hw4soln - Practice Problem 11.5 Combining elements in...

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1 Practice Problem 11.5 8 5 j10 -j4 2A L Z 5 j10 8 -j4 17.9 26.6 V 5 1.79 63.4 A 8 j6 Using a series of source transformations we get the Thevenin equivalent circuit: 1.79 63.4 A Combining elements in parallel: 3.4146+j0.7317 3.4146+j0.7317 6.25 51.34 V Thevenin equivalent circuit: See MATLAB code next page () ( ) 2 2 Th max 22 Th Th L L LL VR PI R RR X X == +++ L Z L Z L Z L Z MATLAB CODE: clear; R=[0:0.1:5]; X=[-5:0.1:5]; for a=1:1:51 for b=1:1:101 P(a,b) = abs(3.9025-4.8781i)^2*((a-1)/10)/((3.4146+(a-1)/10)^2+(0.7317+(b-51)/10)^2); end end mesh(X,R,P) xlabel('Reactive Load'); ylabel('Resistive Load'); zlabel('Power'); Reactive Load: X L for maximum power transfer = -j 0.7317 Resistive Load: X L for maximum power transfer = 3.415
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2 MATLAB OUTPUT: Irms = 2.3152 Iave = 1.9048 Power = 48.2400 MATLAB CODE: for t=0:1:10 i(t+1)=4*t/10; end for t=11:1:20 i(t+1)=-4*t/10+8; end t =[0:0.1:2]; plot(t,i); xlabel('time'); ylabel('i(t)'); Irms = sqrt(0.1*trapz(i.^2)) Iave = mean(i) Power = Irms^2*9 One period of i(t) Practice Problem 11.7 11.12 a) b) () 0 Th 0 2 Th max L j2 40 8j 2 8 31.98 W 4 I VI V P R = = == Th 2 Th max L 4j 3 10 30 9j 3 2.778 W 4 V V P R ⎛⎞ = ⎜⎟ ⎝⎠ 40 A -j3 8 + - + - 8 -j2 -j2 4 4 5 5 j2 j2 -j3 Th V 0 I Th Z Th Z Th V A 30 10 = = + = + = 67 . 1 5 . 2 67 . 1 5 . 2 ) 3 4 ( 5 2 * Th L Th j Z Z j j j Z + = = = = 882 . 1 471 . 0 882 . 1 471 . 0 2 8 * Th L Th j Z Z j j Z
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3 11.14 () ( ) Th * Th Th 2 2 Th 2 j10
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This note was uploaded on 01/19/2012 for the course EECE 253 taught by Professor Shuo,tang during the Spring '11 term at The University of British Columbia.

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hw4soln - Practice Problem 11.5 Combining elements in...

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