Solved Numerical Problem No. 15.14 ( Unit-15: Electromagnetic Induction); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

The back emf in a motor is 120 V when the motor is turning at 1680 rev/min. What is the back emf when the motor turns at 3360 rev/min? (Ans: 240 V). 

Data Given:

1st Back emf in motor at angular speed "`\ധ_1`"  = `\Ɛ_1` = 120 V

1st Angular speed = `\ധ_1` = 1680 rev per min 

= `\frac {1680 ｘ2π rad}{60s}` =  `\frac {1680 ｘ2 ｘ3.1416 rad}{60s}` = 175.93 rad s⁻¹

2nd Angular speed = `\ധ_2` = 3360 rev per min 

= `\frac {3360 ｘ2π rad}{60s}` =  `\frac {3360 ｘ2 ｘ3.1416 rad}{60s}` = 351.86 rad s⁻¹

To Find:

Back emf in the motor at angular speed "`\ധ_2`" = `\Ɛ_2` = ?

Solution:

As the angular speed and back emf are proportional and their ratio will be equal, thus

`\frac {Ɛ_1}{Ɛ_2}`  = `\frac {ധ_1}{ധ_2}`

or

`\Ɛ_2` = `\frac {Ɛ_1ധ_2}{ധ_1}`

Putting the corresponding values 

`\Ɛ_2`  = `\frac {120 V ｘ351.86 rad s⁻¹}{175.93 rad s⁻¹}`

`\Ɛ_2`  = `\frac {42223.2V}{175.93}`

`\Ɛ_2`  = 240 V ----------------Ans.

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