Practice_Homework_Prob_Sol_EE210_Ch-1

Practice_Homework_Prob_Sol_EE210_Ch-1 - 2 1.5 1 Voltage in...

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0 1 2 3 4 5 6 -1 -0.5 0 0.5 1 1.5 2 Voltage in V TextEnd Time in ms Solution 1.12 (a) V A = P/I = 20/4 = 5 V (b) P B = VI = 2 × 7 = 14 W (c) V C = P/I = -3W/3A = -1V (d) I D = P/V = -27W/3V = -9A (e) I E = P/V = 2/1 = 2A (f) P F = VI = -4 × 5 = -20W In all of the above, note that the direction of the current flow relative to the polarity of the voltage across a device determines whether power is delivered or absorbed. Power is absorbed when current flows from the positive terminal of the device to the negative one. Solution 1.13 By inspection: Circuit Element (CE) 1 absorbs –5W, and CE 2 absorbs 6W. Compute power absorbed by all elements including independent sources: I 3A : -15 CE1: -5 V 3V : -12 CE2: +6 V 5V : 10 I 2A : 16 ----------
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Sum: 0 (Verifies conservation of power.) Solution 1.14 (a) Compute power absorbed: I 5A : -85 CE1: 98 V 3V : 33 CE2: 16 V 7V : -42 I 2A : -20 ------------- Sum: 0 (b) Add all terms: I-source: P absorbed = - 3 1 - e - t ( 29 3 + 3 e - t watts V-source: P 2 3 e - t - 1 6 e - t + 2 watts CE1: P = 3 e - t × - e - t = 9 e - t - 9 e - 2 t watts CE2: P = 3 e - t - 1 3 e - t - 1 = 9 e - 2 t - 6 e - t + 1 watts Simple algebraic manipulation of the the sum of all the above terms reveals that the result is zero. Solution 1.15 When I L = 1, P = V L I L = (16-4) × 1 = 12 W. When I L = 2, P = V L I L = (16-16) × 1 = 0. P = (16-4I L 2 )I L . Differentiate this w.r.t. I L and set to zero: 16 – 12I L 2 = 0. Therefore, I L = 1.155A. Solution 1.16 When I L = 2, P = (16-4) × 2 = 24W. When I L = 3, P = (16 - 9) × 3 = 21 W. Maximum occurs in the interval from 0 to 4: P = (16 - I L 2 ) I L Differentiate w.r.t. I L and set to zero: 16 – 3I L 2 = 0. Therefore, I L = 2.31 A. Solution 1.17 (a) Power is the product of the current and voltage. We can compute the product graphically:
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0 0.5 1 1.5 2 2.5 3 0 2 4 6 8 10 Power in Watts TextEnd Time in s (b) W t ( 29 = p t ( 29 dt 0 t = 10 - 10 e - 7 τ ( 29 d τ 0 t = 10 τ] 0 t - - 10 7 e - 7 τ [ ] 0 t = 10 t + 10 7 e - 7 t - 10 7 This can be used as an aid to plot the work function: 0 1 2 0 1 2 3 4 5 6 7 8 9 Time in s Energy in J Solution 1.18 (a) Since, i t ( 29 = 115 - 23 t mA ,
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q 7 ( 29 = i τ ( 29 d τ 0 7 = 115 t - 23 t 2 2 0 7 × 10 - 3 = 0.2415 C (b) Energy is the integral of power: E = p t ( 29 dt 0 7 = v t ( 29 × i t ( 29 dt 0 7 = 25 i t ( 29 dt 0 7 = 25 × 0.2415 = 6.0375 C Solution 1.19 (a) t = 100 o F, Rate of temp. increase is 2.5 Wh/ o F per gallon: Energy = 2.5Wh/ o F/gallon × 100 o F × 30gallons = 7500 Wh = 2.7 × 10 7 J.
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This note was uploaded on 04/30/2008 for the course ECE 2100 taught by Professor N/a during the Fall '06 term at Missouri (Mizzou).

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Practice_Homework_Prob_Sol_EE210_Ch-1 - 2 1.5 1 Voltage in...

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