ECE2040D_HW1_Soln

ECE2040D_HW1_Soln - Chapter 1 – Electric Circuit...

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Unformatted text preview: Chapter 1 – Electric Circuit Variables g11 g12 g11 g12 g11 g12 4 4 4sin 3 cos3 cos3 C 3 3 t t t q t i d q t d t g87 g87 g87 g87 g32 g14 g32 g14 g32 g16 g32 g16 g14 g179 g179 4 3 P1.2-3 Problems P1.2-1 dt i t e e d g32 g16 4 1 g32 20 A g16 5 5 t t g16 g11 g12 g11 g12 P1.2-4 g179 g179 t t 8 8 g11 g12 g11 g12 g11 g12 q t g32 g14 i g87 g87 d q 8 g32 d g87 g14 g32 0 C for 8 < t . g179 g179 t t t q t g32 g14 i g87 g87 d q 4 1 g32 g16 d g87 g14 4 g32 g16 g87 g14 4 g32 8 g16 t C for 4 < t < 8. In particular, q (8) = 0 C. 4 4 4 g11 g12 g11 g12 g11 g12 g179 g179 t t t q t g32 g14 i g87 g87 d q 2 2 g32 d g87 g32 2 g87 g32 2 t g16 4 C for 2 < t < 4. In particular, q (4) = 4 C. 2 2 2 g11 g12 g11 g12 g11 g12 g179 g179 g16g102 g16g102 t t g11 g12 g11 g12 q t g32 g32 i g87 g87 d d g87 g32 0 C for t < 2 so q (2) = 0. P1.5-4 charged). a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully g179 g179 g179 g117 g32 g169 g185 g168 g184 g167 g183 W Pdt g32 g32 vi dt g14 dt g32 t g14 t t = 441 10 J 441 kJ 3600 3600 = 2 11 22 0.5 .5 3 2 5 3600 t 5 3600 g11 g12 g11 g12 3600s kWhr Cost = 441kJ g117 g117 g32 1.23¢ 1 hr 10¢ b.) Page 1 / 9 ECE 2040D Solutions to HW#1 - Spring 2009 g11 g12 g11 g12g11 g12 1 1 cos3 sin3 sin 6 3 6 g32 g32 g11 g12 1 6 g32 g32 g11 g12 1 1 sin 6 0.0466 W 6 p g32 g32 g16 Here is a MATLAB program to plot p ( t ): clear t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time v=4*cos(3*t); % device voltage i=(1/12)*sin(3*t); % device current for k=1:length(t) p(k)=v(k)*i(k); % power end plot(t,p) xlabel('time, s'); ylabel('power, W') P1.5-5 P1.5-8 The element voltages and currents do not satisfy conservation of energy and cannot be correct. The element voltages and currents do not satisfy conservation of energy and cannot be correct....
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ECE2040D_HW1_Soln - Chapter 1 – Electric Circuit...

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