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# Physics III PH 113

RHIT

GPA 3.69

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This 33 page Class Notes was uploaded by Jalon Willms on Monday October 19, 2015. The Class Notes belongs to PH 113 at Rose-Hulman Institute of Technology taught by Renat Letfullin in Fall. Since its upload, it has received 11 views. For similar materials see /class/225113/ph-113-rose-hulman-institute-of-technology in Physics 2 at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15

Electmmcgm sm Lecture 09 Chapter 30 Inductance and LR Circuits B B gt gt gt gt gt TTTTTTTT v gt gtquot gt gt TMTlTTTT B xxxxxx 69 F F0 Fq q17gtlt 39IIED fade Energy in Induc ror39s and Magnetic Fields The LR Cir39cui r Help Session for39 Tes r 2 Physics 113 Dr Renat Letfullin We define the inducfance L of a coil of wire producing flux m as I The unit of inductance is the hem39y 1 henr39y 1 H 1 T mZA 1 WbA The circuit diagram symbol used to represent inductance is W Example The inductance of a long solenoid with N turns of cross sectional area A and length l is B uONI Cppmum BA 1 2 CDquot NCDPmum NBA M1 L g ONZA l solenoid I 1 Physics 113 Dr Renat Letfullin 3 Potential Across an Inductor 2 Induced current Induced field i ReSIstor I AV IR lt R A VL F Decreasmg current The induced 39cun39enl carries positive charge carriers to the right The potential difference is opposite that of Figure 3339 The potential always decreases d1 Em AV L Physics 113 Dr Renal Letfullin Inductor ll A d AVL T L dt The potential decreases if the current is increasing The potential increases if the current is decreasing Energy in Inductors and Magnetic Fiie ds A magnetic field stores considerable energy Therefore an inductor which operates by creating a magnetic field stores energy Let39s consider how much energy UL is stored in an inductor L carrying current I d1 d I PIAVI L LI U L djzlLIZ all dl L J 2 A solenoid of length 1 area A and N turns has L poNZAl So NZA NI 2 UL LIZ 0 12 1 Al u0 21 2 1 0 U 1le 1 2 AIB2 L 5 MB B 2 u UL 1 Bz 2110 Physics 113 Dr Renat Letfullin LR circuit IniTially when a swiTch closes an inducTor appears To have an infiniTe resisTance and has a maximum poTenTial drop across iT UlTimaTely The inducTor reaches a sTeady currenT flow and There is no poTenTial drop across iT Therefore aT T0 The inducTor behaves like aT open circuiT 0200 and aT T The inducTor behaves like a shorT circuiT R0 This behavior is opposiTe ThaT of a conducTor Example 10 n 10 n 10 a WW J Wv v lt ltgt ltgt 30V 20 1 10 mH 30V 20 0 30V 20 0 lt ltgt ltgt CircuiT aT T0 aT 1oo CalculaTe iniTial poTenTials CalculaTe final currenTs Physics 113 Dr Renat Letfullin 6 The LR Circuit d a I 0 The swrtch has been in this pOSition AVR AVL 0 RI L for a long time Att 0 it is moved dt to position b t d dt d R J d 1 10 I L 0 A lbat I I Current In IO 1 0 L R The current has decreased to AVR 37 of its initial value at t T I The current has R decreased to 13 of I I tLR 05010 initial valgue at t 2 L AVL 0 06 03710 I i j e j 39 01310 ThlS is the Circuit With the swnch in position b The inductor prevents the current from stopping instantly Physics 113 Dr Renat Letfullin 7 m Exponential Decoy in an LR Cm The switch mpves from a to b at O The swiTch shown has been in posiTion a for a long Time IT changes To posiTion b aT T0 a WhaT is The cur39r39enT in The cir39cuiT aT T 5 us b AT whaT Time has The cur39r39enT decayed To 1 of iTs original value 10VR10 V100 12100 mA zLRR 3 I 2108 2000 mAe5SOo s 61 mA 20gtlt10 H200 240 us amp us 100 mA Physics 113 Dr Renat Letfullin 8 t 7 1n 10 us1n 0 AV hal The plumber39s analogyquot of an RL circuiT is a pump baTTery pumping waTer in a closed loop of pipe ThaT includes a valve swiTch a consTricTion resisTor and roTaTing Pump Baf fery flywheel Turned by The flow When The Valve SW39TCh ConsTrIcTIon ReSISTor valve sTarTs The flow The flywheel begins To Flywheel Indudor spin unTi a sTeady flow is achieved and The Pressure PoTenTiaI pressure difference across The flywheel Wa39re39 FI W Currenquot PEP3 goes To zero Physics 113 Dr Renat Letfullin 9 R Rank the order of the time constants in these circuits 5 15127522173 b 11lt12lt13 d Tlgt172gt173 c r2ltrlltr3 e r2gtrlgtr3 Physics 113 Dr Renat Letfullin L6 immm A charged capaciTor bears a cerTain resemblance To a sTreTched spring remember The rubber diaphragm sToring energy even when The charge is noT moving An inducTor similarly resembles a moving mass remember The flywheel sToring energy only when charge is in moTion We know ThaT a mass and spring can make an oscillaTor WhaT abouT a capaciTor and inducTor Consider The circuiT shown in The diagram WhaT happens when The swiTch is closed The capaciTor discharges by creaTing a currenT in The inducTor BuT where does The energy go ThaT had been sTored in The inducTor There are no dissipaTive elemenTs in The sysTem Therefore when The charge of The capaciTor goes To zero all of iTs previous energy musT reside in The inducTor The currenT in The inducTor falls while charging The capaciTor in The opposiTe direcTion And so on Physics 113 Dr Renat Letfullin 11 Switch closes at 0 The Oscillation Cycle mQ rl he mm mme we wm am Cc he 0 cccc m mm 0 mM A5 I mm cu mm mm Mu Physics 113 Dr Renat Letfullin LC Oscillations thtml but only m Rae What is the oscillation frequency coo I Begin with the loop rule 2 Q L L d 99 0 39 39 dt C I Guess solution just harmonic oscillator I remember a x QZQOCOSwol mWkx20 where 00 determined from initial conditions Procedure differentiate above form for Q and substitute into loop equation to find wo Note Dimensional analysis wo 2 LC Oscillations Mathew General solution Q Q0 005wot Differentiate dQ sz Q EQ0Qo smwot L dtz 620 d2 air a QO cosa0t Substitute into loop eqn 1 L ngO cosa0t QO cosa0t 0 gt QOZL E 0 Therefore which we could have determined from the mass on a spring result k lC 1 a JE LE The current in a LC circuit is a sinusoidal oscillation with frequency 03 5 If the inductance of the circuit is increased what will happen to the frequency on a increase c doesn t change 6 If the capacitance of the circuit is increased what will happen to the frequency a increase c doesn t change Physics 113 Dr Renat Letfullin 15 quot s m Admit AT 10 The capaciTor has charge Q0 The resuITing oscillaTions have frequency mo The maximum currenT in The circuiT during These scillaTions has value 10 WhaT is The relaTion beTween mo and 02 The frequency of oscillaTions when The iniTiaI charge ZQO a a2 12 we b a2 we c a2 2 What is the relation between IO and 12 the maximum current in the circuit when the initial charge 2Q0 312 IO 0 2 2Io Q12 43910 AT t0 The capaciTor39 has charge Q0 Th r39esulTing oscillaTions have frequency 020 The maximum cur39r39enT in The cir39cuiT during These scillaTions has value 10 WhaT is The r39elaTion beTween mo and 022 The frequency of oscillaTions when The iniTial charge 2Q0 a wz 12 we b 2 00 0 QQ0 c wz 2 Q0 determines the amplitude of the oscillations initial condition The frequency of the oscillations is determined by the circuit parameters L C just as the frequency of oscillations of a mass on a spring was determined by the physical parameters k m AT 20 The c r39esulTing oscillaTions have frequency 020 The maximum cur39r39enT in The circuiT during These scillaTions has value 10 WhaT is The r39elaTion beTween I0 and 12 The maximum cur39r39enT in The circuiT when The iniTial charge ZQO 312 Io M5 210 0 QQ0 012 4IO The initial charge determines the total energy in the circuit U0 Q022C The maximum current occurs when Q0 At this time all the energy is in the inductor U 12 L102 Therefore doubling the initial charge quadruples the total energy To quadruple the total energy the max current must double OscillaTion frequencya0 1 loop equaTion LC The oTher unknowns Q0 5 are found from The iniTial condiTions Eg in our original example we assumed iniTial values for The charge Qi and currenT 0 For These values QO Q 0 Question Does this solution conserve energy UEltzgtQ cos2ltwoz gt 1 1 UB t ELI39ZU ELaJOZQO2 s1n 2wot Energy in Capacitor UEI Q02 0052wot Energy in Inductor UBtL 0 Qozsmza0t 0 1 0 U U UBZ Q sin2wo 2 Therefore UEIUBI 3 0 S m Actmlzt AT t0 The cur39r39enT flowing Through The l cir39cuiT is 12 of iTs maximum value Which of The following pIoTs besT r39epr39esenTs U3 The energy sTor39ed in The inducTor39 as o funcTion of Time I bi c 0 time time time time U3 ng Through Th 7 circuiT is 12 of iTs maximum value Which of The following ploTs besT represenTs U3 The energy sTored in The inducTor as o funcTion of Time b time 9 Qif Lg I c U3 time The key here is to realize that the energy stored in the inductor is proportional to the CURRENT SQUARED Therefore if the current at 10 is 12 its maximum value the energy stored in the inductor will be 14 of its maximum value m M AM Radio Oscillator You have a 10mH induc l or Wha139 capaci l or should you use with it To make an oscillator with a frequency of 920 kHz This frequency is near The cen139er of The AM radio band a 27zf227z920gtlt105 s39l578gtlt106 s391 1 1 C27 6 12 72 30x10 F30pF aJL 578gtlt10 s3910gtlt10 H Physics 113 Dr Renat Letfullin 23 Chapters2930 Summary GENERAL PRINCIPLES Faraday39s Law IIDDEL Make assump um wsuaun Use lLenzfs law to deem ne the direction of are induced mmL some The induced unris dd 5 Ml Multiply by N in an N m n mi size uflhe mm cmntisf EIR 1555 ls the mall mus Lentshw n J J a andonlyiflilemagm c ux llzmughlb 009139s imaging Mdhccduuufmeimhmdmuentissmhlhatlheinwmd mgwlic elduwosesmedmtgein u u Magnetic ux 5 Magnetic ux mums L a 1 small of mgm c eldpassing Waugh 3 suf ce 3 Aaooss Physics 113 Dr Renat Letfullin 24 Max930 Sammy IMPORTANT CONCEPTS Three ways to change the ux Tm ways In createan induced cum I Aloapnnvesiunorcnlof c ekl 1 Amolinnal eml39dne tonngnelic fawn moving dung twins 7 Tlalowchangesaxea errataIns P Anindnuedelemic zelddm on changing magnum eld 3 Tlemgne c eldnrmgzme loop increases deans Physics 113 Dr Renat Letfullin 25 cucpiem2930 Summary APPLICATIONS Indianms W yoNzA Edmund mm 11 madam AV L 39 we quot a m drcu39rt Emlgj slmdULU2 Wammmng Magmtic energy demilyua if Physics 113 Dr Renat Letfullin 26 Faraday39s Law and Lenz39s Law Mo rional Emf LR cir cui rs Physics 113 Dr Renat Letfullin 28 difference of 005 V is developed across a 10cm long wire as it moves through a magnetic field at 5 ms What are the direction and strength of the magnetic field 4 100m 50 ms om Minion Figure below shows a five 39l39urn 10cmdiame139er coil with R 01 9 inside a 20cmdime139er solenoid The solenoid 80cm long has 120 Turn and carries The current shown in The graph A positive current is cw when seen from The left De39l39ermine The current in The coil as a function of Time Give your answer as a currentversus l39ime graph from 139 0 5 To 139 002 s g1 39 a j 39 i i k quot A 50turn 40cmdiameter coil with R 05 Q surrounds a 20 cm diameter solenoid The solenoid is 20cm long and has 200 turns The 60 Hz current through the solenoid is 1wl O50Asin27ft Find an expression for Icon the induced current in the coil as a function of time hi 4 quot1 J 39 u R V i 7 i 4 V V H v 7 j 1 x lt39 1 1 i as 1 r 1 I I s N l V l r quot V y i M39 J ll i V 3 A I quotU 1 39 I 39 a i a The current through inductance L is given by I 106 a Find an expression for the potential difference across the inductor b Evaluate AVL at t 0 1 2 and 3 ms if L 20 mH Io 50 mA and C 10 ms c Draw a graph of AVL versus time from t 0 to 3 ms a for a long Time IT is closed aT T 0 s AfTer39 The swiTch has been closed for a long Time whaT is The currenT in The circuiT Call This currenT Io Find an expression for The currenT I as a funcTion of Time Wr39iTe your expression in Terms Io R and L SkeTch a currenTversusTime graph from T 0s unTil The currenT is no longer changing R H AV bar

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