Notes on Inductance and Circuit Transients Joe Wolfe, Physics UNSW. Circuits with R and C. τ = RC = time constant
|
|
- Brittney Stone
- 5 years ago
- Views:
Transcription
1 Nes n Inducance and cu Tansens Je Wlfe, Physcs UNSW cus wh and - Wha happens when yu clse he swch? (clse swch a 0) - uen flws ff capac, s d Acss capac: Acss ess: c d s d d ln + cns. 0, ln cns. ln ln ln ln a ln b ln a b e / /c e / () me cnsan Uns: ΩF A. s. s - d e - / e - / e - / /e /e /e /e
2 Example A eal uses delecc wh delecc cns. 4.0 and cnducvy σ 0-5 Ω-m-. I s chaged hen lef dscnneced f hu. Wha facn f chage emans? I A K, σ d - (σ /ρ) A d K A d ρ A d Aσ d K Aσ. A d K σ 4.0 x Fm-.78 x 04 s Ω-m- e - / Q Q e / e 3600/ 0.8 Example Elecnc applance eues 0 W a 40 D, wh max vaan f 0%. Wha capacance needed se enegy beween cycles? A + m 90%* m A has 50 cycles pe secnd, uns ff 00 mes/secnd, e evey 0 ms. m e / ms,?, wha s? s he effecve essance f he ccu. We knw and P, s use P /P 80 Ω. e / 0.90 ln (0.90) 0 ms 80 Ω * ln µf Example he membane f a "pacemake" neune f he hea has a capacance f 0 pf and a cnducance G 3.0 pω. Inally he penal dffeence acss hs membane s 80 m. The neune begns a pulse when he penal dffeence eaches 60 m. Hw lng befe begns a pulse? Neglecng he duan f he eleccal mpulse self, wha s he pulse ae? () ln ln ln s The "leakness" f he membane s nw suddenly nceased by a fac f. Hw lng beween eleccal mpulses? and wha s he appxmae pulse ae, agan neglecng he duan f he ndvdual mpulses. - ln 0.48 s - ln...
3 3 chagng Wha happens when yu clse he swch? (clse swch a 0, 0, 0) - uen flws n capac, s I + d Kchff + ake devaves, use + d 0 d + eaange: d - ln - + cns nal cnn ln cns e -/ e-/ () whee / and /e /e ( /e) ( /e ) Inducance Self-nducn B flux hugh each un f cl φ B d(nφ Faaday Β ) Inducance, NΦ B (3) ( s he al flux lnkage pe un cuen) S d (4) Ne: n cl ae f change n cf acss negal f Uns: heny l secnd/amp
4 4 enz's law: acs n decn manan cuen (and B) + f nceases + f deceases esss change n Example Aumble gnn cl (schemac) (n n syllabus) n cnac n ce spng spng mves n gh clse cnac. flws whch causes B B aacs n lef, whch pens ccu d 0 uckly s lage Ideal Slend A l n uns/m NΦ B n l BA bu B µ n NΦ B nlµ na µ nla (5) Ne: f nn-magnec maeals, µ µ ο 4π 0-7 Hm- magnec maeals, µ >> µ n ces f lage. Example l n kele has 60 Ω. I s 0 cm lng, has un/mm, and s 4 mm n damee. ) wha s? ) s cnneced 40 ms. Swch has beakdwn penal f 000 and s uned ff a peak f cycle ( 340 ). Esmae he duan f spak. (Neglec he ). Peak p A. (5) µ ο n l A 4π 0-7.(03 ). (0.). π( 0-3) SI 3. µh. - d ~ 000 dung spak d ~ cns. 000 d 5.7 x (In pacce s lage because f he 's.) 8 ns.
5 5 - Tansens ecall - d (enz's law: back emf) Ths pevens sudden change n, s cnnuus. a ) swch a a. Afe a lng me, d b 0, c/ ) swch b a 0 a 0, c / Kchff: - d d ln - + ln - e -/ c e-/ whee / (6) Uns: H Ω A/s. A s /e /e ) swch a b f lng me ( 0). Swch a a 0. c - d c slve ( - e- / ) whee / (7) ( /e) ( /e ) ' Example 6 64 H 3k ( 3 kω) 47k Swch clsed f lng me. Open a 0. Wha des he O shw? Swch clsed, afe a lng me d 0 Kchff's law: and T + O shws 6
6 6 Swch pened fbds abup change n a 0 (0) /, 0 al sees essance s + s - ( + )/ e hee + - s O - - O - e -(+)/ 6 - Enegy n B ~ f Pwe suppled.e. du B U B du B d. ' d' U B (8) c.f. U E Enegy densy U B vl U B Al Al (5) µ n la (5) (fm mag.) B µ n U B vl B µ (9) (c.f. U E ο E ) Muual Inducn N B N N φ G Muual Inducance M (f w... ) M N Φ c.f. N Φ s M N Φ M d N dφ - (Faaday's law) (lng pf) M M M s: M d, M d (0) (S.I. un heny s A - )
7 7 Example Tw lng cncenc slends. M? x x x x x x x x x x x x n / un lengh and n / un lengh M M N Φ n l Φ B µ n, and Φ b A M n l (µ ο n ) π M π l µ n n () scllans (assume deal : 0, deal : ) Kchff's law: - d d, s - d d d - - ω say slve: cns sn (ω + cns) m sn (ω + φ) whee ω () - d - ω m cs (ω + φ) - m cs (ω + φ) Enegy U E. [...cs...] U B [...sn...] U T U E + U B m [ ω cs ( ) + sn ( )] m [ cnsan] als m m ω s U T m ω m Example Fn end f a ad: max mn. pf max (3 n //) ο A 3 d 3 π d pf ) If we wan une he AM band (54-60 khz) wha values f and? Hw d we make he? f max π ω max π ( mn + ) f mn π ( max + ) f max f... max + mn mn + (f max - f mn ) max f mn - mn f max pf f π. 4π f max +. 4p f mn 490 mh (5) µ ο n la say n 000/m, A 5 cm, l? l µ na 00 m. Ops, y n ce: µ µ µ ~ 000
Field due to a collection of N discrete point charges: r is in the direction from
Physcs 46 Fomula Shee Exam Coulomb s Law qq Felec = k ˆ (Fo example, f F s he elecc foce ha q exes on q, hen ˆ s a un veco n he decon fom q o q.) Elecc Feld elaed o he elecc foce by: Felec = qe (elecc
More information11. HAFAT İş-Enerji Power of a force: Power in the ability of a force to do work
MÜHENDİSLİK MEKNİĞİ. HFT İş-Eneji Pwe f a fce: Pwe in he abiliy f a fce d wk F: The fce applied n paicle Q P = F v = Fv cs( θ ) F Q v θ Pah f Q v: The velciy f Q ÖRNEK: İŞ-ENERJİ ω µ k v Calculae he pwe
More informationNote: Please use the actual date you accessed this material in your citation.
MIT OpenCuseWae hp://cw.m.edu 6.03/ESD.03J Elecmagnecs and Applcans, Fall 005 Please use he fllwng can fma: Makus Zahn, Ech Ippen, and Davd Saeln, 6.03/ESD.03J Elecmagnecs and Applcans, Fall 005. (Massachuses
More informationβ A Constant-G m Biasing
p 2002 EE 532 Anal IC Des II Pae 73 Cnsan-G Bas ecall ha us a PTAT cuen efeence (see p f p. 66 he nes) bas a bpla anss pes cnsan anscnucance e epeaue (an als epenen f supply lae an pcess). Hw h we achee
More informationR th is the Thevenin equivalent at the capacitor terminals.
Chaper 7, Slun. Applyng KV Fg. 7.. d 0 C - Takng he derae f each erm, d 0 C d d d r C Inegrang, () ln I 0 - () I 0 e - C C () () r - I 0 e - () V 0 e C C Chaper 7, Slun. h C where h s he Theenn equalen
More informationTRANSIENTS. Lecture 5 ELEC-E8409 High Voltage Engineering
TRANSIENTS Lece 5 ELECE8409 Hgh Volage Engneeng TRANSIENT VOLTAGES A ansen even s a sholved oscllaon (sgnfcanly fase han opeang feqency) n a sysem cased by a sdden change of volage, cen o load. Tansen
More informationMarch 15. Induction and Inductance Chapter 31
Mach 15 Inductin and Inductance Chapte 31 > Fces due t B fields Lentz fce τ On a mving chage F B On a cuent F il B Cuent caying cil feels a tque = µ B Review > Cuents geneate B field Bit-Savat law = qv
More informationA) 100 K B) 150 K C) 200 K D) 250 K E) 350 K
Phys10 Secnd Maj-09 Ze Vesin Cdinat: k Wednesday, May 05, 010 Page: 1 Q1. A ht bject and a cld bject ae placed in themal cntact and the cmbinatin is islated. They tansfe enegy until they each a final equilibium
More informationEnergy Storage Devices
Energy Srage Deces Objece f ecure Descrbe The cnsrucn f an nducr Hw energy s sred n an nducr The elecrcal prperes f an nducr Relanshp beween lage, curren, and nducance; pwer; and energy Equalen nducance
More informationA) (0.46 î ) N B) (0.17 î ) N
Phys10 Secnd Maj-14 Ze Vesin Cdinat: xyz Thusday, Apil 3, 015 Page: 1 Q1. Thee chages, 1 = =.0 μc and Q = 4.0 μc, ae fixed in thei places as shwn in Figue 1. Find the net electstatic fce n Q due t 1 and.
More informationPhys-272 Lecture 18. Mutual Inductance Self-Inductance R-L Circuits
Phys-7 ectue 8 Mutual nductance Self-nductance - Cicuits Mutual nductance f we have a constant cuent i in coil, a constant magnetic field is ceated and this poduces a constant magnetic flux in coil. Since
More informationELECTROMAGNETIC INDUCTION PREVIOUS EAMCET BITS
P P Methd EECTOMAGNETIC INDUCTION PEVIOUS EAMCET BITS [ENGINEEING PAPE]. A cnduct d f length tates with angula speed ω in a unifm magnetic field f inductin B which is pependicula t its mtin. The induced
More informationESS 265 Spring Quarter 2005 Kinetic Simulations
SS 65 Spng Quae 5 Knec Sulaon Lecue une 9 5 An aple of an lecoagnec Pacle Code A an eaple of a knec ulaon we wll ue a one denonal elecoagnec ulaon code called KMPO deeloped b Yohhau Oua and Hoh Mauoo.
More information2015 Sectional Physics Exam Solution Set
. Crrec answer: D Ne: [quan] denes: uns quan WYSE cadec Challenge 05 Secnal Phscs Ea SOLUTION SET / / / / rce lengh lengh rce enu ass lengh e a) / ass ass b) energ c) wrk lengh e pwer energ e d) (crrec
More informationP a g e 5 1 of R e p o r t P B 4 / 0 9
P a g e 5 1 of R e p o r t P B 4 / 0 9 J A R T a l s o c o n c l u d e d t h a t a l t h o u g h t h e i n t e n t o f N e l s o n s r e h a b i l i t a t i o n p l a n i s t o e n h a n c e c o n n e
More informationChapter 14: Optical Parametric Oscillators
Qunum Oc f Phnc n Olcnc hn n, Cnll Un Ch : Ocl Pmc Ocll. Inucn In h Ch w wll cu n cl mc cll. A mc cll lm lk l. Th ffnc h h cl n n h c cm n fm uln n mum u fm nnln cl mum whch h h cn h h 3 cl nnln. Cn n
More informationdm dt = 1 V The number of moles in any volume is M = CV, where C = concentration in M/L V = liters. dcv v
Mg: Pcess Aalyss: Reac ae s defed as whee eac ae elcy lue M les ( ccea) e. dm he ube f les ay lue s M, whee ccea M/L les. he he eac ae beces f a hgeeus eac, ( ) d Usually s csa aqueus eeal pcesses eac,
More informationFigure Circuit for Question 1. Figure Circuit for Question 2
Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question
More informationUse 10 m/s 2 for the acceleration due to gravity.
ANSWERS Prjecle mn s he ecrl sum w ndependen elces, hrznl cmpnen nd ercl cmpnen. The hrznl cmpnen elcy s cnsn hrughu he mn whle he ercl cmpnen elcy s dencl ree ll. The cul r nsnneus elcy ny pn lng he prblc
More informationME 3600 Control Systems Frequency Domain Analysis
ME 3600 Cntl Systems Fequency Dmain Analysis The fequency espnse f a system is defined as the steady-state espnse f the system t a sinusidal (hamnic) input. F linea systems, the esulting utput is itself
More informationChapter 3: Vectors and Two-Dimensional Motion
Chape 3: Vecos and Two-Dmensonal Moon Vecos: magnude and decon Negae o a eco: eese s decon Mulplng o ddng a eco b a scala Vecos n he same decon (eaed lke numbes) Geneal Veco Addon: Tangle mehod o addon
More information21.9 Magnetic Materials
21.9 Magneic Maerials The inrinsic spin and rbial min f elecrns gives rise he magneic prperies f maerials è elecrn spin and rbis ac as iny curren lps. In ferrmagneic maerials grups f 10 16-10 19 neighbring
More informationLecture 3. Electrostatics
Lecue lecsics In his lecue yu will len: Thee wys slve pblems in elecsics: ) Applicin f he Supepsiin Pinciple (SP) b) Applicin f Guss Lw in Inegl Fm (GLIF) c) Applicin f Guss Lw in Diffeenil Fm (GLDF) C
More informationLecture 4 ( ) Some points of vertical motion: Here we assumed t 0 =0 and the y axis to be vertical.
Sme pins f erical min: Here we assumed and he y axis be erical. ( ) y g g y y y y g dwnwards 9.8 m/s g Lecure 4 Accelerain The aerage accelerain is defined by he change f elciy wih ime: a ; In analgy,
More informationWYSE Academic Challenge Sectional Mathematics 2006 Solution Set
WYSE Academic Challenge Sectinal 006 Slutin Set. Cect answe: e. mph is 76 feet pe minute, and 4 mph is 35 feet pe minute. The tip up the hill takes 600/76, 3.4 minutes, and the tip dwn takes 600/35,.70
More informationCHAPTER 24 GAUSS LAW
CHAPTR 4 GAUSS LAW LCTRIC FLUX lectic flux is a measue f the numbe f electic filed lines penetating sme suface in a diectin pependicula t that suface. Φ = A = A csθ with θ is the angle between the and
More informationExample
hapte Exaple.6-3. ---------------------------------------------------------------------------------- 5 A single hllw fibe is placed within a vey lage glass tube. he hllw fibe is 0 c in length and has a
More informationLecture 2 Feedback Amplifier
Lectue Feedback mple ntductn w-pt Netwk Negatve Feedback Un-lateal Case Feedback plg nalss eedback applcatns Clse-Lp Gan nput/output esstances e:83hkn 3 Feedback mples w-pt Netwk z-paametes Open-Ccut mpedance
More information5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )
5-1. We apply Newon s second law (specfcally, Eq. 5-). (a) We fnd he componen of he foce s ( ) ( ) F = ma = ma cos 0.0 = 1.00kg.00m/s cos 0.0 = 1.88N. (b) The y componen of he foce s ( ) ( ) F = ma = ma
More informationMass Linear Momentum Moment of Momentum Energy Putting it all together!
inie Cnrl lue nalsis vin fr a Sse a inie Cnrl lue a Linear enu en f enu Ener Puin i all eer! D Cnservain f a B = Tal aun f a in e e b = a er uni a = DB ˆ b b n ˆ n ˆ equain Bu D / =! Cninui Equain a leavin
More informationChapter 2. Kinematics in One Dimension. Kinematics deals with the concepts that are needed to describe motion.
Chpe Kinemic in One Dimenin Kinemic del wih he cncep h e needed decibe min. Dynmic del wih he effec h fce he n min. Tgehe, kinemic nd dynmic fm he bnch f phyic knwn Mechnic.. Diplcemen. Diplcemen.0 m 5.0
More information10. A.C CIRCUITS. Theoretically current grows to maximum value after infinite time. But practically it grows to maximum after 5τ. Decay of current :
. A. IUITS Synopss : GOWTH OF UNT IN IUIT : d. When swch S s closed a =; = d. A me, curren = e 3. The consan / has dmensons of me and s called he nducve me consan ( τ ) of he crcu. 4. = τ; =.63, n one
More informationElectric Fields and Electric Forces
Cpyight, iley 006 (Cutnell & Jhnsn 9. Ptential Enegy Chapte 9 mgh mgh GPE GPE Electic Fields and Electic Fces 9. Ptential Enegy 9. Ptential Enegy 9. The Electic Ptential Diffeence 9. The Electic Ptential
More informationElectric potential energy Electrostatic force does work on a particle : Potential energy (: i initial state f : final state):
Electc ptental enegy Electstatc fce des wk n a patcle : v v v v W = F s = E s. Ptental enegy (: ntal state f : fnal state): Δ U = U U = W. f ΔU Electc ptental : Δ : ptental enegy pe unt chag e. J ( Jule)
More informationChapter 32 Solutions
Chapter 3 Solutions *3. ε = L I t = (3.00 0 3 H).50 A 0.00 A 0.00 s =.95 0 V = 9.5 mv 3. Treating the telephone cord as a solenoid, we have: L = µ 0 N A l = (4π 0 7 T m/a)(70.0) (π)(6.50 0 3 m) 0.600 m
More information5.1 Angles and Their Measure
5. Angles and Their Measure Secin 5. Nes Page This secin will cver hw angles are drawn and als arc lengh and rains. We will use (hea) represen an angle s measuremen. In he figure belw i describes hw yu
More informationG e n e r a l I n s t r u c t i o n s
hyscs (ll nda) G e n e a l n s u c n s. ll quesns ae cpulsy. Thee ae 6 quesns n all.. Ths quesn pape has fve secns: ecn, ecn, ecn C, ecn D and ecn. 3. ecn cnans fve quesns f ne ak each, ecn cnans fve quesns
More informationIntroduction of Two Port Network Negative Feedback (Uni lateral Case) Feedback Topology Analysis of feedback applications
Lectue Feedback mple ntductn w Pt Netwk Negatve Feedback Un lateal Case Feedback plg nalss eedback applcatns Clse Lp Gan nput/output esstances e:83h 3 Feedback w-pt Netwk z-paametes Open-Ccut mpedance
More informationPhysics 20 Lesson 9H Rotational Kinematics
Phyc 0 Len 9H Ranal Knemac In Len 1 9 we learned abu lnear mn knemac and he relanhp beween dplacemen, velcy, acceleran and me. In h len we wll learn abu ranal knemac. The man derence beween he w ype mn
More informationThree charges, all with a charge of 10 C are situated as shown (each grid line is separated by 1 meter).
Three charges, all with a charge f 0 are situated as shwn (each grid line is separated by meter). ) What is the net wrk needed t assemble this charge distributin? a) +0.5 J b) +0.8 J c) 0 J d) -0.8 J e)
More informationZVS Boost Converter. (a) (b) Fig 6.29 (a) Quasi-resonant boost converter with M-type switch. (b) Equivalent circuit.
EEL6246 Pwer Electrnics II Chapter 6 Lecture 6 Dr. Sam Abdel-Rahman ZVS Bst Cnverter The quasi-resnant bst cnverter by using the M-type switch as shwn in Fig. 6.29(a) with its simplified circuit shwn in
More informationThe Buck Resonant Converter
EE646 Pwer Elecrnics Chaper 6 ecure Dr. Sam Abdel-Rahman The Buck Resnan Cnverer Replacg he swich by he resnan-ype swich, ba a quasi-resnan PWM buck cnverer can be shwn ha here are fur mdes f pera under
More informationPhysics Courseware Physics I Constant Acceleration
Physics Curseware Physics I Cnsan Accelerain Equains fr cnsan accelerain in dimensin x + a + a + ax + x Prblem.- In he 00-m race an ahlee acceleraes unifrmly frm res his p speed f 0m/s in he firs x5m as
More informationA) N B) 0.0 N C) N D) N E) N
Cdinat: H Bahluli Sunday, Nvembe, 015 Page: 1 Q1. Five identical pint chages each with chage =10 nc ae lcated at the cnes f a egula hexagn, as shwn in Figue 1. Find the magnitude f the net electic fce
More informationElectromagnetic Waves
Chapte 3 lectmagnetic Waves 3.1 Maxwell s quatins and ectmagnetic Waves A. Gauss s Law: # clsed suface aea " da Q enc lectic fields may be geneated by electic chages. lectic field lines stat at psitive
More informationw m -,. t o o p f 0. p 0 we , 44-4 c , 0 0 k 1 0 P ) TIC 0 0 PAM Sc.._ a4C44 IsaA.r a.% Oc Or
S5 l M V5 a 3 3 c a c = 5 ( 5 V 3 J 5 5 5 9 3 p Sx= 5 b S 53 5 8 nmkcc 3G v l a c c w m p f p 5 + 3 + M3 3 5 aca =(M% ccwfv v 5 ac 3 c5ca calavaa w 55 c k 5) 29 5 3 5 3Z`c= calsa c (MM S 3 5 5 3 8 5 G
More informationWork, Energy, and Power. AP Physics C
k, Eneg, and Pwe AP Phsics C Thee ae man diffeent TYPES f Eneg. Eneg is expessed in JOULES (J) 4.19 J = 1 calie Eneg can be expessed me specificall b using the tem ORK() k = The Scala Dt Pduct between
More informationPhysics Exam II Chapters 25-29
Physcs 114 1 Exam II Chaptes 5-9 Answe 8 of the followng 9 questons o poblems. Each one s weghted equally. Clealy mak on you blue book whch numbe you do not want gaded. If you ae not sue whch one you do
More informationReflection and Refraction
Chape 1 Reflecon and Refacon We ae now neesed n eplong wha happens when a plane wave avelng n one medum encounes an neface (bounday) wh anohe medum. Undesandng hs phenomenon allows us o undesand hngs lke:
More informationName of the Student:
Engneeng Mahemacs 05 SUBJEC NAME : Pobably & Random Pocess SUBJEC CODE : MA645 MAERIAL NAME : Fomula Maeal MAERIAL CODE : JM08AM007 REGULAION : R03 UPDAED ON : Febuay 05 (Scan he above QR code fo he dec
More informationSolution: (a) C 4 1 AI IC 4. (b) IBC 4
C A C C R A C R C R C sin 9 sin. A cuent f is maintaine in a single cicula lp f cicumfeence C. A magnetic fiel f is iecte paallel t the plane f the lp. (a) Calculate the magnetic mment f the lp. (b) What
More informationT h e C S E T I P r o j e c t
T h e P r o j e c t T H E P R O J E C T T A B L E O F C O N T E N T S A r t i c l e P a g e C o m p r e h e n s i v e A s s es s m e n t o f t h e U F O / E T I P h e n o m e n o n M a y 1 9 9 1 1 E T
More informationPhysics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant)
ε µ0 N mp T kg Kuwait University hysics Department hysics 0 Secnd Midterm Examinatin Summer Term (00-0) July 7, 0 Time: 6:00 7:0 M Name Student N Instructrs: Drs. bdel-karim, frusheh, Farhan, Kkaj, a,
More informationa. (1) Assume T = 20 ºC = 293 K. Apply Equation 2.22 to find the resistivity of the brass in the disk with
Aignmen #5 EE7 / Fall 0 / Aignmen Sluin.7 hermal cnducin Cnider bra ally wih an X amic fracin f Zn. Since Zn addiin increae he number f cnducin elecrn, we have cale he final ally reiiviy calculaed frm
More informationThe Components of Vector B. The Components of Vector B. Vector Components. Component Method of Vector Addition. Vector Components
Upcming eens in PY05 Due ASAP: PY05 prees n WebCT. Submiing i ges yu pin ward yur 5-pin Lecure grade. Please ake i seriusly, bu wha cuns is wheher r n yu submi i, n wheher yu ge hings righ r wrng. Due
More informationLecture 11 Inductance and Capacitance
ecure Inducance and apacance EETRIA ENGINEERING: PRINIPES AND APPIATIONS, Fourh Edon, by Allan R. Hambley, 8 Pearson Educaon, Inc. Goals. Fnd he curren olage for a capacance or nducance gen he olage curren
More informationElectrical Machines I Week 3: Energy Storage
Electrical Machines Week 3: Energy Storage RECALL REMEMBER.!! Magnetic circuits and electrical circuits are co-related ngredients What is hystresis Magnetic Losses?? WHY DO WE NEED ALL OF THS ANYWAY!!!!
More information1 Constant Real Rate C 1
Consan Real Rae. Real Rae of Inees Suppose you ae equally happy wh uns of he consumpon good oday o 5 uns of he consumpon good n peod s me. C 5 Tha means you ll be pepaed o gve up uns oday n eun fo 5 uns
More informationExercises for Frequency Response. ECE 102, Fall 2012, F. Najmabadi
Eecses Fequency espnse EE 0, Fall 0, F. Najabad Eecse : Fnd the d-band an and the lwe cut- equency the aple belw. µ n (W/ 4 A/, t 0.5, λ 0, 0 µf, and µf Bth capacts ae lw- capacts. F. Najabad, EE0, Fall
More informationSections 3.1 and 3.4 Exponential Functions (Growth and Decay)
Secions 3.1 and 3.4 Eponenial Funcions (Gowh and Decay) Chape 3. Secions 1 and 4 Page 1 of 5 Wha Would You Rahe Have... $1million, o double you money evey day fo 31 days saing wih 1cen? Day Cens Day Cens
More informationNeutron Slowing Down Distances and Times in Hydrogenous Materials. Erin Boyd May 10, 2005
Neu Slwig Dw Disaces ad Times i Hydgeus Maeials i Byd May 0 005 Oulie Backgud / Lecue Maeial Neu Slwig Dw quai Flux behavi i hydgeus medium Femi eame f calculaig slwig dw disaces ad imes. Bief deivai f
More informationDishonest casino as an HMM
Dshnes casn as an HMM N = 2, ={F,L} M=2, O = {h,} A = F B= [. F L F L 0.95 0.0 0] h 0.5 0. L 0.05 0.90 0.5 0.9 c Deva ubramanan, 2009 63 A generave mdel fr CpG slands There are w hdden saes: CpG and nn-cpg.
More informationINDUCTANCE Self Inductance
NDUTANE 3. Self nductance onsider the circuit shown in the Figure. When the switch is closed the current, and so the magnetic field, through the circuit increases from zero to a specific value. The increasing
More informationElectrical Machines I. Week 3: Energy Storage
Electrical Machines Week 3: Energy Storage RECA REMEMBER.!! Magnetic circuits and electrical circuits are co-related ngredients What is hystresis Magnetic osses?? WHY DO WE NEED A OF THS ANYWAY!!!! Energy
More informationPhysics 102. Final Examination. Spring Semester ( ) P M. Fundamental constants. n = 10P
ε µ0 N mp M G T Kuwit University hysics Deprtment hysics 0 Finl Exmintin Spring Semester (0-0) My, 0 Time: 5:00 M :00 M Nme.Student N Sectin N nstructrs: Drs. bdelkrim, frsheh, Dvis, Kkj, Ljk, Mrfi, ichler,
More informationLecture 4. Electric Potential
Lectue 4 Electic Ptentil In this lectue yu will len: Electic Scl Ptentil Lplce s n Pissn s Eutin Ptentil f Sme Simple Chge Distibutins ECE 0 Fll 006 Fhn Rn Cnell Univesity Cnsevtive Ittinl Fiels Ittinl
More informationLet s treat the problem of the response of a system to an applied external force. Again,
Page 33 QUANTUM LNEAR RESPONSE FUNCTON Le s rea he problem of he response of a sysem o an appled exernal force. Agan, H() H f () A H + V () Exernal agen acng on nernal varable Hamlonan for equlbrum sysem
More informationnecessita d'interrogare il cielo
gigi nei necessia d'inegae i cie cic pe sax span s inuie a dispiegaa fma dea uce < affeandi ves i cen dea uce isnane " sienzi dei padi sie veic dei' anima 5 J i f H 5 f AL J) i ) L '3 J J "' U J J ö'
More information( ) [ ] MAP Decision Rule
Announcemens Bayes Decson Theory wh Normal Dsrbuons HW0 due oday HW o be assgned soon Proec descrpon posed Bomercs CSE 90 Lecure 4 CSE90, Sprng 04 CSE90, Sprng 04 Key Probables 4 ω class label X feaure
More informationChapter 2 Linear Mo on
Chper Lner M n .1 Aerge Velcy The erge elcy prcle s dened s The erge elcy depends nly n he nl nd he nl psns he prcle. Ths mens h prcle srs rm pn nd reurn bck he sme pn, s dsplcemen, nd s s erge elcy s
More informationConsider the simple circuit of Figure 1 in which a load impedance of r is connected to a voltage source. The no load voltage of r
1 Intductin t Pe Unit Calculatins Cnside the simple cicuit f Figue 1 in which a lad impedance f L 60 + j70 Ω 9. 49 Ω is cnnected t a vltage suce. The n lad vltage f the suce is E 1000 0. The intenal esistance
More information(C) Pavel Sedach and Prep101 1
(C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach
More informations = rθ Chapter 10: Rotation 10.1: What is physics?
Chape : oaon Angula poson, velocy, acceleaon Consan angula acceleaon Angula and lnea quanes oaonal knec enegy oaonal nea Toque Newon s nd law o oaon Wok and oaonal knec enegy.: Wha s physcs? In pevous
More informationConvection and conduction and lumped models
MIT Hea ranfer Dynamc mdel 4.3./SG nvecn and cndcn and lmped mdel. Hea cnvecn If we have a rface wh he emperare and a rrndng fld wh he emperare a where a hgher han we have a hea flw a Φ h [W] () where
More informationAP Physics 1 MC Practice Kinematics 1D
AP Physics 1 MC Pracice Kinemaics 1D Quesins 1 3 relae w bjecs ha sar a x = 0 a = 0 and mve in ne dimensin independenly f ne anher. Graphs, f he velciy f each bjec versus ime are shwn belw Objec A Objec
More informationRadiometric Dating (tap anywhere)
Radiometric Dating (tap anywhere) Protons Neutrons Electrons Elements on the periodic table are STABLE Elements can have radioactive versions of itself called ISOTOPES!! Page 1 in your ESRT has your list!
More informationThe Gradient and Applications This unit is based on Sections 9.5 and 9.6, Chapter 9. All assigned readings and exercises are from the textbook
The Gadient and Applicatins This unit is based n Sectins 9.5 and 9.6 Chapte 9. All assigned eadings and eecises ae fm the tetbk Objectives: Make cetain that u can define and use in cntet the tems cncepts
More information2. The units in which the rate of a chemical reaction in solution is measured are (could be); 4rate. sec L.sec
Kineic Pblem Fm Ramnd F. X. Williams. Accding he equain, NO(g + B (g NOB(g In a ceain eacin miue he ae f fmain f NOB(g was fund be 4.50 0-4 ml L - s -. Wha is he ae f cnsumpin f B (g, als in ml L - s -?
More informationMarks for each question are as indicated in [] brackets.
Name Student Number CHEMISTRY 140 FINAL EXAM December 10, 2002 Numerical answers must be given with appropriate units and significant figures. Please place all answers in the space provided for the question.
More informationDisplacement, Velocity, and Acceleration. (WHERE and WHEN?)
Dsplacemen, Velocy, and Acceleraon (WHERE and WHEN?) Mah resources Append A n your book! Symbols and meanng Algebra Geomery (olumes, ec.) Trgonomery Append A Logarhms Remnder You wll do well n hs class
More informationA L A BA M A L A W R E V IE W
A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N
More informationChapter 31 Faraday s Law
Chapte 31 Faaday s Law Change oving --> cuent --> agnetic field (static cuent --> static agnetic field) The souce of agnetic fields is cuent. The souce of electic fields is chage (electic onopole). Altenating
More informationElectric Charge. Electric charge is quantized. Electric charge is conserved
lectstatics lectic Chage lectic chage is uantized Chage cmes in incements f the elementay chage e = ne, whee n is an intege, and e =.6 x 0-9 C lectic chage is cnseved Chage (electns) can be mved fm ne
More information5 questions, 3 points each, 15 points total possible. 26 Fe Cu Ni Co Pd Ag Ru 101.
Physical Chemistry II Lab CHEM 4644 spring 2017 final exam KEY 5 questions, 3 points each, 15 points total possible h = 6.626 10-34 J s c = 3.00 10 8 m/s 1 GHz = 10 9 s -1. B= h 8π 2 I ν= 1 2 π k μ 6 P
More informationOutline. Steady Heat Transfer with Conduction and Convection. Review Steady, 1-D, Review Heat Generation. Review Heat Generation II
Steady Heat ansfe ebuay, 7 Steady Heat ansfe wit Cnductin and Cnvectin ay Caett Mecanical Engineeing 375 Heat ansfe ebuay, 7 Outline eview last lectue Equivalent cicuit analyses eview basic cncept pplicatin
More information- Double consonant - Wordsearch 3
Wh 3 Kn, Kn. Wh' h? Hpp. Hpp h? Hpp hy yu, Hpp hy yu! A h f h pg f. Th hn n h pu. Th h n p hny (ng ) y (ng n). Whn yu fn, n un. p n q q h y f h u g h q g u g u n g n g n q x p g h u n g u n y p f f n u
More informationThe Finite Element Method for the Analysis of Non-Linear and Dynamic Systems
Swss Federal Insue of Page 1 The Fne Elemen Mehod for he Analyss of Non-Lnear and Dynamc Sysems Prof. Dr. Mchael Havbro Faber Dr. Nebojsa Mojslovc Swss Federal Insue of ETH Zurch, Swzerland Mehod of Fne
More informationQ1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?
Term: 111 Thursday, January 05, 2012 Page: 1 Q1. A string f length L is fixed at bth ends. Which ne f the fllwing is NOT a pssible wavelength fr standing waves n this string? Q2. λ n = 2L n = A) 4L B)
More informationSection 5.8 Notes Page Exponential Growth and Decay Models; Newton s Law
Sectin 5.8 Ntes Page 1 5.8 Expnential Grwth and Decay Mdels; Newtn s Law There are many applicatins t expnential functins that we will fcus n in this sectin. First let s lk at the expnential mdel. Expnential
More informationChemistry 431 Practice Final Exam Fall Hours
Chemistry 431 Practice Final Exam Fall 2018 3 Hours R =8.3144 J mol 1 K 1 R=.0821 L atm mol 1 K 1 R=.08314 L bar mol 1 K 1 k=1.381 10 23 J molecule 1 K 1 h=6.626 10 34 Js N A = 6.022 10 23 molecules mol
More informationCh. 3: Inverse Kinematics Ch. 4: Velocity Kinematics. The Interventional Centre
Ch. : Invee Kinemati Ch. : Velity Kinemati The Inteventinal Cente eap: kinemati eupling Apppiate f ytem that have an am a wit Suh that the wit jint ae ae aligne at a pint F uh ytem, we an plit the invee
More informationINDUCTANCE Self Inductance
DUCTCE 3. Sef nductance Cnsider the circuit shwn in the Figure. S R When the switch is csed the current, and s the magnetic fied, thrugh the circuit increases frm zer t a specific vaue. The increasing
More informationMains LasagneAlForno LinguineCarbonara PenneGorgonzolawithRocket VPenneSorrentino(Addchickenfor 1) VLasagneVerdura
EyBMnu AbMn-S FOnbf6pm 2Cu 3 3Cu 6 S VTmnm VTmnm VMuhmbknmnh whwhupu VMn VMuhmng VCbbn VGB VGBM VGBM VBuh Mn LgnAFn LngunCbn PnnGgnwhRk VPnnSnnAhknf VLgnVu VAubgnPmgn VAubgnPmgn CnnnVuSpnhGgn LngunBgn
More informationFI 2201 Electromagnetism
FI 2201 Electomagnetism Alexande A. Iskanda, Ph.D. Physics of Magnetism and Photonics Reseach Goup Electodynamics ELETROMOTIVE FORE AND FARADAY S LAW 1 Ohm s Law To make a cuent flow, we have to push the
More informationLecture 5. Plane Wave Reflection and Transmission
Lecue 5 Plane Wave Reflecon and Tansmsson Incden wave: 1z E ( z) xˆ E (0) e 1 H ( z) yˆ E (0) e 1 Nomal Incdence (Revew) z 1 (,, ) E H S y (,, ) 1 1 1 Refleced wave: 1z E ( z) xˆ E E (0) e S H 1 1z H (
More informationKinematics Review Outline
Kinemaics Review Ouline 1.1.0 Vecrs and Scalars 1.1 One Dimensinal Kinemaics Vecrs have magniude and direcin lacemen; velciy; accelerain sign indicaes direcin + is nrh; eas; up; he righ - is suh; wes;
More informationPhysics 321 Solutions for Final Exam
Page f 8 Physics 3 Slutins fr inal Exa ) A sall blb f clay with ass is drpped fr a height h abve a thin rd f length L and ass M which can pivt frictinlessly abut its center. The initial situatin is shwn
More informationσ = neμ = v D = E H, the Hall Field B Z E Y = ee y Determining n and μ: The Hall Effect V x, E x I, J x E y B z F = qe + qv B F y
Detemining n and μ: The Hall Effect V x, E x + + + + + + + + + + + --------- E y I, J x F = qe + qv B F y = ev D B z F y = ee y B z In steady state, E Y = v D B Z = E H, the Hall Field Since v D =-J x
More information5/20/2011. HITT An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement:
5/0/011 Chapte 5 In the last lectue: CapacitanceII we calculated the capacitance C f a system f tw islated cnducts. We als calculated the capacitance f sme simple gemeties. In this chapte we will cve the
More informationSolutions 1F and 2F. 1) correct 12) 2 8 = 4 13) Q = 22 4Q = 13 Q = 13/4 = 3 1/4. 2) correct R 6 S 9. 13) R 6 S -9 or 2) 1 YX YX.
F F ) ) ) 7 ) Q Q Q Q / / ) correct ) correct ) ) ) R S or R S ) ()P P P ) 0 ) ( 0 ) 0 ) 7) 7 ) A D C ) [ 0 7] 00 0 00 ) 7Q( ) ) A B( A B) ) ) ) 0 () () 0 Z() () Z ) [ 0 07C] 000 ) 0 70C 7 70C C 7/70 7/70
More information