R-L Circuit problem.?

Question

A series R-L circuit takes 371.2 watts at a power factor of 0.8 from a 116V 60 cycle source.What are the values of R and L?.
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ANS. R=23.2ohms, L=0.0462 henry

Answer 1

P = 371.20 watt

Q = P*tan φ = 371.20*0.6/0.8 = 278.40 Var

A = √P^2+Q^2 = √371.20^2+ 278.40 = 464.00 Va

I = A/V = 464/116 = 4.00 A

R = P/I^2 = 371.20/16 = 23.20 ohm

X = Q/I^2 = 278.40/16 = 17.40 ohm

L = X/(2*PI*f) = 17.40/(6.2832*60) = 0.04615 H (46.15 mH)



additional question: What additional inductance should be inserted in series in the circuit of the problem above if the overall power-factor is to be reduced to 0.5? what will be the current and power under this condition?



sin φ1 = √1-0.5^2 = √0.75 = √3/2

tan φ1 = sin φ1/cos φ1 = √3/2*2 = √3

X1 = R*tan φ = 23.20*√3 = 40.1836 ohm

ΔX = X1-X = 40.1836-17.40 = 22.7836 ohm

ΔL = ΔX/(2*PI*f) = 22.7836/(6.2832*60) = 0.06044 H (60.44 mH)

I1 = V/Z1 = 116/√R^2+X1^2 = 116/√23.20^2+40.1836^2 = 2.500 A

P1 = R*I1^2 = 23.20*2.5^2 = 145.00 watt

A1 = Z1*I1^2 = √23.20^2+40.1836^2 *6.25 = 290.00 Va

cos φ1 = P1/A1 = 145.00/290.00 = 0.50 ...verified !!!!!

Answer 2

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cos( θ ) = [ true power / apparent power ] = 0.8 ... [ value given ] ;



⇒ apparent power = true power / cos( θ ) = 371.2 / 0.8 = 464 VA ;



apparent power = Vs Is = 464 .... [as obtained immediately above] ;



⇒ Is = 464 / Vs = 464 /116 ..... [ using given value of Vs ] ;

⇒ Is = 4 A



true power ( P ) = (I^2) R ;

⇒ R = P / ( I^2 )

⇒ R = [ 371.2 / ( 4^2 ) ] = [ 371.2 / 16 ] = 23.2 Ω ;



... Z = Vs / Is

⇒ Z = 116 / 4 = 29 Ω .... [ Vs : as given ; Is : as calculated above ] ;



... Z = √ ( R^2 + X^2 ) ;



⇒ X = √ ( Z^2 - R^2 ) ... [ transposing for X : inductive reactance ] ;

⇒ X = √ ( 29^2 - 23.2^2 ) = 17.4 Ω ... [ use Z & R : as calculated ] ;



where (inductive) reactance :



... X = 2π f L ;

⇒ L = X / ( 2π f ) ..... [ transposing for L ] ;

⇒ L = 17.4 / ( 120 π ) = 46.2 mH .



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Answer 3

VR = 116 x cos phi

VR = 116 x 0.8 = 92.8 v

R = V^2 / P = 92.8^2 / 371.2

R = 23.2 ohms



cos phi = 0.8 ----> tan phi = 0.75

tan phi = XL / R

XL = 23.2 x 0.75 = 17.4 ohms

L = 17.4 / 2 pi 60 = 0.0461 H or 46.1 mH