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54 Pages

### ch10_ism

Course: ELEN 214, Spring 2008
School: Texas A&M
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Word Count: 4928

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Sinusoidal 10 Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/- 45 V, I = 20/15 A Therefore 1 P = (100)(20) cos[-45 - (15)] = 500 W, 2 Q = 1000 sin -60 = -866.03 VAR, [b] V = 100/- 45 , I = 20/165 BA AB AB BA P = 1000 cos(-210 ) = -866.03 W, Q = 1000 sin(-210 ) = 500 VAR, [c] V = 100/- 45 , I = 20/- 105 AB P = 1000 cos(60 ) = 500 W, Q = 1000 sin(60 ) = 866.03 VAR, [d] V = 100/0 , I =...

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Texas A&M - ELEN - 214
11 Balanced Three-Phase CircuitsAssessment ProblemsAP 11.1 Make a sketch:We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VN
Texas A&M - ELEN - 214
12 Introduction to the Laplace TransformAssessment ProblemsAP 12.1 [a] cosh t = et + e-t 2 Therefore, 1 -(s-)t [e + e-(s+)t ]dt L{cosh t} = 2 0- = = [b] sinh t = 1 e-(s-)t 2 -(s - ) 1 2 0-+e-(s+)t -(s + ) = s2 0-1 1 + s- s+s - 2et - e
Texas A&M - ELEN - 214
13 The Laplace Transform in CircuitAnalysisAssessment ProblemsAP 13.1 [a] Y = 1 1 C[s2 + (1/RC)s + (1/LC) + + sC = R sL s 1 = 25 108 LC106 1 = = 80,000; RC (500)(0.025) Therefore Y = [b] -z1,225 10-9 (s2 + 80,000s + 25 108 ) s = -40,000 1
Texas A&M - ELEN - 214
14Introduction to Frequency-Selective CircuitsAssessment ProblemsAP 14.1 fc = 8 kHz, c = 1 ; RC c = 2fc = 16 krad/s R = 10 k;. C =1 1 = = 1.99 nF c R (16 103 )(104 )AP 14.2 [a] c = 2fc = 2(2000) = 4 krad/s L= 5000 R = = 0.40 H c 4000 4000
Texas A&M - ELEN - 214
Active Filter Circuits15Assessment ProblemsAP 15.1 H(s) = -(R2 /R1 )s s + (1/R1 C) R1 = 1 , . C = 1 F1 = 1 rad/s; R1 C R2 = 1, R1 .. R2 = R1 = 1 -s s+1Hprototype (s) =AP 15.2 H(s) =-20,000 -(1/R1 C) = s + (1/R2 C) s + 5000 C = 5 F1
Texas A&M - ELEN - 214
Fourier Series16Assessment ProblemsAP 16.1 av = 1 T 2 T2T /3 0Vm dt +1 TT 2T /3Vm 3T7 dt = Vm = 7 V 9 Vm cos k0 t dt 3ak = = bk = =2T /3 0Vm cos k0 t dt + sin 4k 3 =2T /34Vm 3k0 T 2 T2T /3 06 4k sin k 3T 2T /3Vm sin k
Texas A&M - ELEN - 214
The Fourier Transform17Assessment ProblemsAP 17.1 [a] F () = =0 - /2(-Ae-jt ) dt + /2 0Ae-jt dtA [2 - ej /2 - e-j /2 ] j ej /2 + e-j /2 2A 1- = j 2 -j2A [1 - cos ] = 2 [b] F () = AP 17.2 f (t) = = = = 1 2 0te-at e-jt dt = 4ejt d +
Texas A&M - PHYS - 218
44.2: The total energy of the positron is E K mc 2 5.00 MeV 0.511 MeV 5.51 MeV. We can calculate the speed of the positron from Eq. 37.38Emc 2 1v2 c2v c1mc 2 E210.511 MeV 5.51 MeV20.996.
Texas A&M - PHYS - 218
44.1:ma) Kmc 2311 1 v c2 210.1547mc 2149.109 10kg, so K1.27 10Jb) The total energy of each electron or positron is E K mc2 1.1547mc2 9.46 10 14 J. The total energy of the electron and positron is converted into the total energ
Texas A&M - PHYS - 218
44.4: a) hc Ehc m c 2h m c(6.626 10 34 J s) (207)(9.11 10 31 kg) (3.00 108 m s)1.17 10 14 m 0.0117 pm. In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths would mean higher photon energy, and the muons wo
Texas A&M - PHYS - 218
44.5: a)mmm270 me207 me63 meE 63(0.511 MeV) 32 MeV. b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.
Texas A&M - PHYS - 218
44.3: Each photon gets half of the energy of the pion 1 1 1 E m c2 (270 me )c 2 (270)(0.511 MeV) 69 MeV 2 2 2 E (6.9 107 eV)(1.6 10 19 J eV) f 1.7 1022 Hz 34 h (6.63 10 J s)c f3.00 108 m s 1.7 1022 Hz1.8 10 14 m gamma ray.
Texas A&M - PHYS - 218
44.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27 10 23 Hz and a wavelength of 1.32 10 15 m. b) The energy of each photon will be 938.3 MeV 830 MeV 1768 MeV, with frequency 42.8 10 22 Hz and wavelengt
Texas A&M - PHYS - 218
44.7:E( m)c 2(400 kg400 kg )(3.00 108 m s) 27.20 1019 J.
Texas A&M - PHYS - 218
44.9:1 0n10 5B4 27 3Li4 2He4.002603u 11.018607 um( n m( Li7 31 010 5B) 1.008665 u 10.012937 u 11.021602u He) 7.016004 um 0.002995 u; (0.002995u) (931.5 MeV u ) 2.79 MeV The mass decreases so energy is released and the rea
Texas A&M - PHYS - 218
1 44.8: 4 He 9 Be 12 C 0 n 2 4 6 We take the masses for these reactants from Table 43.2, and use Eq. 43.23 Q (4.002603 u 9.012182 u 12.000000 u 1.008665 u) (931.5 MeV u )5.701 MeV. This is an exoergic reaction.
Texas A&M - PHYS - 218
44.10: a) The energy is so high that the total energy of each particle is half of the available energy, 50 GeV. b) Equation (44.11) is applicable, and Ea 226 MeV.
Texas A&M - PHYS - 218
44.11:a) BqB mBm 2mf q q2 (2.01 u) (1.66 10 27 kg u )(9.00 106 Hz) 1.60 10 19 C B 1.18 Tq 2 B 2 R 2 (1.60 1019 C) 2 (1.18 T) 2 (0.32 m) 2 b) K 5.47 1013 J 27 2m 2(2.01 u )(1.66 10 kg u ) 3.42 106 eV 3.42 MeV and v 2K 2(
Texas A&M - PHYS - 218
eB eBR 3.97 107 s. b) R 3.12 107 m s. c) For threem m figure precision, the relativistic form of the kinetic energy must be used, ( 1 )mc 2 eV ( 1 )mc 2 , so eV ( 1 )mc 2 , so V 5.11 106 V. e44.12: a) 2 f
Texas A&M - PHYS - 218
1000 103 MeV 1065.8, so v 938.3 MeV b) Nonrelativistic: eB 3.83 108 rad s . m Relativistic: eB 1 3.59 105 rad s . m 44.14: a) 0.999999559 . c
Texas A&M - PHYS - 218
44.13: a) Ea22mc 2 ( Emmc 2 )Ea2 Em mc 2 2 2mc The mass of the alpha particle is that of a 4 He atomic mass, minus two electron masses. 2 But to 3 significant figures this is just M ( 4 He) 4.00 u (4.00 u) (0.9315 GeV u ) 3.73 GeV. 2(16.0 GeV)
Texas A&M - PHYS - 218
44.15: a) With Em So Emmc , Em2Ea2 Eq. (44.11). 2mc 2[2(38.7 GeV)] 2 3190 GeV. 2(0.938 GeV) b) For colliding beams the available energy E a is that of both beams. So two proton beams colliding would each need energy of 38.7 GeV to give a tota
Texas A&M - PHYS - 218
44.16: The available energy E a must be (m02m )c 2 , so Eq. (44.10) becomes( m 0 Et2m p ) 2 c 4 ( m 02m p c 2 ( E t 2m p c 22m p c 2 ), or2m p ) 2 c 2 2m p(547.3 MeV 2(938.3 MeV) 2 2(938.3 MeV)2(938.3 MeV) 1254 MeV.
Texas A&M - PHYS - 218
44.17: Section 44.3 says m( Z0 ) 91.2 GeV c 2 .E91.2 109 eV 1.461 10 8 J; m 97.2E c21.63 10 25 kgm(Z0 ) m(p)
Texas A&M - PHYS - 218
44.18: a) We shall assume that the kinetic energy of the 0 is negligible. In that case we can set the value of the photon's energy equal to Q. Q (1193 1116) MeV 77 MeV Ephoton . b) The momentum of this photon is Ephoton (77 106 eV) (1.60 10 18 J eV)
Texas A&M - PHYS - 218
44.20: From Table (44.2), (mme2mv )c 2105.2 MeV.
Texas A&M - PHYS - 218
44.19: m M ( ) m p m 0 . Using Table (44.3): E (m)c 2 1189 MeV 938.3 MeV 135.0 MeV 116 MeV.
Texas A&M - PHYS - 218
44.22: a) Conserved: Both the neutron and proton have baryon number 1, and the electron and neutrino have baryon number 0. b) Not conserved: The initial baryon number is 1 +1 = 2 and the final baryon number is 1. c) Not conserved: The proton has bary
Texas A&M - PHYS - 218
44.21: Conservation of lepton number. a) e ve v Lu : 1 1, Le : 0 so lepton numbers are not conserved. b) e ve v Le : 0 1 1 L : 1 1 so lepton numbers are conserved.1 1e . Lepton numbers are not conserved since just one lepton is c) produced from
Texas A&M - PHYS - 218
44.24: a) Using the values of the constants from Appendix F, e2 1 7.29660475 10 3 , 4 0 c 137.050044 or 1 137 to three figures. b) From Section 38.5 e2 v1 2 0 h but notice this is juste2 h c as claimed rewriting as . 4 0c 2
Texas A&M - PHYS - 218
44.25:f2 cand thus(J m) 1 (J s)(m s 1 )f2 is dimensionless. (Recall f 2 has units of energy times distance.) c
Texas A&M - PHYS - 218
44.23: Conservation of strangeness: a) K v . Strangeness is not conserved since there is just one strange particle, in the initial states. b) n K p 0 . Again there is just one strange particle so strangeness cannot be conserved. c) K K
Texas A&M - PHYS - 218
44.26: a)The particle has Q 1 (as its label suggests) and S 3. Its appears as a &quot;hole&quot;in an otherwise regular lattice in the S Q plane. The mass difference between each S row is around 145 MeV (or so). This puts the mass at about the right spot. As
Texas A&M - PHYS - 218
44.27: a)uds :Q eb)2 1 1 0; 3 3 3 1 1 1 B 1; 3 3 3 S 0 0 ( 1) 1 C 0 0 0 0. Q 2 2 cu : 0; e 3 3 1 1 B 0; 3 3 S 0 0 0; C 1 0 1.Q 1 1 3 1; B 3 e 3 3 S 3(0) 0; C 3(0) 0. Q 1 e 3 S 0 0 2 3 0; C 1; B 0 ( 1) 1 3 1. 1;c)ddd:d)dc :1 30;
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Texas A&M - CVEN - 305
Texas A&M - CVEN - 305
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Wisconsin - MATH - 222
CHAPTER 1 PRELIMINARIES1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, 2. Executing long division,&quot; 9 &quot; 11oe 0.1,2 9oe 0.2,2 113 9oe 0.3,3 118 9oe 0.8,9 119 9oe 0.911 11oe 0.09,oe 0.18,oe 0.27,oe 0.81,
Wisconsin - MATH - 222
CHAPTER 2 LIMITS AND CONTINUITY2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get ar
Wisconsin - MATH - 222
CHAPTER 3 DIFFERENTIATION3.1 THE DERIVATIVE OF A FUNCTION 1. Step 1: f(x) oe 4 c x# and f(x b h) oe 4 c (x b h)# Step 2: oe c2x c h Step 3: f w (x) oe lim (c2x c h) oe c2x; f w (c\$) oe 6, f w (0) oe 0, f w (1) oe c2h!# # # # # \$ # # # #f(x b h)
Wisconsin - MATH - 222
CHAPTER 4 APPLICATIONS OF DERIVATIVES4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x oe c# , an absolute maximum at x oe b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [a b]. 2. An absolute minimu
Wisconsin - MATH - 222
CHAPTER 5 INTEGRATION5.1 ESTIMATING WITH FINITE SUMS 1. faxb oe x# Since f is increasing on ! &quot;, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.(a) ~x oe (b) ~x oe (c) ~x oe (d) ~x oe 2. faxb oe x\$&quot;c! # &quot;c! %
Texas A&M - CVEN - 306
Cornell - MATH - 2940
ISM: Linear AlgebraSection 1.1Chapter 1 1.11. x + 2y x + 2y = 1 -y 2x + 3y = 1 -2 1st equation x + 2y = 1 -2 2nd equation x y=1 y 2. 4x + 3y 7x + 5y3 x + 4y 1 -4y 3 x + 4y = 2 4 =3 7x + 5y=1 = -1 (-1)= -1 , so that (x, y) = (-1, 1). =