An L-R-C series circuit. R = 140 Ohm, L = 0.770 H and C = 1.70 Times 10^-2 mu F

Question

An L-R-C series circuit. R = 140 Ohm, L = 0.770 H and C = 1.70 Times 10^-2 mu F . The source has a voltage amplitude of 160 V and a frequency equal to the resonance frequency of the circuit. What is the power factor? What is the average power delivered by the source? Express your answer using two significant figures. The capacitor is replaced by one with a capacitance of C = 3.50 Times 10^-2 mu F and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source? Express your answer using two significant figures.

Explanation / Answer

1) as they R and L are in resonance, the voltage will be maximum ,

which means that pf=Cos|0|=1
b)power = 1602 /140=182.86

c)on adjusting again the system will be in resonace, that means again pf=1

power delvered = V2/R = 1602 /140=182.86

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