基本電學問題一條

(a) Since Iav = 0.636.(Ip)
where Iav is the average cirrent and Ip is the peak current

thus, 5 = 0.036.Ip
Ip = 5/0.636 A = 7.86 A
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(b) The root-mean-square current Irms = 0.707.Ip
therefore, Irms = 0.707 x 7.86 A = 5.56 A

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(c) Angular frequency = 2 x pi x 50 s^-1
= 2 x 3.14159 x 50 s^-1 = 314 s^-1