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| 2017/06/17 09:53:54瀏覽55｜回應0｜推薦0 | |
| 標題: 電學題目!!!急急急20點 發問: Given that an alternating current i(l)=20sin(314.15t) Determine the following: (a) the peak value of i(l); (b) the instantaneous value of i(l) at 15 and 20 ms. (c)the peak-to-peak of i(l); (d) the frequency of i(l); (e) the period of i(l); (f) the r.m.s of i(l) (g) the average of i(l)
此文章來自奇摩知識+如有不便請留言告知 最佳解答:We can express i(t) = 20 sin 100πt a) Peak value = 20 A b) When t = 0.015, i = 20 sin 1.5π = -20 A When t = 0.02, i = 20 sin 2π = 0 A c) peak-to-peak value = 40 A d) Frequency = 100π/(2π) = 50Hz e) Period = 1/50 = 0.02 s f) r.m.s. = 20/√2 = 10√2 A g) Time average of i = 0 A (Over a period) 其他解答: |
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