A 10 mH , 20 Ω coil is connected across 240 V and 180/π Hz source. How much power does it dissipate? (Ans: 2778 W ) 


Data Given:

Inductance = L = 10 mH = 10 ｘ10⁻³ H = 0.01 H

Resistance of the coil = R =  20 Ω

Voltage (rms) = Vrms = 240 V

Frequency of A.C. supply = f = 180ㄫ Hz




To Find:


 The power dissipated  = P = ?


Solution:

We know that Power consumed  P is calculated by formula

 P = IrmsVrms cos θ -------(1)

But the value of Irms and angle θ  is unknown So, to find these values we have


Z = VrmsIrms 

Irms = VrmsZ

where Z = √R2+X2L 

Irms = Vrms√R2+X2L

where XL = ယ L = 2ㄫf L 

Irms = Vrms√R2+(2ㄫfL)2

by putting values


Irms = 240V√(20Ω)2+(2ｘㄫｘ0.01hzｘ180ㄫ)2

or

Irms = 240V√400Ω2+(3.6Hz H)2

Irms = 240 V√400Ω2+12.96Ω2

Irms = 240 V√412.96Ω2

Irms = 240 V20.321Ω

Irms = 11.810 A 


The phase difference θ can be calculated as

θ = tan⁻¹ (ധLR)

or

θ = tan⁻¹ (2ㄫfLR)

By putting values

θ = tan⁻¹ (2ㄫｘ0.01Hzｘ180ㄫH20Ω)

θ = tan⁻¹ (3.6Ω20Ω)

θ = tan⁻¹ (0.18)

θ = 10.2° 


Now putting all the corresponding values in Eqn (1)

 P = 11.8 A ｘ240 V ｘ cos (10.2°) 

 P = 2832 A V ｘ 0.984 

 P = 2786.688 w

 P = 2787 w  -----------------Ans. 





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