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# Non PHYS 2020

MTSU

GPA 3.61

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This 40 page Class Notes was uploaded by Gilda Wisozk on Wednesday September 23, 2015. The Class Notes belongs to PHYS 2020 at Middle Tennessee State University taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/213136/phys-2020-middle-tennessee-state-university in Physics (Mtsu And Rodp) at Middle Tennessee State University.

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Math Review Page 1 of 5 Mathematical Equations This page provides links to various web pages containing review equations for several areas of mathematics These pages are not intended to explain and do examples of the associated mathematics please see a mathematics text for that However the links below will provide lists of useful relations and de nitions Click on the topic below to go to the relevant page The BACK button on your browser will then return you to this page Alg bra geometry Trigonometry Exponential Functions expx eX NaturalLanLi Lms lnx Basic Physics Equations from First Semester Physics ALGEBRA The quadratic equation bi 2 aZZbZcD gt 2 lb 43 2a GEOMETRY Triangle Sum of angles 1800 Area Math Review Page 2 of 5 Circle Circumference C 211R Area A 91R Sphere Surface Area A 411ng Volume v i3nR3 Cylinder Q Surface Area A 2 R2 2aIRH Volume V nR 2H TRIGONOMETRY rOpp side opposite angle 8 radj Slde adjacent to angle 9 Math Review rhyp hypotenuse of right triangle ropp rhyp sin 8 radj rhyp COSB tan9 l gt Htanl 93 rad adj sin2 8 cos2 8 1 Page 3 of 5 KINEMATICS LINEAR Assume a constant ANG ULAR Assume on constant I X1 Xi Vixt ZLath 1 2 8f Bi itI 50 Xf Xi Vix V it 8f 6i cofjt VfX ViX axt f i06t 2 2 Vs Vix 23xXf Xi 2 2 03f 03a 2U9f 9i NEWTON S 2nd LAW Linear Zgm2m2 m Angular Math Review Page 4 of 5 WORK AND ENERGY where WF work done by the force F x displacement of the object under consideration and FX component of the force F in the direction of the displacement x how much of F points in the direction of the object s motion WorkEnergy Theorem W total AKE KEf KEi Kinetic Energy Energy of Motion 2 2 KEtranslation mv KErotation I Potential Energy Energy of Position 2 PEgravity mg PE spring kx Conservation of Energy KEi PEi KEf PEf Wnc where WHO is the work done by any nonconservatz39ve forces such as friction or explosive forces CIRCULAR MOTION 2 ZFC mt R where Fc component of force in the radially inward centripetal direction Math Review CONSERVATION OF MOMENTUM Linear m i Zpix Zpeg and Zpiy pry Angular iIo393 zLiz 2sz MOMENTS OF INERTIA Description of Object Moment of Inertia cylindrical shell or ring of radius R and I m 2 mass M about an axis through its center and along the axis of symmetry solid cylinder or disk of radius R and l m 2 mass M about its axis of symmetry thinshelled hollow sphere of radius R I 2 MR 2 and mass M about an axis through its center solid sphere of radius R and mass M I 2 m 2 about an axis through its center thin rod of length L and mass M about I i ML an axis perpendicular to the rod at one 3 end thin rod of length L and mass M about I T1i ML2 an axis perpendicular to the rod through its center AAAAI I 1 139quot n Page 5 of 5 A 1 IAAAA L15a dc Circuits II Capacitors and RC Circuits Conceptual Page 1 of 1 L153 Capacitors and RC Circuits In the previous lectures we have seen how the presence of an electric eld results in the movement of charge within a conductor When the relationship between the potential difference across the ends of the conductor and the resulting current is a linear one then we say that the conductor is a resistor In the circuits that we have studied thus far a power source pushed current through one or more loops in the circuit Other than the power source the circuit has consisted solely of resistors Our approach to nding the current in the circuit has been to reduce the resistors to one equivalent resistance and then effectively to apply Kirchhoff S voltage law to solve for the resulting current In this formalism it is implicitly assumed that the current remains constant not only in direction but also in magnitude In this lecture we shall discuss circuits which contain a new circuit element called a capacitor In such circuits the current magnitude will not be constant with time but will rather vary in a very speci c mathematical manner This lecture will introduce us to the capacitor and to the qualitative description of circuits containing a capacitor The next lecture will introduce us to the quantitative description of such circuits hffnHmtcn prinnl1vc7n7nn pr htrpcT 17T 152 lqnhnriv H lt51 hfml W qnno Capacitors Page 1 of 2 Introduction to Capacitance A capacitor simply consists of two conductors which are electrically isolated from one another This means that no current can readily ow from one conductor to the other Figure 151 shows the generic set up of a capacitor The most common depiction of a capacitor is one in which the two conductors are in the form of two parallel planes or plates For this reason we will often refer to one or both of the conductors in a capacitor as plates It is imponant keep in mind however that capacitors come in shapes other than the parallel plate capacitor For example cylindrical capacitors are very common Lama 33 W Figure 151 Capacitors store charge No current ows through a capacitor but rather it just builds up or is stored on the conductors in the capacitor The two conductors or the two plates in a capacitor are initially uncharged that is they have a total charge of zero To charge a capacitor we remove charge from one of the plates and place it on the other This happens in such a way that the net charge on the capacitor is always equal to zero and that the amount of charge on one plate is equal in magnitude and opposite in sign to the charge on the other plate Figure 152 below shows two forms of the generic circuit symbol for the capacitor The rst circuit symbol in the gure represents the simplest geometry for a capacitor the parallelplate capacitor It is this symbol that we will use in our circuit diagrams Figure 152 It is found experimentally that the potential difference magnitude between the two conductors in a capacitor is proportional to the magnitude of charge on either one of the conductors This means that we can write that the magnitude of the potential difference is equal to a constant times the magnitude of charge on either conductor where the constant is a constant proportionality Writing the constant of proportionality in the form 1C where C is called the capacitance of the capacitor it then follows that 1 avg E Q 151 LWllmfnn curlquot nla rnqnqnT nnhnnnT 1 0 T 19 1 qanoru39fnrnknrdxy nnnnru39iu vn Lm1 3 l5 Innno Capacitors Page 2 of 2 where we have written the subscript C on AV to emphasize the fact that the potential difference is measured across the capacitor Solving for the capacitance C we get that C lQAVC so that the units of the capacitance must equal one coulomb per volt which is de ned to be one farad F onefaradzonew 131 152 v0 t Typical capacitances are usually in units OfJF 10 6 F 11F 10 9 F orpF 10 12 F The value of the capacitance is basically determined by three factors the area of the plates the separation between the plates and the material sandwiched between the plates This material is called a dielectric Typical dielectrics for capacitors are ceramic polystyrene or oil Again we re using the terminology of the generic parallel plate capacitor in this discussion but the fundamental ideas are similar for all geometries of capacitors To make the capacitance larger the area can be increased so that effectively the charges put on the conductors have a larger area to spread themselves over the separation can be decreased for a given charge on the plates a smaller distance between the plates decreases the voltage between them or the dielectric between the plates can be changed The ceramic capacitors often found in labs tend to have a parallelplate geometry encased in the protective ceramic On the other hand the small metal case capacitors tend to consist of layers of conducting material and dielectric rolled up to t into a smaller package Again the case is just to protect the capacitor We note that capacitors can be combined in series and parallel much as was done with resistors However the rules for calculating the equivalent capacitance are switched from the rules for combining resistors Series Combination i 12 1 012 Cl 02 Parallel Combination 39 Q4 Q C4 154 153 The method of approach for combining capacitors is exactly the same as that for combining resistors even though the equations for combining them are reversed and will be addressed no further in these lectures Some simple exercises on combining capacitors will be included in the homework problems just to make sure that you get the idea of how it is done hftnImtcn Priiinllc n nT pntancT l7T 1 527 1 lt9 qnnr itnrclnnrlv nanam39fnrc lam Ol Onno RC Circuits Page 1 of 2 RC Circuits The following sections will address the qualitative description of circuits containing resistors and capacitors which are charging and discharging In this section we simply set up and describe the generic RC circuit that will be used in the discussions of the following sections The generic circuit consisting of a power source AVS a resistor R and a capacitor C as shown below in Fig 153 The switch in the circuit can be thrown to either position A or to position B It is assumed unless stated explicitly otherwise that the capacitor is initially uncharged before the switch is thrown to position B at the time t 0 Switch R Figure 153 Let s take a moment to examine the circuit in these two con gurations the switch at positions A and B Unless stated otherwise we will assume the switch is thrown to position B at the time t 0 When the switch is at the position B the circuit consists of the single loop which contains starting at point B and moving around the circuit clockwise the resistor R the capacitor C and nally the voltage source AVS In this con guration the voltage source attempts to push charge around the circuit in a clockwise direction remember that the power source tries to push current out of its positive terminal After some time at position B we will throw the switch to position A The time since the switch was thrown to position A is called a new time t The prime on a symbol is used to denote the fact that this is the value of the quantity under consideration since the switch was thrown to position A It therefore follows that the switch is thrown to the position A at the time t z 0 After the switch has been thrown to position A the circuit consists solely of the resistor and the capacitor with no voltage source In this case there is no external energy being used to move charges around the circuit loop The following section will qualitatively discuss the behavior of the voltage across the capacitor as a function of time and the current around the circuit and in particular through the resistor as a function of time for the case in which the switch has been thrown to position B This is the case of the charging capacitar The section after that will discuss the case in which the switch has been thrown to position A the case of LHaJmhu oAnL Lunononn anf11vnnT 1quot T 1QT 1RaDF f irnnx t39nlanrlu vn nvn 14n Lm1 nc nno RC Circuits Page 2 of 2 the discharging capacitor ht mmtsuedunhvs 2070TenmrmT17115211 an Pirm tckndu rn Mfrmun mm ognnno Charging Voltage Page 1 of 3 Charging Capacitor Voltage Consider again the circuit shown in Fig 153 At the time t 0 the switch in the circuit is thrown from an intermediate position between points A and B to the position B thereby including the voltage source in the circuit We wish to examine the qualitative behavior of the potential difference across the capacitor and the current through the resistor as functions of time Switch R q I av A 03 TJ lli i Figure 153 We rst recall that from the de nition of capacitance IQ C AVG This tells us that if the charge on the capacitor increases then the potential difference across the capacitor will increase Remember that the capacitance C depends only on constants such as the size and shape of the conductors and any material or dielectric between the conductors Unless stated explicitly otherwise we shall always assume that the capacitor is initially uncharged Therefore at the time t 0 when the switch is thrown to position B the voltage source immediately introduces an electric eld throughout the circuit this electric eld tries to push charge in a clockwise fashion around the circuit Initially the charge is very easily moved around the circuit since the capacitor consists of wide open conductors at least from the point of View of the tiny electrons being moved onto which the charges can readily be pushed It is very important to keep in mind that as positive charge is pushed onto the top plate of the capacitor an equal amount of positive charge is repelled om the bottom plate of the capacitor thereby giving the bottom conductor a charge equal to negative that on the top conductor See Fig 154 below From our view outside of the circuit it appears as if the charge has moved all the way around the circuit from the top of the battery through the resistor down through the capacitor and back into the negative terminal of the power source It is important to remember that such is not the case as positive charge moves out the positive terminal of the battery clockwise around the circuit through the resistor it piles up on the top conductor of the capacitor and a di erent but equal charge leaves the bottom plate of the capacitor and continues the journey around the bottom portion of the circuit and up into the negative terminal of the battery LHmnn grinA n1nm fff nnhnonT 1quot T 1QT 1ltnh1anrn1nn Inlimnn l11 ALA Ir Ann Charging Voltage Figure 154 No charge jumps om the top conductor in the capacitor to the bottom conductor Charge just builds up on one plate and is repelled from the other Once again immediately after the switch is thrown to position B in Fig 153 it is very easy for the power source to push charge around the circuit Soon thereafter however the charges that have already been pushed onto the top conductor in the capacitor start repelling new charges that the power source is tryin g to push onto that top conductor As a result as time goes on the rate at which charge is moved around the circuit by the power source decreases that is the current decreases However even with a smaller current positive charge is still being moved clockwise around the circuit and therefore the magnitude of charge on each plate of the capacitor continues to increase with time just not at as large a rate as was initially the case immediately after the switch was thrown to position B It therefore follows that we expect the potential difference across the capacitor to increase with time from an initial value of 0 since the capacitor was initially uncharged The rate of increase of potential difference across the capacitor is relatively large at rst but as time goes on the rate of increase declines Eventually the power source will have pushed as much charge onto the conductors of the capacitor as it possibly can at which point the potential difference across the capacitor will stop increasing and will reach a constant value Theoretically this occurs at a time t in nity We will call this maximum value of the potential difference across the capacitor AVmaX This corresponds to an amount of charge Qmax on the capacitor plates The expected behavior of the potential difference across the capacitor as a function of time is therefore as shown in Fig 155 below Don t worry about the labels for t 0 and throwing the switch to position Awwe ll get to this in a bit lnlfanp knr x7 nkarn inn laHnHmfcn n 11Inhxrc9n7nn 3thch 1 7T TRT l a knroino Page 2 of 3 01400 Charging Voltage Page 3 of 3 ii i n m N quotha WWWWWWW WW MWMWW Figure 155 htmmtsuedunhv92020IecturesT1Q TJ81Sa Thmoino anmoe mndv ohmm39no ORonnR Charging Current Page 1 of 1 Charging Capacitor Current From the point of View of current in the circuit when the switch is initially thrown to position B the capacitor acts like a short circuit that is it acts as if the capacitor were removed from the circuit and replaced by a straight wire At this instant the instant after the switch is thrown to position B the circuit effectively consists of only a power source and a resistor so that the potential difference across the resistor is simply equal to the potential difference of the power source From Ohm s law we then get that AVS IR It then follows that the initial value of the current in the circuit is equal to I AVSR This is the value of the current in the circuit only immediately after the switch is thrown to position B The current in the resistor then starts decreasing with time since it gets increasingly harder for the power source to push more charge onto the capacitor plates until a long time later the current approaches zero as the charge on the capacitor reaches the maximum value the power source is capable of placing on it When the capacitor has become fully charge that is when the power source has pushed as much charge into the capacitor as it can and the Coulomb repulsive force between the positive charges piled on the top conductor of the capacitor is so large that the force due to the electric eld from the voltage source is not great enough to overcome it and push additional charge on to the upper conductor the current in the circuit stops 1 0 and the capacitor effectively acts like an open circuit That is the capacitor acts as if it were simply removed from the circuit and not replaced with anything at all so that the two wires in the circuit one from the resistor at top and one from the lower end of the battery at the bottom are just left hanging Recalling the current needs a closed loop around which to ow we can once again see that no current will ow through the circuit in this case From the above discussion it follows that we would expect the current in the circuit as a function of time to behave as follows C acts like sham circuit C acts iike open circuit Figure 156 ht mlmtqn minnhvc707nlentm ecl17TlRll 52Chnroino nrrpnf an u nhorm nn n Olglonno Discharging Page 1 of 2 Discharging Capacitor Switch R avg A T c 39lH l Figure 153 After some amount of time at position B in Fig 153 so that the voltage across the capacitor has reached a certain value which we shall call AV 039 the 0 and the remind us that this is the voltage at the new time t 0 we throw the switch to position A See Fig 155 This effectively removes the voltage source from the circuit leaving us only with the resistor and the capacitor AV which has some amount of positive charge piled onto its upper conductor and an equal magnitude of negative charge on its lower 531535 conductor These two charges would like nothing more than to be reunited and now that the power source has been removed from the circuit they see a way to achieve that goal As soon as the switch is thrown to position A the charge on the top plate f immediately starts moving counter 0 t clockwise around the circuit and recombining with the negative charge on the Figure 1 55 bottom plate This rush of charge that is current starts off very rapidly since the Coulomb repulsive force is initially very large since there is lots of positive charge crammed onto that top plate But as time goes on the repulsive force decreases since there is not as much charge piled on the upper plate anymore and therefore so does the current until eventually the current reaches zero after all the charge has been removed from the upper plate of the capacitor and recombined with the negative charge on the lower plate Again it will theoretically take an in nite amount of time for this happen 1 Thus we would expect the voltage across the capacitor and the current through the resistor to behave as shown in Figs 157 and 158 below Note that the current through the resistor is now negative since it is owing in the opposite direction compared to when the switch was at position B thlmtcn primnl1vcon 7nT Pl fHTPQI 1 7J 1 RN lRQmicr hnroinohnrlv Aicnhqroina html 9147an Discharging Page 2 of 2 mwm M si hiak m W mwxwww W t Jx ir l m w m MM w M G W M M Mw wm 3 g x i rmv switch g g v A i s x rt 3 x t j W W 4 Figure 157 Figure 158 h hJmen 9dnnhvc39n7n Pr herQT 1quotT 1R 1ltam crhar0innnn u Aicnkmorn39nn khan OIK mnno 801151 Page 1 of2 Solution to Example 151 Consider the circuit shown in Fig 159 below At t 0 the switch is thrown to position A and the capacitor is initially uncharged Data AVS 50V R1 10 k9 39R2 50 kg A 1 D AVE C R l T Figure 159 a Find the current through each resistor immediately after the switch is thrown to position A Immediately after the switch is closed to position A any uncharged capacitor in the circuit acts like a wire that is a short circuit Therefore to see how the circuit behaves at this time all we have to do is to take out the capacitor and put a wire in its place If we do this we end up with a wire shorting out the resistor R2 This means that all of the current owing out of resistor R1 ows down through the Wire that replaces the capacitor and no current ows through the resistor R2 Current owing out of R1 has two choices when it reaches the junction point in the circuit after R1 it can either go into the branch with the resistance R2 or it can go into the branch with zero resistance where the capacitor is acting like a wire All of the current will ow into any region with zero resistance Thus the current through resistor R2 is 12 0 A To nd the current through resistor R1 we simply redraw the circuit with R2 removed since it does not participate in the circuit at this time and with the capacitor replaced by a wire as shown in the gure ht mHmtqn Rd11nhVQ7070lRCillTRQTl7Tl RTl 2Fhrl 4 lRnll 5 lhnriv gnll 5 l html 767002 801151 Page20f2 The voltage across the resistor R1 in this circuit is simply the source voltage AVS If the current owing out of the source voltage is denoted IS then we simply have that AVS ISRI or IS AVSR1 50 mA The current owing through the resistor R1 is just the current IS Therefore 11 50 mA b Find the current through each resistor a very long time after the switch was thrown to position A A very long time after the switch has been thrown to position A the fully charged capacitor can take no more charge and so stops any current from owing through its branch of the circuit This means that the capacitor acts like an open circuit which effectively means that we cut that branch right out of the circuit If we do this the circuit then looks like that shown in the gure This is a simple series circuit at least it should be simple at this point The current in both resistors must be the same and has the value make sure you can show this 11 12 AVSRl R2 33 mA LH JM mn nAn ammononn nnhucnnT 1quot T 1QT 1nEr1lt 1C AT1C 1 AAA quot11 1 L1 AirAnna Homework 15a Page 1 of 2 Homework 153 1 Describe the behavior of the capacitor in an RC circuit immediately after the switch is closed and the power source is suddenly connected in the circuit and a very long time after the switch is closed Explain why the capacitor behaves this way 2 The circuit shown below initially contains an uncharged capacitor At time t O the switch is thrown to Position A thereby connecting the SO V power supply into the circuit What is the current in the resistor a immediately after the switch is thrown to Position A and b a long time after the switch is thrown to Position A 33 k0 J sV v A l Switch 1 015 ILF 50 V 3 The circuit shown below contains two initially uncharged capacitors At time t 0 the switch is thrown to Position A What is the current in each of the resistors a immediately after the switch is thrown to Position A and b a long time after the switch is thrown to Position A 33 m l l mm Al 4 Switch 17 F 22 km 9 V T 1 m VV V 4 The circuit shown below contains two initially uncharged capacitors At time t t1 Switch 1 is thrown to Position A Switch 2 is left open At some later time t t2 a very long time after the time t t1 Switch 2 is thrown to position B Find the current kmlmtnn nr 11m1ntlnqnfnT nnhnnaT 1 T 1QT 1ltofLTX71ltoknr1n land in Lml cnnno Homework 15a Page 2 of 2 in each of the resistors 21 just after time t b just before the time t2 and c a long time after the time t2 1 Is the current in any of the resistors ever nonzero between the times t2 and a long time after t2 Explain Switch 1 A 3 m W r B J 1 m Switch 2 j 12 v 22 km 15 IF 95 LF T An SW8 t 13 lamaHmequot prl11an1ve7070T pnfnrpcT 17T lRT lltQMWl ltal1nrlv l nxl 49 km oAonno HWAnslSa Page 1 of 1 Answers to Homework 15a 1 See the lecture notes 2 a 152 mA to right b o 3 a 22 kg resistor 41 mA down 33 kQ resistor 21 mA to right 1 k9 resistorzs 21 mA to left b O in all resistors 4 a 1 kg resistor 12 mA up 3 k9 resistorz O 22 kg resistor O b 0 in all resistors c 0 in all resistors 1 Yes For times after t2 but not very long after current will be owing counterclockwise around the right hand loop this current will be decreasing with time as the 95 11F capacitor charges LJmnn nr 11 nkunqn H39XT anhrrnon 1quot T lQT 1RnUXX714oT I X7Ano1 Cnknrln 1n n1Cn 1dr o nnno L15b More with RC Circuits Quantitative L15b More with RC Circuits In the last lecture we discussed the expected qualitative behavior of an RC circuit containing in general a resistor a capacitor and a power source We found that when the voltage source is switched into the RC circuit the voltage across the capacitor increases with time toward a constant maximum value AVmax while the current in the circuit decreases from the maximum initial value given by AVsR to zero We also found that when the voltage source is switched out of the complete circuit both the current and the voltage across the capacitor are seen to decay In this lecture we will discuss the quantitative features of this circuit and we will nd that the behavior of the quantitative description matches our expectations as discussed in the previous lecture It will be important for you to have care tlly studied the Math Re ejg sections on expone11t and logarithms as the properties of these functions will be eely exploited in this lecture h n39ml39 editnhvs 20 20l lecturesLl 392 T l RT l Shhnrlv ll Shhtml Page 1 of l 7670052 Charging C I Page 1 of 2 Charging Capacitor Current We start our quantitative description of the RC circuit by examining the generic RC circuit from the last lecture as shown in Fig 1510 below Switch R E C 39 l AVE A AVE I T4 I Figure 1510 At t 0 the switch is thrown to position B and the capacitor starts charging Let s apply Kirchhost voltage law the voltage must have the same value after going around a closed loop as it did at the beginning of the loop to the circuit some time after t 0 Starting at point B and going around the circuit clockwise and keeping in mind that a current is owing clockwise around the circuit we rst cross the resistor in the same direction as the current ow then we cross the capacitor from the positive conductor to the negative conductor and then we cross the voltage source from its negative terminal to it positive terminal We are then back at point B where we started In equation form this trip around the circuit loop looks like the following VB an avc mtg vB 155 or 1R avcavsn 156 From Eq 156 we get the following expression for the current I I iii mtg avc 15 Equation 157 shows us that if we know the voltage of the power source and the voltage across the capacitor then we immediately know the current through the resistor in the circuit which since this is a singleloop circuit so that all of the circuit elements are in series with one another must also be the current through the capacitor and through the voltage source Remember that no current really ows through the capacitorwcharge piles up on one plate as charge is repelled off of the other plate In what follows we will therefore only give you the equation for the voltage across the capacitor as a function of time If you want to nd the current in the circuit at any time you can rst nd the ht rnHmtcn editnhvc7n7nTenhirRRTl7Tl RI qhnharainc r Ihndv nharoina n 139 hfml 0ROnn9 Charging C I Page 2 of 2 voltage and then follow through the steps for Eqs 155 through 157 to nd the desired current InfraNmequot prhINnhvu397nonT Pt hn PQT 1 7T 152 IREPharohna r Tknr v nkarr nn n km1 0mman Charging C V Page 1 of 1 Charging Capacitor Voltage The equation for the potential difference between the conductors in the capacitor as a function of time is given by avctavm 1 6 quot 153 Where AVmax IS the maxnnum voltage across the capacrtor after a very long time wind will be just the voltage of the power source for the generic circuit of Fig 1510 with the switch at position B mm ms 159 and the constant 1 the Greek letter tan is called the RC time constant This time constant has the units of seconds S and is given by 1 RC RC Time Constant 151 D If there is only one capacitor in a circuit as in the circuits that we will be dealing with in this course but there is more than one resistor in the circuit then the time constant of the circuit is equal to the equivalent resistance Re times the capacitance i REC If you wish to know the value of the voltage across the capacitor at any time simply plug the numbers for the source voltage the desired time and the time constant into Eq 158 and you will have it A plot of the function for AVCt as a function of time is given below ax WMm vnmwe wwWumpmm w m k Figure 1511 kfhnmfon n n ihl xroqnonT anhrrooT 10 1ltkr l ovninn F flan xy nknrr ri nn n quot 1n alt nno Discharging Cap I Page 1 of 1 Discharging Capacitor Current Consider again the generic RC circuit Switch R 3 AV A 03 TJ Ill ltgt i Figure 1510 We assume now that the switch has been at position B for some time so that there is a nonzero potential difference across the capacitor as given by Eq 158 and that we suddenly throw the switch to position A When we do this we restart our stop watch and start counting a new time which we shall denote t 39 Thus the switch is thrown to position A at the time t 0 In this case the only circuit elements remaining in the circuit are the resistor and the capacitor Thus Kirchhost voltage law applied to this circuit is the same as Eq 155 written down for the charging circuit in the last section except that the voltage source is missing and the switch is now at position A instead of position B The equation for this discharging circuit is thus vA avR avpvg 1511 from which we get that using Ohm s law AVR IR EVC T 151 2 Note that this equation shows us explicitly that the current in the circuit must be negative that is it ows counterclockwise around the circuit instead of clockwise Can you explain Why this is physically Again from this equation we see that we can nd the current in the discharging circuit if we know the voltage across the capacitor kHnHmfnn grinineroqn397nT anturine 1 T 19 1ltlnmicnl10rninn r on Tknrlxy r1nn anvnnnr ol mnno Discharging Cap V Page 1 of l Discharging Capacitor Voltage The equation for the voltage across the capacitor AVC as a function of time since the switch was thrown to position A t for the discharging capacitor circuit is given by we t39av0 39e t39 1513 In the equation above the voltage AVO is the voltage across the capacitor when the switch is thrown from B to A that is at t 0 and t is the time constant in the discharging circuit which need not be the same value as the time constant in the charging circuit The voltage AVO can be calculated from Eq 158 in the previous section for the charging capacitor if we know how long the capacitor was initially charging before the switch was thrown to position A A graph for the function in Eq 1513 is shown below the dashed curve shows the behavior that would have resulted if the switch had not been thrown to position A Wmaa W w m M W fe V w w at a M i1 xxxA switch i i to A l l l f A g L i 1 WWW t Figure 1512 k llmiu39ni cannulanaononT nnfxlvonT 1 T 1 QT 1 kh139nnlanv n139nn FA XMLAA AinnLh hL nlc Innno 801152 Page 1 of2 Solution to Example 152 A series RC circuit is built with AVS 10 V R 12 kQ and C 47 uF At t 0 the switch is thrown to position B Switch R E AVS A CE TJ Ill F l Figure 1510 a What is the maximum voltage that will appear across the capacitor after the switch is thrown to position B The maximum voltage to appear across the capacitor is the voltage of the voltage course connected into the circuit when the switch is closed to position B Once the capacitor is fully charged it will stop any current ow in the circuit There will therefore be no current through the resistor so there will be no voltage drop across the resistor The voltage across the capacitor will then simply be the voltage of the power source Thus AVmax AVS 10 V b How much charge is on the capacitor plates when the voltage across the capacitor is the maximum value We have that Q C AVC When AVC is a maximum so will be the charge Q since C is a constant Thus Q C AVmax 047 mC max 0 What is the time constant of this circuit The time constant is simply given by T RC 56 ms 1 How long does it take after the switch is thrown to position B for the charge on the capacitor to reach 030 mC We want to nd the time twhen Q 035 mC From this value of Q we can nd the corresponding value of the voltage Q C AVC so that AVC QC 745 V The equation for the voltage across a charging capacitor as a function of time is h nHmfcn prliinl17c7n7nl Pr hn PCT l7Tl RT lRHFvnmnlp 14 7Qn114 7knrlv nnl1lt quot 0mman 801152 Page 2 of2 MC airm Him To solve for the time we rst have to get the exponential function by itself on one side of the equation Dividing both sides of this equation by AVmax and then solving for the exponential function gives us that e t 13 F 9 40255 max To solve for time we must pull it down from the exponent of the exponential function To do this we take the natural logarithm the logaI ithnz base e of both sides of the equation This then gives us the following In 5m 1n 0255 137 T or t 137 7 77 ms e What is the current through the resistor at the time in part d To nd the current we write out Kirchhost voltage law for the circuit loop Starting in the lower left corner and working our way around clockwise the same direction as the current ow we get that AVS IR AVC0 from which we get solving for the current I I M2i mg R thNm rqn p iinh1c n7nl pnfancT 17139 1 RT 1 ltl1pvnmnlp 14 7inl lt 7knr1v anll lt 7 9147an 801153 Page 1 of1 Solution to Example 153 Take the natural logarithm of both sides of Eq 1513 for the discharging capacitor and discuss the mathematical fonn of the resulting equation in relation to a graph of AVC and t data The equation for the discharging capacitor is AVG t M0 3954 1513 Taking the natural logarithm of both sides and applying the rules for working with logarithms gives us that In MC 1navo 391ne t 1nthVC1n V0 3 3 Since our data are of AVC and the time t and since the voltage is already on one side of the equation by itself so it will be associated with the role of the y Variable it follows that the variable t will play the role of the x variable We thus rewrite this equation as follows In MC t 391n an 39 Note that this has the generic form y slope x vertical intercept Comparison of these two equations then shows us that if we plot not AVC but rather 1n AVC as the yavariable and the time t as the x variable then we should get a linear behavior with a slope equal to 11 and a vertical intercept equal to 1nAVO Make sure that you understand this important reasoning LH Jmnn nAnn1 nn 3nT anhnonnT 1quot T 191quot lik E v 1C QC n11E Qknrlquot quotA11 9 141 nI nno Half Life Page 1 of 1 HalfLife There are a number of phenomena in the physical world that can grow or decay exponentially Bacterial population growth radioactive emission the decrease in light intensity as we dive deeper beneath the surface of water and of course the voltage and charge on a capacitor in an RC circuit are all examples of such phenomena In any such case the halflife of the exponential growth or decay denoted by the symbol t this does not mean to multiply or divide t by wthe whole thing is just a symbol 1 is de ned to be the amount of time for the growth to reach oneIzalfof its limiting value for example the maximum voltage AVmax for the case of the voltage across the capacitor in a charging RC circuit or for the decay to drop to one halfof its initial value no matter where along the decay curve you want to call the initial point it does not have to be the instant the decay starts That is once the decay has started you can start your stop watch at any time and note the value of the quantity of interest at that time It will take the same amount of time for the value to drop by a factor of 2 no matter when you started your stop watch For RC circuits this half life is determined by the time constant for the circuit which itself is determined by the values of R and C for the circuit as the next example shows Growth eagmmmtt39 i Rwy i 1quot 5 an 39x LGW r 103m M E 10 t 2 e c t i 539 v 1 as an u at u Mm j a g 35 a St lt4 t quot W i 0 t3 tm ztm a 5 Figure 1513 hftn39Hmtcn editnhvn707nlienhirenTl711RTl QBl lan Tif ehndv hal li a lq rml OAonno Sol154 Page 1 of2 Solution to Example 154 3 Find the expression for the halflife in terms of the RC timeconstant for the case of the discharging capacitor circuit For the discharging circuit the voltage as a function of time is given by Mo Mo 39e t39 When the time t is equal to the half life t 2 the voltage AVC must equal one half of the initial voltage AVO I I t I gm avo e 2 T Dividing both sides by AV 0 then gives us that 16 tugi39122 1 2 am Cross multiplying then gives us that 8111211 2 lnet 2 rtln2 5121 2 I n i This nally gives us that t l 1112 1 l b Find the expression for the half life in terms of the RC timeconstant for the case of the charging capacitor circuit For the charging capacitor the voltage time relationship is given by we warm 1 e t When the time t equals the halflife t1 2 the voltage AVC must equal onehalf of the maximum voltage InfhaImfnii onluLnlum T nnhnanT 1quot T lQT 1C1nEv1i ACnllc AILAAquot quot11 A 141 nitnnno 801154 Page 2 of2 151va ravmax1 equott112 Dividing both sides and rearranging gives us that etlrz39f r l 2 which is exactly the same equation as was obtained above The following steps are therefore exactly the same as shown above make sure you can reproduce them as is the answer t 1112 T 12 thNmfcn pdnNrshvc n7 T pnhirpcT 1 7T lRT 141113le Ainl 4 A anlv onl1lt A 11ml omnnno 801155 Page 1 of3 Solution to Example 155 Consider the RC circuit of Fig 165 below The capacitor is initially uncharged and the switch is thrown to position A at time t 0 Data AVs 100 V R1 20 kW R2 50 k9 R3 50 kW and C 10 HF Figure 165 a What is the time constant of the circuit with the switch at position A The time constant for an RC circuit is equal to the equivalent resistance times the capacitance for a onecapacitor circuit With the switch at position A we see that the circuit contains the two resistors R1 and R2 in a series combination The equivalent resistance is therefore RC R1 R2 70 k9 The time constant for this circuit is therefore I ReC 070 s b What is the charge on the capacitor one second after the switch is thrown to position A We wish to nd the charge on the capacitor at the time t 10 s Remember the relationship between charge and voltage Q CAVC Therefore if we can nd the voltage at any time we can immediately nd the charge on the capacitor at that time if we know the capacitance of that capacitor Likewise we can nd the voltage if we know the charge We have an equation for the voltage across a charging capacitor as when the switch is thrown to position A since this includes the voltage source in the circuit as a function of time Substituting the values we are given in the problem statement into this equation along with the desired time gives us the following we nvm 1 e t 7611 We then nd the charge on the capacitor at t 10 s Q CAVC 076 mC nlmfcn nAnvl39xrc 7n nT onhivncT 171 19 14kmv1lt ltinllt flArlxy anl1C C Lm1 nmnno 801155 Page 2 of3 c What is the current in the circuit at the time in part b With the switch at position A current is owing out of the positive terminal of the voltage source and around the circuit clockwise To find the current we simply apply Kirchhoff s voltage law Starting at point A and working our way around clockwise with the current we get that VA I R1AVC IR2DVSVA Remember that the voltage change as we cross the capacitor with the current ow is negative since current is the ow of positive charge so a clockwise current ow means that positive charge is building up on the top capacitor plate and negative charge on the bottom plate Since the electrostatic potential is higher near positive Charge as we cross the capacitor from top to bottom we are going from a region of higher potential to one of lower potential the change in potential is negative This accounts for the sign in front of the AVG in the equation above Subtracting V A from both sides and solving for the current gives us vs ve R1 R2 I 034mA Note that this is the value of the current at the time t 10 s since we used the value for AVC at t 10 s If we used the value of this voltage at any other time then we would get the value for the current at that same time d At the time in part b the switch is suddenly thrown to position B How long does it take for the voltage across the capacitor to drop to one fourth of its value when the switch was thrown to B We now throw the switch to position B When we do this the voltage across the capacitor has the value calculated in part b above Since we are now discharging the capacitor we will call this voltage that was on the capacitor plates when the discharge began AVO We want to nd the time t it takes for the voltage across the capacitor to drop to l4 its value when the switch was thrown AVO 76 V We could use the value of AVO to compute the desired voltage AVO 4 but we won t The reason for this is that in some problems of this sort you won t even know the value of the initial voltage when discharge begins that is you may not know the value of AVO But we can Still solve the problem Let s see how without using the value of AV 0 although it s certainly not wrong to use it for this problem The equation for the discharging capacitor voltage is ave AVG 39e tquot 139 where the new time constant for the discharging circuit is equal to 13116 R1R2 R3C 12 s laHnlmfcn nAnnl1vc39n nT pnhn pcn 1 7T lRT 14kmv1 inl 4 ltfknr1v an11lt 4 kfml OIKOHHQ 801155 Page 3 of3 since all three resistors are in series when the switch is in position B We want to solve for the time t when the voltage AVC is equal to AVO M Using this expression for AVC in the equation above gives us that VU39 4 at 39e39tquot 1quot We can now see why the value of AVO does not matter we can simply divide both sides by AVO to get 1 e t39f t39 4 lnverting both sides then gives us that BWTI4 or L1n4 The previous equation came from taking the natural logarithm of both sides and remembering that the natural logarithm and the exponential function are inverses of one another and therefore cancel each other out Solving for the time and substituting in numerical values then yields the nal answer t ln4 r 139 t 17 s e What is the current in the circuit at the time in part 1 To nd the current we again go to Kirchhoff s voltage law Just as in part c we treat the current as owing in the clockwise direction around the circuit Starting at point B and working our way around clockwise with the current ow then gives us that Subtracting VB from both sides and solving for the desired current I then gives us that VC 016mA R1R2 33 The negative sign in our answer of course means that the current is owing in the negative or counterclockwise direction LH I1M4n11 minr nernqnqnT nnhrrncT 1 T 19 1EkL v1lt QnHC flu 1quot n11 14 1 nInnno Sample Quiz 15 Page 1 of 1 Sample Quiz 15 l A capacitor basically consists of a a conductor and a dielectric b a battery and a conductor 0 two isolated conductors d a resistor and a conductor e any of these can count as a capacitor 2 Like resistors capacitors can be combined in series and parallel However you must be careful that a capacitors cannot always be placed in series b capacitors can not always be placed in parallel c capacitors are not power sources d resistors are ohmic devices e the rules are switched from combining resistors 3 When a capacitor initially starts charging the ow of charge is at rst relatively fast because a the wideopen capacitor plates readily accept charge b the resistor in the circuit is small 0 the power source is so strong 1 both I and c e all of a b and c 4 After a very long time of charging the capacitor acts like a a resistor b a short circuit c an open circuit 1 a power source e a switch 5 In today s activity we will be a computing the time constant for an RC circuit b graphing the charging and discharging of an RC circuit c guring out how to correctly connect a switch in a circuit 1 both a and b e all of a b and c Auswa rs L44ll4mn Alinl Lumnnnnn Anh unnT 1 T 1QT 11lc n1c1vv1quot quotn1 141 nirlnnnn SQlSAns Page 1 of1 Answers to Sample Quiz 15 Luudlmbnn Jul Lnn n nT AAunnT 1quot T 1 OIT 1 CLC n1 zlon1 C ALAL 1 5 LLHI II nnn Homework 15b Page 1 of 2 Homework 15b 1 An RC circuit has AVS 25 V R 12 M0 and C 100 JR At t 0 the switch is thrown to position B The capacitor is initially uncharged a What is the time constant for this circuit b What is the voltage across the capacitor at t 80 s c Would it be necessary to study the voltages as a function of time in this circuit with an oscilloscope Why or why not 12 MD M Switch A B L1 2 100 pf 25 939 2 Consider the RC circuit shown in Problem 1 above a What is the voltage across the resistor at time t 80 s b What is the current in the circuit at t 80 s c What is the magnitude of the charge on either capacitor plate at t 80 s 3 The capacitor in the circuit shown below is initially uncharged At time t 0 the switch is thrown from Position A to Position B a What is the time constant of this circuit b What is the charge on the capacitor plates if we wait for a very long time after t 0 This is Qmax c What is the charge on the capacitor plates at time t 12 s d What is the current in the circuit at t 12 s magnitude and direction e At 12 s the switch is thrown back to Position A What is the current in the circuit magnitude and direction at time t 22 s that is 10 s after the switch is thrown to A f What is the charge on the capacitor plates at t 22 s 150 m m Switch A B Lj 20 11F L6 9quot 4 The circuit shown below has been sitting with the switch at Position A for a very long time At some later time t 0 the switch is thrown to Position B a What is the significance of the rst sentence Is it important for the solution of the following parts in this problem b What is the time constant in the circuit immediately after the switch is LHalmfnrl AA11 L mononn nnfvirncT 1 T 1QT 1 kUHH kknAn Lun Ck 14 1 nmnno Homework 15b Page 2 of 2 thrown to Position B c What is the current in the circuit immediately after the switch is thrown to Position B d What is the voltage across the capacitor at time t 15 ms e What is the polarity of the capacitor that is which end is positive at time t 15 ms i What is the voltage across the resistor at time t 15 ms g What is the current in the circuit at time t 15 ms Hi I 22 LF 5 This is a continuation of Problem 4 above At time t 25 ms in the problem above the switch is thrown back to Position A The clock is immediately reset and the new time is called t 0 when the switch is thrown back to A a What is the charge on the capacitor plates irmnediately after the switch was thrown back to Position A that is at time t O b What is the time constant of the circuit after the time t 0 c How long after t 0 does it take for the voltage across the capacitor to drop to one third of its initial value at t 0 d How long after t 0 does it take for the current in the circuit to drop to one fourth of its initial value at t 0 tin ewe rs LHnImfniu naninhunononn mnhivnDT 1quot T 1 9n 1 RkLT XT1 Iahark 14111 KL 14 1 rIz Innno

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