(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

Efficiency = o/p divided by i/p

I/P KW divided by input voltage gives input current

Field Current = voltage divided by field resistance

armature current = total current - field current

Back Emf = V - IaRa

T proportional to Ia

Back Emf proportional to flux and N

3. The attempt at a solution

For 1st part:

we have efficiency = 0.75 = o/p divided by input

o/p = 18KW

So we get i/p = 24KW

So Vt * (Total current) = 24kW

Vt = 200V

So Total current = 24000/200 = 120A

Field current = voltage / Rf = 200/100 = 2A --- (constant all the time)

Now to limit starting current to three times we have,

Total current must be 120 * 3 = 360A

Field current = 2A

Armature current = 358A

At start Eb = 0

Terminal Voltage = Armature Current times Ra

So 200 = 358 times Ra

This gives Ra = 0.558

Given Ra = 0.3

So we need to add extra 0.26 ohm

So for first part answer is B

But I am not able to solve second part.

At full load Torque is same irrespective of starting resistance since torque depends on load.

In shunt motor Torque proportional to Ia

So Ia is same in both cases

Back Emf is proportional to Speed times flux

Flux is same in both cases of armature resistance as Field Current is same.

So Eb1/Eb2 = 1500/(Nx)

Eb1 = 200 - 118 * 0.3 = 164.6

Eb2 = 200 - 188 * 0.56 = 133.92

So Eb1 / Eb2 = 1.23

So 1.23 = 1500 / Nx

Nx = 1220 rpm

But that is not even in the options.

Where am I wrong, or is the question wrong?

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# Homework Help: DC Shunt Motor Starting Current

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