing flux). For an ideal inductor (its wire has negligible resistance), the mag- current changes (a) nitude of VL is equal to the magnitude of the self-induced emf ᏱL. the flux, which If, instead, the wire in the inductor has resistance r,we mentally separate the creates an emf i (decreasing) inductor into a resistance r (which we take to be outside the region of changing that opposes flux) and an ideal inductor of self-induced emf ᏱL.As with a real battery of emf the change. Ᏹ and internal resistance r,the potential difference across the terminals of a real inductor then differs from the emf. Unless otherwise indicated, we assume here that inductors are ideal. Self-inductance question L L

CHECKPOINT 5 The figure shows an emf EL induced in a coil. The figure shows an emf Ᏹ induced in a coil. Which of (b) L L the following can describe the current through the coil: (a) Fig. 30-14 (a) The current i is increasing, constant and rightward, (b) constant and leftward, (c) in- and the self-induced emf Ᏹ appears along creasing and rightward, (d) decreasing and rightward, L the coil in a direction such that it opposes (e) increasing and leftward, (f) decreasing and leftward?Which of the following can describe the current through the coil: the increase.The arrow representing ᏱL can (A) constant and rightward be drawn along a turn of the coil or along- (B) increasing and rightward side the coil. Both are shown. (b) The cur- (C) decreasing and rightward rent i is decreasing, and the self-induced emf (D) decreasing and leftward appears in a direction such that it opposes 30-9 RL Circuits the decrease. In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into a single-loop circuit containing a resistor R and a capacitor C,the charge on the capacitor does not build up immediately to its final equilibrium value CᏱ but approaches it in an exponential fashion: q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The rate at which the charge builds up is determined by the capacitive time constant tC,defined in Eq.27-36 as tC ϭ RC. (30-37) If we suddenly remove the emf from this same circuit, the charge does not immediately fall to zero but approaches zero in an exponential fashion:

Ϫt/␶C q ϭ q0e . (30-38)

The time constant tC describes the fall of the charge as well as its rise. An analogous slowing of the rise (or fall) of the current occurs if we introduce an emf Ᏹ into (or remove it from) a single-loop circuit containing a resistor R and an inductor L.When the switch S in Fig.30-15 is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Fig. 30-15 An RL circuit.When switch and the magnitude of the self-induced emf, which is proportional to di/dt, S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically. proaches a limiting value Ᏹ/R. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

CHECKPOINT 5 The figure shows an emf EL induced in a coil. The figure shows an emf Ᏹ induced in a coil. Which of (b) L L the following can describe the current through the coil: (a) Fig. 30-14 (a) The current i is increasing, constant and rightward, (b) constant and leftward, (c) in- and the self-induced emf Ᏹ appears along creasing and rightward, (d) decreasing and rightward, L the coil in a direction such that it opposes (e) increasing and leftward, (f) decreasing and leftward?Which of the following can describe the current through the coil: the increase.The arrow representing ᏱL can (A) constant and rightward be drawn along a turn of the coil or along- (B) increasing and rightward side the coil. Both are shown. (b) The cur- (C) decreasing and rightward ← rent i is decreasing, and the self-induced emf (D) decreasing and leftward appears in a direction such that it opposes 30-9 RL Circuits the decrease. In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into a single-loop circuit containing a resistor R and a capacitor C,the charge on the capacitor does not build up immediately to its final equilibrium value CᏱ but approaches it in an exponential fashion: q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The rate at which the charge builds up is determined by the capacitive time constant tC,defined in Eq.27-36 as tC ϭ RC. (30-37) If we suddenly remove the emf from this same circuit, the charge does not immediately fall to zero but approaches zero in an exponential fashion:

Ϫt/␶C q ϭ q0e . (30-38)

PART 3 30-9 RL CIRCUITS 807

CHECKPOINT 5 The figure shows an emf Ᏹ induced in a coil. Which of (b) L L the following can describe the current through the coil: (a) Fig. 30-14 (a) The current i is increasing, constant and rightward, (b) constant and leftward, (c) in- and the self-induced emf Ᏹ appears along creasing and rightward, (d) decreasing and rightward, L the coil in a direction such that it opposes (e) increasing and leftward, (f) decreasing and leftward? the increase.The arrow representing ᏱL can be drawn along a turn of the coil or along- side the coil. Both are shown. (b) The current i is decreasing, and the self-induced emf appears in a direction such that it opposes 30-9 RL Circuits the decrease. In Section 27-9 we saw that if we suddenly introduce an emf Ᏹ into a single-loop circuit containing a resistor R and a capacitor C,the charge on the capacitor does not build up immediately to its final equilibrium value CᏱ but approaches it in an exponential fashion: q ϭ CᏱ(1 Ϫ eϪt/␶C). (30-36) The rate at which the charge builds up is determined by the capacitive time constant tC,defined in Eq.27-36 as tC ϭ RC. (30-37) If we suddenly remove the emf from this same circuit, the charge does not immediately fall to zero but approaches zero in an exponential fashion:

Ϫt/␶C q ϭ q0e . (30-38)

The time constant tC describes the fall of the charge as well as its rise. RL Circuits An analogous slowing of the rise (or fall) of the current occurs if we introduce Just like circuits with capacitors and resistor, circuits with an emf Ᏹ into (or remove it from) a single-loop circuit containing a resistor Rinductorsand and resistors have time-dependent behavior. an inductor L.When the switch S in Fig.30-15 is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Fig. 30-15 An RL circuit.When switch Initially, an inductor acts to oppose changes in the current through and the magnitude of the self-induced emf, which is proportional to di/dt, it. S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically. proaches a limiting value Ᏹ/R. A long time later, the current stabilizes and it acts like ordinary connecting wire. 32.2 RL Circuits 973 element that has a large inductance is called an inductor and has the circuit symbol When switch S1 is thrown . We always assume the inductance of the remainder of a circuit is negligi- closed, the current increases ble compared with that of the inductor. Keep in mind, however, that even a circuit and an emf that opposes the without a coil has some inductance that can affect the circuit’s behavior. increasing current is induced Because the inductance of an inductor results in a back emf,RL an Circuits inductor in Questiona cir- in the inductor. cuit opposes changes in the current in that circuit. The inductor attempts to keep R the current the same as it was before the change occurred. If the battery voltage in a S2 the circuit is increased so that the current rises, the inductor opposes this change b and the rise is not instantaneous. If the battery voltage is decreased, the inductor S causes a slow drop in the current rather than an immediate drop. Therefore, the 1 inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage. L ϩ Consider the circuit shown in Figure 32.2, which contains a battery of negligible e internal resistance. This circuit is an RL circuit because the elements connected to Ϫ the battery are a resistor and an inductor. The curved lines on switch S2 suggest this switch can never be open; it is always set to either a or b. (If the switch is connected to neither a nor b, any current in the circuit suddenly stops.) SupposeConsider S theis set circuit to a shownWhen with the switchS1 open S2 is thrown and S2 at position a. 2 to position b, the battery is no and switch S1 is open for t , 0 and then thrown closed at t 5 0.Switch The currentS1 is in now the thrown longer closed. part of the circuit and circuit begins to increase, and a back emf (Eq. 32.1) that opposes the increasing the current decreases. current is induced in the inductor. At the instant it is closed, across which circuit element is the voltage equal to the emf of the battery? With this point in mind, let’s apply Kirchhoff’s loop rule to this circuit, travers- Figure 32.2 An RL circuit. ing the circuit in the clockwise direction: (A) the resistor When switch S2 is in position a, the battery is in the circuit. di e 2 iR 2 L 5 0 (B) the inductor(32.6) dt (C) both each of the inductor and the resistor where iR is the voltage drop across the resistor. (Kirchhoff’s rules were developed for circuits with steady currents, but they can also be applied to(D) a circuitneither in which the current is changing if we imagine them to represent the circuit at one instant of time.) Now let’s find a solution to this differential equation, which is similar to that for the RC circuit (see Section 28.4). A mathematical solution of Equation 32.6 represents the current in the circuit as a function of time. To find this solution, we change variables for convenience, letting x 5 (e/R) 2 i, so dx 5 2di. With these substitutions, Equation 32.6 becomes L dx x 1 5 0 R dt Rearranging and integrating this last expression gives x dx R t 52 dt 3 x 3 x0 L 0 x R l n 52 t x 0 L where x0 is the value of x at time t 5 0. Taking the antilogarithm of this result gives

2Rt/L x 5 x0e

Because i 5 0 at t 5 0, note from the definition of x that x0 5 e/R. Hence, this last expression is equivalent to e e 2 i 5 e2Rt/L R R e i 5 1 2 e2Rt/L R This expression shows how the inductor affects1 the 2current. The current does not increase instantly to its final equilibrium value when the switch is closed, but instead increases according to an exponential function. If the inductance is removed from the circuit, which corresponds to letting L approach zero, the exponential term 32.2 RL Circuits 973 element that has a large inductance is called an inductor and has the circuit symbol When switch S1 is thrown . We always assume the inductance of the remainder of a circuit is negligi- closed, the current increases ble compared with that of the inductor. Keep in mind, however, that even a circuit and an emf that opposes the without a coil has some inductance that can affect the circuit’s behavior. increasing current is induced Because the inductance of an inductor results in a back emf,RL an Circuits inductor in Questiona cir- in the inductor. cuit opposes changes in the current in that circuit. The inductor attempts to keep R the current the same as it was before the change occurred. If the battery voltage in a S2 the circuit is increased so that the current rises, the inductor opposes this change b and the rise is not instantaneous. If the battery voltage is decreased, the inductor S causes a slow drop in the current rather than an immediate drop. Therefore, the 1 inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage. L ϩ Consider the circuit shown in Figure 32.2, which contains a battery of negligible e internal resistance. This circuit is an RL circuit because the elements connected to Ϫ the battery are a resistor and an inductor. The curved lines on switch S2 suggest this switch can never be open; it is always set to either a or b. (If the switch is connected to neither a nor b, any current in the circuit suddenly stops.) SupposeConsider S theis set circuit to a shownWhen with the switchS1 open S2 is thrown and S2 at position a. 2 to position b, the battery is no and switch S1 is open for t , 0 and then thrown closed at t 5 0.Switch The currentS1 is in now the thrown longer closed. part of the circuit and circuit begins to increase, and a back emf (Eq. 32.1) that opposes the increasing the current decreases. current is induced in the inductor. At the instant it is closed, across which circuit element is the voltage equal to the emf of the battery? With this point in mind, let’s apply Kirchhoff’s loop rule to this circuit, travers- Figure 32.2 An RL circuit. ing the circuit in the clockwise direction: (A) the resistor When switch S2 is in position a, the battery is in the circuit. di e 2 iR 2 L 5 0 (B) the inductor(32.6)← dt (C) both each of the inductor and the resistor where iR is the voltage drop across the resistor. (Kirchhoff’s rules were developed for circuits with steady currents, but they can also be applied to(D) a circuitneither in which the current is changing if we imagine them to represent the circuit at one instant of time.) Now let’s find a solution to this differential equation, which is similar to that for the RC circuit (see Section 28.4). A mathematical solution of Equation 32.6 represents the current in the circuit as a function of time. To find this solution, we change variables for convenience, letting x 5 (e/R) 2 i, so dx 5 2di. With these substitutions, Equation 32.6 becomes L dx x 1 5 0 R dt Rearranging and integrating this last expression gives x dx R t 52 dt 3 x 3 x0 L 0 x R l n 52 t x 0 L where x0 is the value of x at time t 5 0. Taking the antilogarithm of this result gives

2Rt/L x 5 x0e

Because i 5 0 at t 5 0, note from the definition of x that x0 5 e/R. Hence, this last expression is equivalent to e e 2 i 5 e2Rt/L R R e i 5 1 2 e2Rt/L R This expression shows how the inductor affects1 the 2current. The current does not increase instantly to its final equilibrium value when the switch is closed, but instead increases according to an exponential function. If the inductance is removed from the circuit, which corresponds to letting L approach zero, the exponential term 32.2 RL Circuits 973 element that has a large inductance is called an inductor and has the circuit symbol When switch S1 is thrown . We always assume the inductance of the remainder of a circuit is negligi- closed, the current increases ble compared with that of the inductor. Keep in mind, however, that even a circuit and an emf that opposes the without a coil has some inductance that can affect the circuit’s behavior. increasing current is induced Because the inductance of an inductor results in a back emf,RL an Circuits inductor in Questiona cir- in the inductor. cuit opposes changes in the current in that circuit. The inductor attempts to keep R the current the same as it was before the change occurred. If the battery voltage in a S2 the circuit is increased so that the current rises, the inductor opposes this change b and the rise is not instantaneous. If the battery voltage is decreased, the inductor S causes a slow drop in the current rather than an immediate drop. Therefore, the 1 inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage. L ϩ Consider the circuit shown in Figure 32.2, which contains a battery of negligible e internal resistance. This circuit is an RL circuit because the elements connected to Ϫ the battery are a resistor and an inductor. The curved lines on switch S2 suggest this switch can never be open; it is always set to either a or b. (If the switch is connected to neither a nor b, any current in the circuit suddenly stops.) SupposeConsider S theis set circuit to a shownWhen with the switchS1 open S2 is thrown and S2 at position a. 2 to position b, the battery is no and switch S1 is open for t , 0 and then thrown closed at t 5 0.Switch The currentS1 is in now the thrown longer closed. part of the circuit and circuit begins to increase, and a back emf (Eq. 32.1) that opposes the increasing the current decreases. current is induced in the inductor. After a very long time, across which circuit element is the voltage equal to the emf of the battery? With this point in mind, let’s apply Kirchhoff’s loop rule to this circuit, travers- Figure 32.2 An RL circuit. ing the circuit in the clockwise direction: (A) the resistor When switch S2 is in position a, the battery is in the circuit. di e 2 iR 2 L 5 0 (B) the inductor(32.6) dt (C) both each of the inductor and the resistor where iR is the voltage drop across the resistor. (Kirchhoff’s rules were developed for circuits with steady currents, but they can also be applied to(D) a circuitneither in which the current is changing if we imagine them to represent the circuit at one instant of time.) Now let’s find a solution to this differential equation, which is similar to that for the RC circuit (see Section 28.4). A mathematical solution of Equation 32.6 represents the current in the circuit as a function of time. To find this solution, we change variables for convenience, letting x 5 (e/R) 2 i, so dx 5 2di. With these substitutions, Equation 32.6 becomes L dx x 1 5 0 R dt Rearranging and integrating this last expression gives x dx R t 52 dt 3 x 3 x0 L 0 x R l n 52 t x 0 L where x0 is the value of x at time t 5 0. Taking the antilogarithm of this result gives

2Rt/L x 5 x0e

Because i 5 0 at t 5 0, note from the definition of x that x0 5 e/R. Hence, this last expression is equivalent to e e 2 i 5 e2Rt/L R R e i 5 1 2 e2Rt/L R This expression shows how the inductor affects1 the 2current. The current does not increase instantly to its final equilibrium value when the switch is closed, but instead increases according to an exponential function. If the inductance is removed from the circuit, which corresponds to letting L approach zero, the exponential term 32.2 RL Circuits 973 element that has a large inductance is called an inductor and has the circuit symbol When switch S1 is thrown . We always assume the inductance of the remainder of a circuit is negligi- closed, the current increases ble compared with that of the inductor. Keep in mind, however, that even a circuit and an emf that opposes the without a coil has some inductance that can affect the circuit’s behavior. increasing current is induced Because the inductance of an inductor results in a back emf,RL an Circuits inductor in Questiona cir- in the inductor. cuit opposes changes in the current in that circuit. The inductor attempts to keep R the current the same as it was before the change occurred. If the battery voltage in a S2 the circuit is increased so that the current rises, the inductor opposes this change b and the rise is not instantaneous. If the battery voltage is decreased, the inductor S causes a slow drop in the current rather than an immediate drop. Therefore, the 1 inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage. L ϩ Consider the circuit shown in Figure 32.2, which contains a battery of negligible e internal resistance. This circuit is an RL circuit because the elements connected to Ϫ the battery are a resistor and an inductor. The curved lines on switch S2 suggest this switch can never be open; it is always set to either a or b. (If the switch is connected to neither a nor b, any current in the circuit suddenly stops.) SupposeConsider S theis set circuit to a shownWhen with the switchS1 open S2 is thrown and S2 at position a. 2 to position b, the battery is no and switch S1 is open for t , 0 and then thrown closed at t 5 0.Switch The currentS1 is in now the thrown longer closed. part of the circuit and circuit begins to increase, and a back emf (Eq. 32.1) that opposes the increasing the current decreases. current is induced in the inductor. After a very long time, across which circuit element is the voltage equal to the emf of the battery? With this point in mind, let’s apply Kirchhoff’s loop rule to this circuit, travers- Figure 32.2 An RL circuit. ing the circuit in the clockwise direction: (A) the resistor When switch S2 is in position a, ← the battery is in the circuit. di e 2 iR 2 L 5 0 (B) the inductor(32.6) dt (C) both each of the inductor and the resistor where iR is the voltage drop across the resistor. (Kirchhoff’s rules were developed for circuits with steady currents, but they can also be applied to(D) a circuitneither in which the current is changing if we imagine them to represent the circuit at one instant of time.) Now let’s find a solution to this differential equation, which is similar to that for the RC circuit (see Section 28.4). A mathematical solution of Equation 32.6 represents the current in the circuit as a function of time. To find this solution, we change variables for convenience, letting x 5 (e/R) 2 i, so dx 5 2di. With these substitutions, Equation 32.6 becomes L dx x 1 5 0 R dt Rearranging and integrating this last expression gives x dx R t 52 dt 3 x 3 x0 L 0 x R l n 52 t x 0 L where x0 is the value of x at time t 5 0. Taking the antilogarithm of this result gives

2Rt/L x 5 x0e

808 CHAPTER 30 INDUCTION AND INDUCTANCE

We can generalize these results as follows:

Initially, an inductor acts to oppose changes in the current through it.A long time later, it acts like ordinary connecting wire.

Now let us analyze the situation quantitatively.With the switch S in Fig. 30-15 thrown to a,the circuit is equivalent to that of Fig.30-16.Let us apply the loop rule, starting at point x in this ﬁgure and moving clockwise around the loop along with current i. 1. Resistor. Because we move through the resistor in the direction of current i, the electric potential decreases by iR.Thus,as we move from point x to point y,we encounter a potential change of ϪiR.

2. Inductor. Because current i is changing, there is a self-induced emf ᏱL in the inductor.The magnitude of ᏱL is given by Eq. 30-35 as L di/dt.The direction of ᏱL is upward in Fig. 30-16 because current i is downward through the inductor and increasing. Thus, as we move from point y to point z,opposite the direc-

tion of ᏱL,we encounter a potential change of ϪL di/dt. 3. Battery. As we move from point z back to starting point x,we encounter a potential change of ϩᏱ due to the battery’s emf. Thus, the loop rule gives us di ϪiR Ϫ L ϩ Ᏹ ϭ 0 dt

di or L ϩ Ri ϭ Ᏹ (RL circuit). (30-39) dt

Equation 30-39 is a differential equation involving the variable i and its first derivative di/dt.To solve it,we seek the function i(t) such that when i(t) and its first derivative are substituted in Eq. 30-39, the equation is satisfied and the initial condition i(0) ϭ 0 is satisfied. Equation 30-39 and its initial condition are of exactly the form of Eq. 27-32 for an RC circuit, with i replacing q, L replacing R,and R replacing 1/C.The solution of Eq. 30-39 must then be of exactly the form of Eq. 27-33 with the same replacements.That solution is Ᏹ i ϭ (1 Ϫ eϪRt/L), (30-40) R which we can rewrite as

Ᏹ i ϭ (1 Ϫ eϪt/␶L) (rise of current). (30-41) R

RL Circuits Here tL,the inductive time constant, is given by L i ␶L ϭ (time constant). (30-42) x y R R Let’s examine Eq. 30-41 for just after the switch is closed (at time t ϭ 0) + L – L and for a time long after the switch is closed (t : ϱ) . If we substitute t ϭ 0 into Eq. 30-41, the exponential becomes eϪ0 ϭ 1.Thus, Eq. 30-41 tells us that the cur- z rent is initially i ϭ 0, as we expected. Next, if we let t go to ϱ,then the exponen- Ϫϱ Fig. 30-16 The circuit of Fig. 30-15 tial goes to e ϭ 0. Thus, Eq. 30-41 tells us that the current goes to its equilib- with the switch closed on a.We apply rium value of Ᏹ/R. Loop rule: the loop rule for the circuit clockwise, We can also examine the potential differences in the circuit. For example, Fig. starting Eat −x. EL − iR = 0 30-17 shows how the potential differences VR (ϭ iR) across the resistor and

di E − L −iR = 0 dt This is a diﬀerential equation. Solution:

E L i = (1 − e−t/τL ) ; τ = R L R Current varies with time

di E − L −iR = 0 dt

Rearranging: di E R = − i dt L L di R E = − i dt L R 1 R di = dt E/R − i L Z Z

The limits of our integral will be determined by the initial conditions for the situation we are considering. RL Circuits: current rising When charging an initially uncharged capacitor: i = 0 at t = 0

i 1 t R di’ = dt’ E/R − i 0 L Z0 Z0 R − ln(E/R − i) + ln(E/R − 0) = t L E/R Rt ln = E/R − i L E/R = eRt/L E/R − i

The solution is:

E i(t) = 1 − e−Rt/L R 974974 ChapterChapter 32 Inductance32 Inductance

AfterRL switchAfter Circuits: Sswitch1 is thrown S1 currentis thrownclosed closed rising The timeThe rate time of change rate of of change of at t ϭ 0,at tthe ϭ current0, the current increases increases current iscurrent a maximum is a maximum at t ϭ 0, at t ϭ 0, towardtoward its maximum its maximum value value which is thewhich instant is the at instant which at which e/R.Currente/R. in loop: Derivativeswitch S1 ofswitch is current:thrown S1 is closed. thrown closed.

i i di di dt dt e e e R e R L L L 0.632 e e t ϭ L R0.632 R t ϭ R R

t t t t t t Figure 32.3 Plot of the current versus Figuretime for 32.3the RLE Plot circuit of the current Figure 32.4di PlotE of di/dt versus versusi( timet) = for the1 − RLe− circuitRt/L Figure= 32.4e−Rt Plot/L of di/dt versus shown in Figure 32.2.R The time time for thedt RL circuitL shown in Fig- constantshown t is the in Figuretime interval 32.2. The time ure 32.2.time The forrate the decreases RL circuit exponen shown- in Fig- requiredconstant for i to t reach is the 63.2% time intervalof its tially withure time 32.2. as iThe increases rate decreases toward exponen- maximumrequired value. for i to reach 63.2% of its its maximumtially value.with time as i increases toward maximum value. its maximum value. becomes zero and there is no time dependence of the current in this case; the current increasesbecomes zeroinstantaneously and there isto no its time final dependence equilibrium ofvalue the incurrent the absence in this ofcase; the the cur- inductance.rent increases instantaneously to its final equilibrium value in the absence of the Weinductance. can also write this expression as We can also write this expressione as i 5 1 2 e2t/t (32.7) R e i 5 1 2 e2t/t (32.7) 1 R 2 where the constant t is the time constant of the RL circuit: 1 2 where the constant t is the time constantL of the RL circuit: t5 (32.8) R L t5 (32.8) Physically, t is the time interval required for theR current in the circuit to reach 21 (1 2Physically, e ) 5 0.632 t is5 the 63.2% time of intervalits final requiredvalue e/ R. for The the time current constant in the is circuita useful to reach parameter(1 2 efor21) comparing 5 0.632 5 the 63.2% time responsesof its final of value various e /circuits.R. The time constant is a useful Figureparameter 32.3 shows for comparing a graph of the the time current responses versus oftime various in the circuits. RL circuit. Notice that the equilibrium value of the current, which occurs as t approaches infinity, is Figure 32.3 shows a graph of the current versus time in the RL circuit. Notice e/R. That can be seen by setting di/dt equal to zero in Equation 32.6 and solving that the equilibrium value of the current, which occurs as t approaches infinity, is for the current i. (At equilibrium, the change in the current is zero.) Therefore, the e/R. That can be seen by setting di/dt equal to zero in Equation 32.6 and solving current initially increases very rapidly and then gradually approaches the equilib- for the current i. (At equilibrium, the change in the current is zero.) Therefore, the rium value e/R as t approaches infinity. Let’scurrent also investigate initially increases the time veryrate ofrapidly change and of thethen current. gradually Taking approaches the first timethe equilib- derivativerium ofvalue Equation e/R as 32.7 t approaches gives infinity. Let’s also investigate the time rate of change of the current. Taking the first time di e derivative of Equation 32.7 gives5 e2t/t (32.9) dt L di e This result shows that the time rate of change5 of the e2 currentt/t is a maximum (equal to (32.9) e/L) at t 5 0 and falls off exponentially todt zero asL t approaches infinity (Fig. 32.4).

NowThis consider result shows the RL that circuit the time in Figure rate of 32.2 change again. of Supposethe current switch is a maximumS2 has been (equal to set ate position/L) at t 5a long0 and enough falls off (and exponentially switch S1 remains to zero asclosed) t approaches to allow infinitythe current (Fig. 32.4). to reach Now its equilibrium consider the value RL ecircuit/R. In inthis Figure situation, 32.2 the again. circuit Suppose is described switch by S the2 has been outerset loop at inposition Figure a32.2. long If enoughS2 is thrown (and from switch a to S b1, remainsthe circuit closed) is now todescribed allow the by current only theto reach right-hand its equilibrium loop in Figure value 32.2. e/R. Therefore, In this situation, the battery the has circuit been iseliminated described by the 5 fromouter the circuit. loop in Setting Figure e 32.2. 0 in If EquationS2 is thrown 32.6 from gives a to b, the circuit is now described by only the right-hand loop in Figure 32.2.di Therefore, the battery has been eliminated from the circuit. Setting e 5iR 0 1in LEquation 5 0 32.6 gives dt di iR 1 L 5 0 dt halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 809

PART 3 30-9 RL CIRCUITS 809 halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 809

VL (ϭ L di/dt) across the inductor vary with time for particular values of Ᏹ, L, The resistor's potential and R.Compare this ﬁgure carefully with the corresponding ﬁgure for an RC difference turns on. circuit (Fig. 27-16). The inductor's potential To show that the quantity tL (ϭ L/R) has the dimension of time, we convert difference turns off. from henries per ohm as follows: PART 3 H H 1 Vи s 1 ⍀иA 80910 1 ϭ 1 ϭ 1 s.30-9 RL CIRCUITS 8 ⍀ ⍀ ΂ 1 HиA ΃΂ 1 V ΃ (V) 6 R

V 4 The first quantity in parentheses is a conversion factor based on Eq. 30-35, and VL (ϭ L di/dt) across the inductor vary with time for particular values of Ᏹ, L, RL Circuits:The resisto currentr's poten risingtial 2 the second one is a conversion factor based on the relation V ϭ iR. and R.Compare this figure carefully with the corresponding figure for an RC 0 2 4 6 8 The physical significance of the time constant follows fromdiffe Eq.ren 30-41.ce turn Ifs owen. circuit (Fig. 27-16). The inductor's potential t (ms) put t ϭ tL ϭ L/R in this equation, it reduces to Potential drop across resistor: emf across inductor:(a) To show that the quantity tL (ϭ L/R) has the dimension of time, we convert difference turns off. from henries per ohm as follows: Ᏹ Ᏹ i ϭ (1 Ϫ eϪ1) ϭ 0.63 . (30-43) H H 1 Vи s 1 ⍀иA R R 10 10 1 ϭ 1 ϭ 1 s. 8 8

⍀ ⍀ ΂Thus,1 Hи theA ΃ time΂ 1constant V ΃ t is the time it takes the current in the circuit to reach (V) 6 (V) L 6 L R V

V 4 about 63% of its final equilibrium value Ᏹ/R.Since the potential4 difference VR The first quantity in parentheses is a conversion factor based on Eq. 30-35, and 2 2 across the resistor is proportional to the current i,a graph of the increasing the second one is a conversion factor based on the relation V ϭ iR. 0 2 4 6 8 current versus time has the same shape as that of V in Fig. 30-170 a. 2 4 6 8 R t (ms) The physical significance of the timeIf the constant switch Sfollows in Fig. from 30-15 Eq. is closed 30-41. on If wea long enough for the equilibriumt (ms) put t ϭ t ϭ L/R in this equation, it reduces to (b) L current Ᏹ/R to be established and then is thrown to b,the effect will be( toa) remove di |EL(t)| = L Ᏹ the battery from theᏱ circuit. (The connection to b must actuallyV beR = madeiR an Fig. 30-17 The variationdt with time of i ϭ (1 Ϫ eϪ1) ϭ 0.63 . (30-43) R instant before the Rconnection to a is broken. A switch that10 does this is called a (a) VR,the potential difference across the make-before-break switch.) With the battery gone, the current8 through the resis- resistor in the circuit of Fig. 30-16, and (b)

Thus, the time constant t is the timetor will it takesdecrease. the However,current in it the cannot circuit drop to immediatelyreach to zero (V) 6 but must decay to VL,the potential difference across the in- L L about 63% of its final equilibriumzero value over Ᏹ time./R.Since The differentialthe potential equation difference that Vgoverns the decayV 4 can be found by ductor in that circuit.The small triangles R 2 across the resistor is proportionalputting to the Ᏹ ϭ current0 in Eq. i,a 30-39: graph of the increasing represent successive intervals of one induc- t ϭ current versus time has the same shape as that of V in Fig. 30-17a. 0 2 4 6 8 tive time constant L L/R.The figure is R di t (ms) plotted for R ϭ 2000 ⍀, L ϭ 4.0 H, and If the switch S in Fig. 30-15 is closed on a long enough for Lthe equilibriumϩ iR ϭ 0. (30-44) dt (b) Ᏹ ϭ 10 V. current Ᏹ/R to be established and then is thrown to b,the effect will be to remove the battery from the circuit. (TheBy connection analogy with to Eqs.b must 27-38 actually and 27-39, be made the solution an ofFig. this 30-17 differentialThe variation equation with time of instant before the connection to thata is broken.satisfies the A switch initial conditionthat does i (0)thisϭ isi 0calledϭ Ᏹ/R ais (a) VR,the potential difference across the make-before-break switch.) With the battery gone, the current through the resis- resistor in the circuit of Fig. 30-16, and (b) Ᏹ V ,the potential difference across the in- tor will decrease. However, it cannot drop immediately to zeroϪt /but␶L mustϪt /␶decayL to L i ϭ e ϭ i0e (decay of current).(30-45) zero over time. The differential equation that governs theR decay can be found by ductor in that circuit.The small triangles putting Ᏹ ϭ 0 in Eq. 30-39: represent successive intervals of one induc- We see that both current rise (Eq. 30-41) and currenttive decay time (Eq. constant 30-45) tL inϭ anL/R RL.The figure is circuitdi are governed by the same inductive time constant,plottedtL. for R ϭ 2000 ⍀, L ϭ 4.0 H, and L ϩ iR ϭ 0. (30-44) Ᏹ ϭ 10 V. dt We have used i0 in Eq. 30-45 to represent the current at time t ϭ 0. In our case that happened to be Ᏹ/R,but it could be any other initial value. By analogy with Eqs. 27-38 and 27-39, the solution of this differential equation that satisfies the initial condition i(0) ϭ i0 ϭ Ᏹ/R is CHECKPOINT 6 Ᏹ Ϫt/␶L Ϫt/␶L i ϭ e ϭ Thei0e figure shows(decay three of current) circuits.(30-45) with identical batteries, inductors, and resistors. Rank R the circuits according to the current through the battery (a) just after the switch is closed and (b) a long time later, greatest first. (If you have trouble here, work through We see that both current rise (Eq. 30-41)the next and sample current problem decay and (Eq. then 30-45)try again.) in an RL circuit are governed by the same inductive time constant, tL. We have used i0 in Eq. 30-45 to represent the current at time t ϭ 0. In our case that happened to be Ᏹ/R,but it could be any other initial value.

CHECKPOINT 6 The ﬁgure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current through the battery(1) (a) just after the switch (2) is (3) closed and (b) a long time later, greatest ﬁrst. (If you have trouble here, work through the next sample problem and then try again.)

(1) (2) (3) RL Circuits: Time Constant

τL = L/R

τL is called the time constant of the circuit.

(Notice that it is deﬁned diﬀerently for RL circuits as opposed to RC circuits.)

This gives the time for the current to reach (1 − e−1) = 63.2% of its ﬁnal value.

Alternatively, it is the time for the potential drop across the inductor to fall to 1/e of its initial value.

It is useful for comparing the “relaxation time” of diﬀerent RL-circuits. halliday_c30_791-825hr.qxd 11-12-2009 12:19 Page 807

PART 3 30-9 RL CIRCUITS 807

Ϫt/␶C q ϭ q0e . (30-38)

The time constant tC describes the fall of the charge as well as its rise. An analogous slowing of the rise (or fall) of the current occurs if we introduce an emf Ᏹ into (or remove it from) a single-loop circuit containingRL a resistor Circuits R and an inductor L.When the switch S in Fig.30-15 is closed on a,for example,the current in the resistor starts to rise. If the inductor were not present, the current would rise rapidly to a steady value Ᏹ/R.Because of the inductor,however,a self- a S induced emf ᏱL appears in the circuit; from Lenz’s law, this emf opposes the rise of b R the current, which means that it opposes the battery emf Ᏹ in polarity. Thus, the + – current in the resistor responds to the difference between two emfs, a constant Ᏹ L due to the battery and a variable ᏱL (ϭϪL di/dt) due to self-induction.As long as ᏱL is present, the current will be less than Ᏹ/R. As time goes on, the rate at which the current increases becomes less rapid Battery switchedFig. out 30-15 - switchAn to RLb.circuit.When switch and the magnitude of the self-induced emf, which is proportional to di/dt, S is closed on a,the current rises and ap- becomes smaller. Thus, the current in the circuit approaches Ᏹ/R asymptotically.Loop rule: proaches a limiting value Ᏹ/R. EL − iR = 0

di −L −iR = 0 dt Solution:

E L i = e−t/τL ; τ = R L R 32.2 RL Circuits 975

It is left as a problem (Problem 22) to show that the solution of this differential ϭ equation is At t 0, the switch is thrown to RL Circuits: currentposition falling b and the current has its maximum value e/R. e 2t/t 2t/t i 5 e 5 Ii e (32.10) R i where e is the emf of the battery and Ii 5 e/R is the initial current at the instant e the switch is thrown to b. R If the circuit did not contain an inductor, the current would immediately decrease to zero when the battery is removed. When the inductor is present, it opposes the decrease in the current and causes the current to decrease exponentially. A graph of the current in the circuit versus time (Fig. 32.5) shows that the current is continuously decreasing with time. t

Q uick Quiz 32.2 Consider the circuit in Figure 32.2 with S1 open and S2 at posi- Figure 32.5 Current versus E L time for the−t /τright-handL loop of tion a. Switch S1 is now thrown closed. (i) At the instant it is closed, across which i = e ; τL = circuit element is the voltage equal to the emf of the battery? (a) the resistor the circuitR shown in Figure R32.2. For t , 0, switch S is at position a. (b) the inductor (c) both the inductor and resistor (ii) After a very long time, 2 across which circuit element is the voltage equal to the emf of the battery? Choose from among the same answers.

Example 32.2 Time Constant of an RL Circuit

Consider the circuit in Figure 32.2 again. Suppose the circuit elements have the following values: e 5 12.0 V, R 5 6.00 V, and L 5 30.0 mH. (A) Find the time constant of the circuit.

SOLUTI O N

Conceptualize You should understand the operation and behavior of the circuit in Figure 32.2 from the discussion in this section. Categorize We evaluate the results using equations developed in this section, so this example is a substitution problem. L 30.0 3 1023 H Evaluate the time constant from Equation 32.8: t5 5 5 5.00 ms R 6.00 V

(B) Switch S2 is at position a, and switch S1 is thrown closed at t 5 0. Calculate the current in the circuit at t 5 2.00 ms.

SOLUTI O N e 12.0 V Evaluate the current at t 5 2.00 ms from i 5 1 2 e2t/t 5 1 2 e22.00 ms/5.00 ms 5 2.00 A 1 2 e20.400 R 6.00 V Equation 32.7: 5 0.6591 A 2 1 2 1 2

SOLUTI O N At the instant the switch is closed, there is no current and therefore no potential difference across the resistor. At this instant, the battery voltage appears entirely across the inductor in the form of a back emf of 12.0 V as the inductor tries to maintain the zero-current condition. (The top end of the inductor in Fig. 32.2 is at a higher electric potential than the bottom end.) As time passes, the emf across the inductor decreases and the current in the resistor (and hence the voltage across it) increases as shown in Figure 32.6 (page 976). The sum of the two voltages at all times is 12.0 V.

WHAT IF ? In Figure 32.6, the voltages across the resistor and inductor are equal at 3.4 ms. What if you wanted to delay the condition in which the voltages are equal to some later instant, such as t 5 10.0 ms? Which parameter, L or R, would require the least adjustment, in terms of a percentage change, to achieve that? continued Summary

• inductance • RL circuits

Homework Serway & Jewett: • Ch 32, onward from page 988. Obj. Qs: 3, 5; Conc. Qs.: 5; Probs: 1, 3, 9