A120.0-v motor draws a current of 7.00 a when running at normal speed. the resistance of the armature wire is 0.720 ω. (a) determine the back emf generated by the motor. (b) what is the current at the instant when the motor is just turned on and has not begun to rotate? (c) what series resistance must be added to limit the starting current to 15.0 a?

back emf = 114.96 V

current = 166.7 A

series resistance R is 7.28Ω

Explanation:

given data

voltage = 120 V

current = 7 A

resistance = 0.720 Ω

to find out

back emf , current at the instant and  What series resistance

solution

we apply formula to calculate back emf  that is

back emf = V - IR

here V = 120 and I = 7 and R = 0.720

put all these value

back emf = 120 - (7) ( 0.720)

back emf = 114.96 V

and

we know emf = 0 when motor is on and no rotation

so

back emf = V - IR

0 = V -IR

so I = V/R

put the value

I = 120 / 0.720

current = 166.7 A

and

series resistance R is

R = 120 / 15 - 0.720

series resistance R  7.28Ω

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

d. air pressure decreases as altitude increases.