What is the phase angle? What is the power factor for this circuit... An L-R-C s

Question

What is the phase angle? What is the power factor for this circuit... An L-R-C series circuit L = 0.120 H, R=240 ohm, and C=7.30 muF Carries an rms current of 0.450 A with a frequency of 400 Hz What is the phase angle? What is the power factor for this circuit? What is the impedance of the circuit? What is the rms voltage of the source? What average power is delivered by the source? What is the average rate at which electrical energy is converted to thermal energy in the resistor? What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?

Explanation / Answer

Given that

The inductor (L) =0.120H

The resistance (R) =240ohm

The capacitor (C) =7.30uF =7.30*10-6F

The r.m.s current in the circuit is (Irms) =0.450A

The frequency (f) =400Hz

The phase angle is given by

Tantheta =(XL-XC/R)

Now XL =wL =2pif*L =2*3.14*400*0.120 =301.44ohm

and XC =1/wC =1/2pifC =1/2*3.14*400*7.30*10-6F=54.532ohm

now phase angle is theta =tan-1(301.44-54.532/240)=45.812degrees

The impedance of the circuit is given by Z =Sqrt(R2+(XL-XC)2=Sqrt(240)2+(301.44-54.532)2 =344.330ohm

We also know that

Vrms =Irms*Z =0.450A*344.330ohm =155V

The power factor of the circuit is given by

P =(1/2)VIcostheta =0.5*0.450*155cos(45.812) =24.308W

The average power delivered by the resistor is given by PR =Irms2R=(0.450)2*240=48.6W

The average power delivered by the capacitor is Pc ==Irms2*XC =(0.450)2*54.532=11.042W

The average power delivered by the capacitor is PL ==Irms2*XL =(0.450)2*301.44=61.041W

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