An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 ) a

Question

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 ) are connected in series. A 44.0-Hz AC generator connected in series to these elements produces a maximum current of 160 mA in the circuit. (a) Calculate the required maximum voltage Vmax. 138.4 Correct: Your answer is correct. V (b) Determine the phase angle by which the current leads or lags the applied voltage. The current Correct: Your answer is correct. the voltage by a magnitude of -54.7 Incorrect: Your answer is incorrect.

Explanation / Answer

(a) The capacitive reactance is 1/(2 pi f C) = 1/(2*3.14*44*4.43*10-6) = 816.5 Ohms.

The inductive reactance = 2 pi f L = 2 *3.14 *44* 400*10-3 = 110.6 Ohms.

The impedeance is then sqrt(R2 + (XL - XC)2) = sqrt(5002 + (110.6 - 816.5)2) = 865 Ohm.

V = IZ = 0.160*865 = 138.4 V.

(b) tan(phi) = (XL - XC)/R = (110.6 - 816.5)/500 = -1.41, giving phi = -54.7 degrees. Which means the current lags the voltage by 35.3 degrees.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.