A 120.0 volt motor draws a current of 7.60 A when running at normal speed. The resistance of the armature wire is 0.610.

(a) Determine the back emf generated by the motor.

115.364 V

(b) What is the current at the instant when the motor is just turned on and has not begun to rotate?

196.72A

(c) What series resistance must be added to limit the starting current to 15.0 A?

(a) and (b) answers are both correct. I am having a problem figuring out (c). Please help!

If you want to limit starting current to 15 amp, then

15= 120/(.610+R)

To determine the series resistance required to limit the starting current to 15.0 A, we need to use Ohm's Law, which states that the voltage across a resistor is equal to the current flowing through it multiplied by its resistance.

In this case, we know the desired current (15.0 A) and the voltage across the resistance (120.0 V - the supply voltage). Our goal is to find the resistance value.

Using Ohm's Law, we can rearrange the formula to solve for resistance:

Resistance = Voltage / Current

Resistance = (120.0 V - back emf) / 15.0 A

Substituting the value for back emf (115.364 V) we found from part (a):

Resistance = (120.0 V - 115.364 V) / 15.0 A

Resistance = 4.636 V / 15.0 A

Resistance = 0.309 Ω

Therefore, to limit the starting current to 15.0 A, a series resistance of approximately 0.309 Ω must be added.