# Homework Help: DC Shunt Motor Starting Current

1. Dec 2, 2017

### jaus tail

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?

2. Dec 4, 2017

### cnh1995

I beileve the starting current is the armature current at the start and not the total current.
Here, the starting current would be 120-2=118A and you need to limit it at 118*3=354A and not 358A. But the answer is still B).

I agree with your calculations for the second part. I believe all the options are incorrect.