Exercise 31.25 A series ac circuit contains a 250-Ohm resistor, a 14.0 mH induct
ID: 1378715 • Letter: E
Question
Exercise 31.25 A series ac circuit contains a 250-Ohm resistor, a 14.0 mH inductor, a 3.90 MuF capacitor, and an ac power source of voltage amplitude 45.0V operating at an angular frequency of 360rad/s. Part A What is the power factor of this circuit? Cos Phi= Part B Find the average power delivered to the entire circuit. P=W Submit My Answers Give Up Part C w What is the average power delivered to the resistor, to the capacitor, and to the inductor? Enter your answers numerically separated by commas. PR,PC,PL= WExplanation / Answer
a) cos phi = cos [ tan-1(XL - XC / R) ]
XL = w L = 360 * 14 * 10-3 = 5.04
XC = 1 / w C = 1 / 360 * 3.90 * 10-6 = 712.25
cos phi = cos [ tan-1(5.04 - 712.25 / 250) ]
= 0.33
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Irms = I / sqrt 2 = E / Z sqrt 2 = E / [sqrt (XL - XC )2 + R2] sqrt 2
= 0.0424 A
P = Irms * Vrms * cos phi
= 0.0424 * [45 / sqrt[2]] * 0.33
= 0.445 W
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PR = P = 0.445 W
PL = 0
PC = 0
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