Answer to Question 5

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 a. Xc = 1/2pfC = 1 (2 p 200 150 10-6) = 5.31 ohms

 

 b. Z2 = 5.312 + 332 = 1117 W2

      Z = 33.4 W

 

 c. I = V/Z = 12 33.43 = 0.359 A

 

 d. tan f = Xc/R = 5.31 33.43 = 0.159

     f = tan-1 0.159 = 9.03o (= 0.158 rad)

 

 e. Power factor = cos f = cos 9.025 = 0.988

 

 f. P = 12 0.359 0.988 = 4.25 W