An circuit with R = 28.0 ohm , L = 397mH , and C= 46.2microF is connected to an

Question

An circuit with R = 28.0 ohm , L = 397mH , and C= 46.2microF is connected to an ac generator with an rms voltage of 23V .

a. Determine the average power delivered to this circuit when the frequency of the generator is equal to the resonance frequency. P=W
b. Determine the average power delivered to this circuit when the frequency of the generator is twice the resonance frequency.
c. Determine the average power delivered to this circuit when the frequency of the generator is half the resonance frequency.

all answer are to two significant figures

Explanation / Answer

R = 28 , L = 397 mH, C= 46.2 F, Vrms = 23 V

a> At resonance frequency, phase angle = 0

Average power = Vrms2 / R = 18.89 W

b> Let freuency of the generator be w.

Resonance frequency = 1/(LC)

w = 2/(LC)

XL = wL = 2L/(LC) = 2(L/C) = 185.40

XC = 1/wC = 1/2 * (L/C) = 46.35

Phase angle = tan inverse ((185.40 - 46.35)/28) = 78.61o

Average power = (Vrms2/R) * cos = (232/R) * cos 78.61 = 3.73 W

c> w = 1/2(LC)

XL = wL =  1/2 * (L/C) = 46.35

XC = 1/wC = 2(L/C) = 185.40

Phase angle = tan inverse ((46.35 - 185.40)/28) = - 78.61o

Average power = (Vrms2/R) * cos = (232/R) * cos -78.61 = 3.73 W

NOTE that the signs of phase angles in b and c are opposite but the magnitude of average power is same.

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