Question

A motor is designed to operate on 117 V and draws a current of 12.3 A when it first starts up. At its normal operating speed, the motor draws a current of 3.38 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

Answer 1

Answer:

a) using

R=V/I =117/12.3

R=9.5 ohms

b)

E=V-I*R =117-3.38*9.5

E=84.8Volts

c)

at (1/3)rd of normal speed ,back emf is (1/3) of its maximum

value

E=(1/3)*84.8=28.3Volts

Current drawn

I=V-Eback/R =117-28.3/9.5

I=9.33A

Explanation:

Answer 2

The resistance is = 9.5 ohms

The back emf developed at normal speed is = 84.90 volts

The current drawn at one-third normal speed =9.33 A.

To calculate the resistance of the armature coil this formula is used;

V = IR

make R the subject of formula,

R = V/I

where R = resistance

V = voltage

I = Current

R = 117/12.3

R = 9.5 ohms

To calculate the back emf developed at normal speed, this formula is used;

E = V + Ir ( for normal emf)

But for back emf which is the difference between the supplied voltage and the loss from the current through the resistance, this formula is used;

E = V - Ir

where V = 117v

I = 3.38

r = 9.5

E = 117 - ( 3.38 × 9.5)

= 117 - 32.11

= 84.90 volts

To calculate the current drawn at one-third normal speed;

1/3 of Emf = 1/3 × 84.90

= 28.3volts

Therefore current (I) = V - E/ R

= 117- 28.3/9.5

= 88.7/9.5

= 9.33 A

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