Question

Q.1 A 500 V. DC shunt motor runs at its normal speed of 250 RPM when the armature current is
200 ampere. The resistance of armature is 0.12 2. Calculate the speed when a resistance is
inserted in the field circuit, reducing the shunt field to 80% of normal value and the armature
current is 100 ampere​

Answer 1

Answer:   Speed when a resistance is  inserted in the field circuit, reducing the shunt field to 80% of normal value and the armature  current is 100 ampere​ is 320 RPM.

Explanation:

Since, it is given in the question that it is a DC shunt motor,

the formula for DC shunt motor is,

$E_{b}$ = $\frac{P.Z.N}{60}$ X Ф , where

$E_{b}$ = back EMF

Ф = flux

P = Number of poles

Z = Number of coils

N = rotational speed

∴ For the same motor, we can rewrite above as,

$E_{b}$  α Ф.N ( Directly proportional )

In question, we have been given two separate conditions,

$E_{b1}$ = 500 -200 X 0.12 = 476 V

$E_{b2}$ = 500 - 100 X 0.12 = 488 V

It is given that when resistance is inserted it reduces the shunt field by 80%, Shunt field is nothing but flux.

∴ Ф2 = 0.8 Ф1

And,

$N_1$ = 250 RPM, $N_2$ = ? (Needs to be found)

∴ $\frac{N_2}{N_1}$ = $\frac{E_{b2} }{E_{b1}}$ X (Ф1 / Ф2)

∴ $N_2$ = $\frac{488}{476}$ X 250 X (Ф1 / 0.8Ф1)

∴ $N_2$ = 320 RPM.

Hence, Speed when a resistance is  inserted in the field circuit, reducing the shunt field to 80% of normal value and the armature  current is 100 ampere​ is 320 RPM.

Answer 2

Given :

Normal speed of shunt motor = 250rpm

Armature current (I₁) = 200A

Resistance of Armature = 0.12 ohm

Armature current (I₂) = 100A

To Find :

The speed of shunt motor when shunt field is reduced to 80%

Solution :

• We know the relation E = V- IR

• E₁ = 500 - 200(0.12)  = 476V

         E₂ = 500 - 100(0.12)  = 488V

• The flux also decreases as electric field is reduced to 80%

           Φ₂ = 0.8Φ₁

• N₂/N₁  = (E₂/E₁)×(Φ₁/Φ₂)

          N₂/250  = (488/476)×(Φ₁/0.8Φ₁)

          N₂ = 320rpm

The speed of shunt motor is 320rpm