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) = `\V_{rms}` = 240 V

Frequency of A.C. supply = f = `\frac {180}{ㄫ}` Hz




To Find:


 The power dissipated  = P = ?


Solution:

We know that Power consumed  P is calculated by formula

 P = `\I_{rms}``\V_{rms}` cos θ -------(1)

But the value of `\I_{rms}` and angle θ  is unknown So, to find these values we have


Z = `\frac {V_{rms}}{I_{rms}}` 

`\I_{rms}` = `\frac {V_{rms}}{Z}`

where Z = `\sqrt {R^2 + X_L^2}` 

`\I_{rms}` = `\frac {V_{rms}}{sqrt {R^2 + X_L^2}}`

where `\X_L` = ယ L = 2ㄫf L 

`\I_{rms}` = `\frac {V_{rms}}{sqrt {R^2 + (2ㄫf L)^2}}`

by putting values


`\I_{rms}` = `\frac {240 V}{sqrt {(20 Ω)^2 + (2 ｘ ㄫ ｘ 0.01 hz ｘ 180/ㄫ)^2}}`

or

`\I_{rms}` = `\frac {240 V}{sqrt {400 Ω^2 + (3.6 Hz  H)^2}`

`\I_{rms}` = `\frac {240 V}{sqrt {400 Ω^2 + 12.96 Ω^2}`

`\I_{rms}` = `\frac {240 V}{sqrt { 412.96 Ω^2}`

`\I_{rms}` = `\frac {240 V}{20.321 Ω}`

`\I_{rms}` = 11.810 A 


The phase difference θ can be calculated as

θ = tan⁻¹ (`\frac {ധL}{R}`)

or

θ = tan⁻¹ (`\frac {2ㄫf L}{R}`)

By putting values

θ = tan⁻¹ (`\frac {2ㄫｘ0.01 Hz ｘ 180/ㄫ H}{20 Ω}`)

θ = tan⁻¹ (`\frac {3.6 Ω}{20 Ω}`)

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. 





************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.