基本電學一條題目

(a) The peak current is 10 A, thus current varies between +10 A to -10 A. The peak-to-peak current is thus 2 x 10 A = 20 A

(b) Since angular frequency = 50 s^-1,
frquency = 50/(2.pi) Hz = 7.96 Hz

(c) Period T = 1/frequency = 1/7.96 s = 0.126 s

(d) Instantaneous current I = 10.sin(50x2) A = 10x(-0.51) A = - 5.1 A

Note that 50x2(=100 ) is expressed in radians, which is equal to 15 x(2.pi) + 5.75 radians
thus, sin(100) = sin(15x2.pi + 5.75) = sin(5.75) = -0.51

i=10sin(50t) A,當中,由於sin(50t)最大是1,所以i的最大值是10A,因此(a)的答案是10A;

(b)50是角速度,以rad^-1作單位,已知頻率的定義是一秒鐘完成循環次數,設t=1,那麼一秒中,電流'轉了'50 rad,由於2兀 rad為一次完整循環,所以頻率=50/(2兀)=7.56 Hz;
(c)已知週期的定義是完成一次循環所需時間,電流一秒完成7.56次循環,所以週期即頻率的倒數,也就是1/7.56=0.132 s;
(d)將t=2代入式中,得50t=50X2=.100 rad^-1,再將rad^-1轉化為角度單位(即360度的'度'),100 rad^-1=100/(2兀)X360=5730 度,將之代入式中,得
i=10sin(5730)=-5 A=5 A(相反方向)

2008-05-21 16:01:27 補充：
'是亂碼,不用管它

2008-05-21 16:02:19 補充：
& # 39 ;是亂碼,不用管它