Question

.A 460-V, series motor runs at 500 r.p.m. taking a current of 40 A. Calculate the speed if the load is reduced so that the motor is taking 30 A. Total resistance of armature and field circuit is 0.8 Ω. Assume flux proportional to the field current.​

Answer 1

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Answer 2

The speed after reduction in load is 679 r.p.m.

Given,

A 460V motor is running at 500 r.p.m.,

Taking current, $I_{1}=40A$

Current after reducing the load, $I_{2}=30A$

To find,

Speed after the load is reduced.

Solution,

As flux is proportional to field current.

$\frac{T_{2}}{T_{1}} =\frac{I_{2}^{2}}{I_{1}^{2}}$

⇒$\frac{T_{2}}{T_{1}}=\frac{30^{2}}{40^{2}}$

⇒$\frac{T_{2}}{T_{1}}=\frac{900}{1600}$

⇒$\frac{T_{2}}{T_{1}}=\frac{9}{16}$

And also,

$E_{b}=460V$

Now,

$E_{b1}=460(40(0.8))V$

⇒$E_{b1}=428V$

Similarly,

$E_{b2}=460(30(0.8))V$

⇒$E_{b2}=436V$

We know that,

$\frac{E_{b2}}{E_{b1}}= \frac{N_{2}}{N_{1}}= \frac{I_{1}}{I_{2}}$

⇒$\frac{N_{2}}{500}=\frac{436}{428}=\frac{40}{30}$

⇒$N_{2}=679r.p.m.$

Therefore, the speed after the load is reduced will be 679r.p.m.