HW1 Solution - Chapter 1, Solution 20. Since P=0 -306+612 +...

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EE302 Homework #1 Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms Chapter 1, Problem 6 Solution. (a) At t=1ms, i = = = 2 80 dt dq 40.0A (b) At t=6ms, i = = dt dq 0.00A (c) At t=10ms, i = = = 4 80 dt dq -20.0A
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Chapter 1, Problem 11. A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h. How much charge can it release at that rate? If its terminals voltage is 1.2 V, how much energy can the battery deliver? Chapter 1, Problem 11 Solution. q= it = 85 x10 -3 x 12 x 60 x 60 = 3,672 C = 3.67 x10 3 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J = 4.41 x10 3 J
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Chapter 1, Problem 18. Find the power absorbed by each of the elements in Fig. 1.29. Chapter 1, Solution 18. P 1 = -V 1 I = -30(10) = -300. W P 2 = V 2 I = 10(10) = 100. W P 3 = V 3 I 3 = 20(14) = 280. W P 4 = -V 4 I 4 = -8(4) = -32.0 W P 5 = -V 5 (0.4I) = -12(4) = -48.0 W
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Chapter 1, Problem 20 Find V 0 in the circuit of Fig. 1.31.
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Unformatted text preview: Chapter 1, Solution 20. Since P=0 -306+612 + 3V 0 + 28 + 282-310 = 0 72 + 84 + 3V 0 = 210 or 3V 0 = 54 V 0 = 18.0 V Chapter 1, Problem 28. A 30-W incandescent lamp is connected to a 120-V source and is left burning continuously in an otherwise dark staircase. Determine: (a) the current through the lamp, (b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh. Chapter 1, Solution 28. (a) i = = = 120 30 V P 0.250 A (b) Number of Watt-hours = pt = 30 365 24 Wh = 262.8 kWh Cost = $0.12 262.8 = $31.54 Chapter 1, Problem 36. A battery may be rated in ampere-hours (Ah). A lead-acid battery is rated at 160 Ah. (a) What is the maximum current it can supply for 40 h? (b) How many days will it last if it is discharged at 1 mA? Chapter 1, Solution 36. (a) i = h Ah 40 160 = 4.00 A (b) t = = = day h h A Ah / 24 000 , 160 001 . 160 6,667 days or 6.67 x 10 3 days...
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HW1 Solution - Chapter 1, Solution 20. Since P=0 -306+612 +...

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