Question

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.

Answer 1

Answer:

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 = $\frac{dE}{dt}$

Now,

when the motor is connected to 120 V power supply, the current through the  motor (I₂) =$\frac{V}{R}$

or

I₂ =$\frac{120}{19.6}A$

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 = $\frac{P_2}{P_1}=\frac{734.693}{11.025}=$66.63