Question

The power input to a 230V dc
shunt motor is 8.477kW. The
field resistance is 230 Q and
armature resistance is 0.28 0.
The back emf of the motor is....

210.25 V
219.961
230V
211.5V​

Answer 1

Answer:

The power input to a 230V DC shunt motor is 8.477kW. The field

resistance is 230Ω and armature resistance is 0.28Ω. Find the input current,

armature current and back EMF.

Answer 2

Answer: The correct answer is (B) 219.96V.

Given:

230 V DC Shunt motor

Power= 8.477 kw,

Field resistance(Rf) =230 ohm

Armature resistance(Ra)= 0.28 ohm.

To find: the back EMF of dc motor.

Explaination: First we calculate the shunt field current,

If=  V/Rf = 230/230 = 1A.

Power =8.477kw

V.Iι = 8.477Kw (Input current = Iι)

230.ILι= 8.477/230*103=36.85

Now, we calculate armature current, Ia=Iι-If=35.85A

Input current = Iι =36.85A.

At last, we calculate Back emf by the formula,

Emf = V - Ia.Ra

=230-35.85×0.28=219.96V.

Armature resistance

The impact of the flux created by the currents in the armature winding on the main field flux is known as armature resistance. Both lap- and wave-wound machines have the same effect.

To learn more about Armature resistance, refer these links.

https://brainly.in/question/16106328

https://brainly.in/question/14481043

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