HW1Solution &amp; grading rules - quantity pass through...

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1) 1.16 Graphic solution: (by Abrams) V 1 =9V V 2 =6V I=20*10 -3 A P=VI P 1 =180*10 -3 W P 2 =120*10 -3 W or the area under the graph W=120*288+1/2*(180-120)*288 W=43200J W=43.2kJ Numerical solution: V 1 =9V V 2 =6V write the equation of V vs t: V = 3 80h t + 9 V = 3 288000s t + 9 i = 20mA so the equation of Power vs t is: P = iV = 20m × 3 288000s t + 9 ± = 3 14400000 t + 0.18 Energy delivered by the battery is: W = ² Pdt 288000 0 = ² 3 14400000 t + 0.18 ± dt 288000 0 = 43.2kJ 0 50 100 150 200 0 100 200 300 400 Power [mW] time [ks] Energy

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2) 1.17 a) p = vi = 30e -500t – 30e -1500t – 40e -1000t + 50e -2000t – 10e -3000t p (1 ms) = 3.1 mW b) w(t) = (30 ? 500 ± 30 ? 1500 ± 40 ? 1000 ± + 50 ? 2000 ± 10 ? 3000 ± ) ²± ³ 0 = 21.67 – 60e -500t + 20e -1500t + 40e -1000t – 25e -2000t + 3.33e -3000t μJ w(1 ms) = 1.24 μJ c) w total = w(infinite) = 21.67 μJ You will get 0.5 if any of them is right. 3) V = 5V i = 250 mA R eq = V i = 5 250m = 20 Ω P = Vi= 5 * 250m = 1.25W In 1 sec, electric
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Unformatted text preview: quantity pass through: Q = it = 250m * 1 = 0.25 Coulomb because each electron has 1.6 * 10-19 coulomb electric quantity, electrons pass through is: n = Q e = 0.25 1.6 × 10 − 19 = 1.56 × 10 18 Because 4 battery working in series, they are working at the same time. t = electric quantity in one battery i = 2300mAhour 250mA = 9.2hour = 33120sec You will get 0.5 if any of them is right. 4) (Figure by Abrams) 3 nodes (Neglect ground, so 4 nodes is OK) 4 nodes 3 loops 3 loops You will get 0.5 if 2 or 3 of them is right. 5) Box a, b, d, f are dissipating loads, because the current directions are from + to - If your answer is c, e without explanation, you will get 0.5. If you show that you know a, b, d, f are absorbing power, c, e are delivering power. You will get full score....
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HW1Solution &amp; grading rules - quantity pass through...

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