The motor in normal operation carries a direct current of 0.750 a when connected to a 120 v power supply. the resistance of the motor windings is 19.6 ω. your supervisor asks you to determine by what factor the rate of change of internal energy in the windings will increase if the rotor seizes while it is operating and the supply voltage is not cut off.

Factor by which the intrnal energy changes is 66.63

Explanation:

Given:

When the is motor operated normally, the current through the motor,I₁ =0.750A

Motor winding resistance, R = 19.6Ω

Thus, the Power dissipated at Normal condition,(P₁):

P₁=I₁²×R

P₁=(0.750)²×19.6= 11.025 W

Also,

Power = Rate of change of energy =

Now,

when the motor is connected to 120 V power supply, the current through the  motor (I₂) =

or

I₂ =

or

I₂ =6.122 A

Thus, the power dissipated due the current I₂ will be

P₂ = I₂² × R

⇒ P₂ = (6.122)² × 19.6 = 734.693 W

Hence, the factor by which the internal energy rate in resistor winding increases is = 66.63

u can delete ur account.

1. go on google and write

2 open this and click on your profile pic and if your profile is not appearing, then u have to log in.

3. when u click on your profile pic, there appears four options on the top.

4. click on the 1st option and scroll down and go at the bottom.

5. click on edit profile.

6. there appears a option 'prefernces' and at that option at the right side , 'edit' is written.

7. click on the edit option.

8. there is a option " i want to delete my account" and click on that.

hope this u

and ur welcm

: )