Answer to Question 3

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 a. 120 A

 

 

 b. T = 2p/w = 2p 100 p = 0.02 s (= 20 ms)

 

 

 c. f = 1/T = 50 Hz.

 

 

 d. 0.36 rad above the zero point

 

 

 e. i = 120 sin (0 + 0.36) = 42.3 A

 

 

 f.  i = 120 sin [(100p 8 10-3) + 0.36] = 31.8 A

 

 

 

 g. 120 = 120 sin [(100pt) + 0.36]

 

 

      sin (100pt + 0.36) = 1

 

 

      (100pt + 0.36) = sin-1 (1) = p/2 = 1.571 rad  (since sin-1 (1) = 90o = p/2 rad)

 

 

 

       t = (1.571 0.36) 100p = 3.85 10-3 s = 3.85 ms