An inductor (L = 405 mH), a capacitor (C = 4.43 µF), and a resistor (R = 400 ) are connected in series. A 50.0 Hz AC generator produces a peak current of 250 mA in the circuit.

Question

An inductor (L = 405 mH), a capacitor (C = 4.43 µF), and a resistor (R = 400 ) are connected in series. A 50.0 Hz AC generator produces a peak current of 250 mA in the circuit.
(a) Calculate the required peak voltage ΔVmax. 
  V
(b) Determine the phase angle by which the current  the applied voltage.
  °

Henry · Answer

a. Xl = 2pi*F*l = 6.28*50*0.405=127.2 Ohms
Xc = 1/(6.28*50*4.43*10-6) = -718.9 Ohms

tan A = (Xl-Xc)/R
tan A = (127.2-718.9)/400 = -1.47925
A = -55.94o

Z=R/cosA = 400/cos(-55.94) = 714.2 Ohms[-55.94o]= The Impedance.

Vmax = Imax * Z=0.25 * 714.2 =178.6 Volts.

b. The negative impedance angle means that the circuit is capacitive. Therefore, the current leads the applied
voltage by 55.94. Imax = 0.25A[55.94o].