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### HW 27 Solutions

Course: EECE 112, Spring 2008
School: Vanderbilt
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Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Eleven Solutions10 March 200631. (a)v = 10 + 9 cos100t + 6sin100t 1 1 Veff = 100 + 81 + 36 = 158.5 = 12.590 V 2 2(b)Feff =1 2 (10 + 202 + 102 ) = 150 = 12.247 4(c)Favg =(10)(1) +
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Eleven Solutions10 March 200649.Vx - 100 V V - j100 + x + x =0 6 + j4 - j10 5 1 100 Vx + j 0.1 + 0.2 = + j 20 6 + j4 6 + j4 Vx = 53.35- 42.66 V 100 - 53.35- 42.66 I1 = = 9.806 - 64.44
Vanderbilt - EECE - 112
Correction: The answer for the power of the two resistors is correct, but the formula used isn't right. Only the voltage should be squared, not both the voltage and the resistance. So it should be: P = .5 * (52.442/5) * cos(50) P = .5 * (52.442/8) *
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Ten Solutions10 March 200641. (a) = 800 : 2F - j 625, 0.6H j 480 Zin = 300(- j 625) 600( j 480) + 300 - j 625 600 + j 480 = 478.0 + j175.65 300(- j 312.5) 300 - j 312.5(b) = 1600 : Zin =
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Ten Solutions10 March 200622.(a) 7 -90o = -j 7 (b) 3 + j + 7 -17o = 3 + j + 6.694 j 2.047 = 9.694 j 1.047 (c) 14ej15 = 14 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3. 623 (d) 1 0o = 1 (
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
it should still be 44.23
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Eleven Solutions10 March 20061.50 (- j80) 106 = - j80 , = 42.40 - 32.01 j 500 25 50 - j80 V = 84.80 - 32.01 V, I R = 1.696 - 32.01 A Zc = I c = 1.060057.99 A ps ( / 2ms) = 84.80 cos (45 - 32.0
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Four Solutions10 March 200617.Choosing the bottom node as the reference terminal and naming the left node &quot;1&quot;, the center node &quot;2&quot; and the right node &quot;3&quot;, we next form a supernode about nodes 1
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Two Solutions10 March 200626.The supply voltage is 110 V, and the maximum dissipated power is 500 W. The fuses are specified in terms of current, so we need to determine the maximum current tha
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Two Solutions10 March 200615.(a) Pabs = (+3.2 V)(-2 mA) = 6.4 mW (b) Pabs = (+6 V)(-20 A) = 120 W(or +6.4 mW supplied)(or +120 W supplied)(d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Three Solutions10 March 20062.(a) six nodes; (b) nine branches.PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for co
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Three Solutions10 March 200645.(a) Applying KCL, 1 i 3 + 3 = 0so i = 1 A.(b) Looking at the left part of the circuit, we see 1 + 3 = 4 A flowing into the unknown current source, which, by
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Eight Solutions10 March 20061.i (t ) = i (0)e-R t L= 2e -4.710 t mA9(a) i(100 ps) = 2e-4.7109 10010-12()= 1.25 mA(b) i(212.8 ps) = 2e (c) vR = iR-4.7109 212.810-12()= 7
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Seven Solutions10 March 200631. C equiv 1 1 in series with 10 in series with 10 + 10 + 1 1 1 1 10 + 10 10 + 10 4.286FPROPRIETARY MATERIAL. 2007 The McGraw-Hill Comp
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Seven Solutions10 March 20066.(a)For VA = 1V, W =-12 2K s 0 . (Vbi - V A ) = 2(11.8) 8-854 10 24 (0.57 + 1) qN 1.6 10 19 1 10(()() )= 45.281 10 Cj =-9m11.8 8.854 10 -12
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Six Solutions10 March 200610.This is again a non-inverting amplifier. Similar to question 9, we have:a) Vout = 200m (1 + 4.7) = 1.14 V b) Vout = (1 + 1) 9 = -18 V c) Vout = 7.8 100m = 0.78
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
(b) Connecting a 10resistor, + 10 = ) = 3.33 AIload = VTH/(RTH+10 ) = 75 V/(12.5 Pload = Iload2 * 10 = (3.33 A)2 * 10 resistor,111 W(c) Connecting a 75Iload = VTH/(RTH+75 ) = 75 V/(12.5 Pload = Iload2 * 75 = (.857 A)2 * 75+ 75 =) = .857
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Five Solutions10 March 200642.(a) Replacing the 7 resistor with a short circuit, we find ISC = 15 (8)/ 10 = 12 A. Removing the short circuit, and open-circuiting the 15 A source, we see that R
Vanderbilt - EECE - 112
Engineering Circuit Analysis, 7th EditionChapter Five Solutions10 March 20063.Open circuit the 4 A source. Then, since (7 + 2) | (5 + 5) = 4.737 , we can calculate v1 = (1)(4.737) = 4.737 V. To find the total current flowing through the 7 r
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
Vanderbilt - EECE - 112
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TEAM LinGBeginning AlgorithmsBeginning AlgorithmsSimon Harris and James RossBeginning AlgorithmsPublished by Wiley Publishing, Inc. 10475 Crosspoint Boulevard Indianapolis, IN 46256 www.wiley.com Published 2006 by Wiley Publishing, Inc., Indi
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1362 Spring 07 Exam 1 YELLOW VERSION 1. Which of the following is not correct concerning interphase? A) A functioning cell that has not received a stimulus to divide spends its time in G1 and may go into G0. B) DNA synthesis is completed prior to the
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Exam 2 Spring 2007 Biol 1362 PINK VERSION 1. Which A) B) C) D) E) Ans: 2. All are A) B) C) D) E) Ans:Bubble in name (last name first). Bubble in last 4 digits of student ID # in boxes G-H-I-J. Write color of exam above your name.of the following
U. Houston - BIOL - 1362
BIOL 1362 Exam 3 Spring 2007 BLUE 1. Which of the following statements is true about natural selection? A) Natural selection acts by preserving traits acquired during an individual's lifetime. B) Natural selection acts on the phenotypic variation in
U. Houston - BIOL - 1362
Spring 07 1362 Exam 4 WHITE 1. Although primary producers can capture light (solar) energy to fix CO2, they are absolutely dependent upon the _ to provide them with usable forms of nitrogen and minerals. A) carnivores B) decomposers C) autotrophs D)
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UCSD - BIBC - 102
UCSD - BIBC - 120
UCSD - BIBC - 102
Stanford - CS - 154
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