An inductor (L 400 mH), a capacitor (C 4.43 ?F), and a resistor (R 500 ) are connected in series. A 42.0 Hz AC generator connected in series to these elements produces a maximum current of 200 mA in the circuit. (a) Calculate the required maximum voltage ΔVmax.?______ V. (b) Determine the phase angle by which the current leads or lags the applied voltage. The current leads or lags? the voltage by? _____ degrees.
The reactance of the inductor is XL2 * Pi * f * L XL 2 * 3.1416 * 42 * 0.4 XL 105.55776 ohms The reactance of the capacitor is XC 1 / (2 * Pi * f * C) XC 1 / (2 * 3.1416 * 42 * 0.00000443) XC 855.3938 ohms The total reactance of the circuit is X XL - XC X 105.55776 - 855.3938 X - 749.83604 ohms The reactance is actually 749.83604 ohms. The minus sign indicates the power factor is a leading power factor. The impedance of the circuit is Z - Sqrt(R^2 + X^2) Z Sqrt(500^2 + 749.83604^2) Z Sqrt(812254.08688) Z 901.2514 ohms The voltage to produce 200 mA in the circuit is E I * Z E 0.2 * 901.2514 E 180.25 V (This is RMS voltage) The peak voltage will be E / 0.707 180.25 / 0.707 254.95 V The power factor of the circuit is PF R / Z PF 500 / 901.2514 PF 0.554784 leading The phase angle is the angle having a cosine value of 0.554784. This is 56.30415 degrees. I suggest you check my calculations. TexMav
