Need help with part A An inductor ( L = 400 mH), a capacitor ( C = 4.43 µF), and

Question

Need help with part A

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 ) are connected in series. A 48.0-Hz AC generator connected in series to these elements produces a maximum current of 365 mA in the circuit.

(a) Calculate the required maximum voltage Vmax.
Determine the impedance of the circuit in terms of the given information and then use the relation between current and AC voltage to determine the maximum voltage Vmax. V

(b) Determine the phase angle by which the current leads or lags the applied voltage.

The current leads lags the voltage by a magnitude of 51.34 °.

Explanation / Answer

L=400 mH

C=4.43 uF

R=500 ohms

f=48 Hz

Io=365 mA

a)

Z=sqrt(R^2 + (WL-Wc)^2)

Z=sqrt(500^2 + (2pi*f*L - 1/2pi*f*C)^2)

Z=sqrt(500^2 + (2pi*48*0.4 - 1/(2pi*48*4.43*10^-6))^2)

Z=802.6 ohms

maximum voltage, Vo=Io*Z

Vo=365*10^-3*802.6

Vo=292.95 v

b)

tan(theta)=WL-WC/R

=(2pi*f*L - 1/2pi*f*C)/R

=((2pi*48*0.4) - 1/(2pi*48*4.43*10^-6))/(500)

====> theta=-51.5 degrees

here,

current leads the voltage by 51.5 degrees

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