Homework 2 ECE 2100_soln

# Homework 2 ECE 2100_soln - ECE 2100: Professor Alyosha...

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ECE 2100: Professor Alyosha Molnar Homework Assignment #2 (rev 2) Due Feb 25, 2010 Areas covered: Node, mesh analysis. Thevenin, Norton equivalents. Maximum power transfer. Superposition. LEDs, Photodiodes, Photovoltaics. 1) Book Problem 4.1 a) 5 currents b) 4 equations c) 4 5 5 3 2 4 3 1 2 1 0 0 0 0 R Vd Vb R Vd Vc ig R Vd Vc R Vc Vb R Vc Va R Vd Vb R Vc Vb R Vb Va R Vc Va R Vb Va ig - + - + - = - - - + - = - - - - - = - - - - = d) 2 equations e) ( 29 ( 29 ( 29 ( 29 5 3 1 2 4 2 3 2 1 2 1 1 1 2 0 0 R ig I R I I R I R I I R ig I R I + + - + = - + + + =

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2) Book Problem 4.5 a) 4 3 1 2 4 3 2 1 0 ) 3 ( 0 ) 2 ( 0 ) 1 ( i i i ig eq i i i eq ig i i eq - - - = - + = - + = b) eq1+eq2=eq3 b eq1=eq3-e2 eq2=eq3-eq1 3) Book Problem 4.16 a) = = = = - + + - + - = n k k n k k n v n v v nv R v v R v v R v v 1 0 1 0 0 0 1 0 1 1 ... 0 b) v 0 =(150+200-50)/3= 100V
4) Book Problem 4.23 a) Vd Vc Vb Vd Vc Vd Vd Vb Vd Vc Va Vc Vd Vc Vc Va Vd Vb Va Vb Vd Vb Va Vb V Vc Vb Va Va Vc Va Vb Va V 4 2 0 2 4 4 3 6 2 0 2 6 3 11 37 22 2000 0 4 2 11 500 4 6 13 1500 0 3 2 4 500 - + = = - + - - + - = = - + - - + - + = = - + - + - + + - = = - + - + - solve using MATLAB: >> A=[-13 6 4 0;22 -37 0 11;2 0 -6 3;0 1 2 -4] A = -13 6 4 0 22 -37 0 11 2 0 -6 3 0 1 2 -4 >> b=[-1500;-2000;0;0] b = -1500 -2000 0 0 >> A^-1*b ans = 300.0000 280.0000 180.0000 160.0000 Which means: Va=300, Vb=280, Vc=180, Vd=160 So the 5 resistor sees V=(500V-280V)* 5 /(5 +6 )=100V b P=V 2 /R= 2kW b) Itot=Vc/6 +Vd/4 =70A Ptot=-VI=-35000W= - 35kW a b c d

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5) Book Problem 4.36 Repeat book problem 4.23 using mesh current analysis ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4 3 2 1 4 3 4 2 4 1 4 3 2 4 3 2 3 3 4 3 2 1 3 2 4 2 1 2 2 4 2 1 2 1 4 1 1 13 6 3 4 500 0 6 3 4 500 6 12 2 0 6 2 4 3 2 11 2 0 2 3 2 4 4 2 17 0 2 4 11 I I I I V I I I I I I V I I I I I I I I I I I I I I I I I I I I I I I I I I I + - - - - = = - - - - - - - - + - = = - + - + - - + - = = - + - + - + - - = = - + - + put into MATLAB: >> A=[17 -2 0 -4;-2 11 -2 -3;0 -2 12 -6;-4 -3 -6 13] A = 17 -2 0 -4 -2 11 -2 -3 0 -2 12 -6 -4 -3 -6 13 >> b=[0;0;0;500] b = 0 0 0 500 >> A^-1*b ans = 20.0000 30.0000 40.0000 70.0000 So, C=20A, I 2 =30A, I 3 =40A, I 4 =70A, b P= I 1 2 *R=(20A) 2 5 = 2kW b) Ptot=-500V* I 4 = -35kW I1 I2 I3 I4
6) Book Problem 4.38 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 2 1 1 3 2 3 3 1 3 2 1 1 2 2 3 2 3 2 2 1 3 1 1 40 10 30 50 10 20 0 : 3 10 30 15 660 15 5 660 10 0 : 2 50 15 90 15 50 25 0 : 1 I I I I I I I I I eq I I I V I I I V I I eq I I I I I I I I eq + - - = - + - + - = - + - = - + + + - = - - = - + - + = Use MATLAB: >> A=[90 -15 -50;-15 30 -10;-30 -10 40] A = 90 -15 -50 -15 30 -10 -30 -10 40 >> b=[0;-660;0] b = 0 -660 0 >> A^-1*b ans = -22.0000 -42.0000 -27.0000 I 1 =-22, I 2 =-42, I 3 =-27 P=I 3 (20 (I 1 -I 3 )) = -2700W I1 I2 I3

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7) Book Problem 4.58 a) For io=0, V1=V3 At the same time, by voltage divider
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## This note was uploaded on 11/21/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Fall '05 term at Cornell University (Engineering School).

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Homework 2 ECE 2100_soln - ECE 2100: Professor Alyosha...

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