A coil having a resistance of 7 Ω and an inductance of 31.8 mH is connected to a 230V, 50Hz supply. Calculate (a) the circuit current (b) the phase angle (c) the power factor. (d) power consumed. Ans (18.85, 55° lag, .573° lag, 2484.24 w) 


Data Given:

Resistance of the coil = R =  7 Ω

Inductance = L = 31.8 mH = 31.8 ｘ10⁻³ H

Frequency of A.C. supply = f = 50 Hz

Voltage (rms) = VrmsVrms = 230 V


To Find:

(a) Current = I = ?

(b) Phase angle = ф = ?

(c) Power factor = cos ф  = ?

(d) Power consumed =<P>= ?


Solution:

(a) Current = I = ?

The current is given by:

IrmsIrms = VrmsZVrmsZ ---------(1)

But the impedance Z is unknown So

Z = √R2+X2L√R2+X2L -----------(2)

Also, we will find the value of XLXL 

XLXL  = ယ L = 2ㄫf L

by putting values

XLXL  = 2 ｘ 3.1416 ｘ 50 Hz ｘ 31.8ｘ10⁻³ H 

XLXL  = 9.985 Ω

Now eqn(2)

Z = √(7Ω)2+(9.985Ω)2√(7Ω)2+(9.985Ω)2

Z = √49Ω2+99.704Ω2√49Ω2+99.704Ω2

Z = √148.704Ω2√148.704Ω2

Z = 12.194 Ω 


Thus equation (1) will be 

Irms = 230V12.194Ω

Irms = 18.86 A -----------------------Ans. (1)



(b) Phase angle = ф = ?

Phase angle can be calculated as

ф = tan⁻¹(XLR)

ф = tan⁻¹(9.985Ω7Ω)

ф = tan⁻¹(1.426)

ф = 55° ---------------------Ans. (2)


(c) Power factor = cos ф  = ?

cos ф  = cos 55° = 0.573 ---------------Ans. (3)



(d) Power consumed = <P> = ?

 <P> = IrmsVrms cos ф 

 <P> = 18.86 A ｘ230 V ｘ 0.573 

 <P> = 2485.559 w  -----------------Ans. (4) 



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