A series circuit is made up of an AC source of variable frequency, a 115 omega r

Question

A series circuit is made up of an AC source of variable frequency, a 115 omega resistor, a 1.25 Mu F capacitors and a 4.50 mH inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (1) the resonance angular frequency (2) twice the resonance angular frequency. If the power source provides a peak voltage of 60V, what are (3) the peak current in the circuit and (4) the average power dissipated in the resistor when the circuit is operated at the same angular frequencies as in part 1 and 2 respectively?

Explanation / Answer

Xc = 1/wC

XL = wL

Impedance = sqrt(R^2 + (XL - Xc)^2)

phi = tan-1((XL - Xc) / R)

1) for resonance frequency,

w0 *L = 1/ w0*C

w0 = sqrt(1/LC) = 13333.33 rad/s

Impedance at Resonance frequency Z = R = 115 ohm

and phi = tan^-1(0) = 0

2) w = 2w0 = 26666.67 rad/s

Z = sqrt[ 115^2 + ( (2666.67x4.50x10^-3 - 1/(26666.67x1.25 x10^-6))) = 146.03 Ohm

and phi = tan-1( (2666.67x4.50x10^-3 - 1/(26666.67x1.25 x10^-6))/115) = 38.05 degree

3) for case 1.

Imax = Vmax / Z = 60 / 115 = 0.522 A

For case 2 :

Imax = 60 / 146.03 = 0.41 A

4) Average power = Vmax * Imax * cosphi / 2

for case 1:

Pavg = 60 x 0.522 x cos0 / 2 = 16.56 W

For case 2

Pavg = 60 x 0.41 x cos38.05 / 2 = 9.69 W

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