A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.

Resistance of the armature coil = 6.61 ohms

Back emf developed at normal speed = 98.62 V (Approx.)

Current drawn by the motor at one-third normal speed = 12.73 A

Explanation:

Given:

Potential difference V = 117 V

Current = 17.7 A

Motor drawn current = 2.78 A

Find:

Resistance of the armature coil

Back emf developed at normal speed

Current drawn by the motor at one-third normal speed

Computation:

A] Resistance of the armature coil R = V/ I

Resistance of the armature coil = 117 / 17.7

Resistance of the armature coil = 6.61 ohms

B] Back emf developed at normal speed  = V- IR

Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)

Back emf developed at normal speed = 117 V - 18.37

Back emf developed at normal speed = 98.62 V (Approx.)

C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)

Current drawn by the motor at one-third normal speed = 17.7 - 4.97

Current drawn by the motor at one-third normal speed = 12.73 A

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.