DC shunt motor

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  • Thread starter nooby
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  • #1
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Homework Statement:
A D.C. shunt motor has an efficiency of 92.21 % when operated under full-load conditions. The motor is supplied by a 120 V D.C. source and has a rating of 1.55 kW at full-load. Given that the armature resistance of the motor is 0.33 Ω, find the resistance of the field winding. (You may neglect all losses other than those due to the armature and field winding resistances.)

Note – the rated power of a motor is the output power under full-load conditions.
Relevant Equations:
efficiency and or power at full load
Hi all, i'm needing help cracking this nut. If someone can help it would be greatly appreciated.

I am trying to find the current in the field or armature, but cant connect the dots.
 

Answers and Replies

  • #2
cnh1995
Homework Helper
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As per forum rules, you need to show your attempt at a solution first.

Start with a circuit diagram.
Label the known voltages and currents. Think of the relavant equations that you can use to find the unknowns.
 
  • #3
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i got the supply current, but cant seem to understand the current and resistor relationship.
 

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  • #4
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the reason why i think i need the current is so i can use KVL. This will assist me in finding current in armature which i can then use KCL to find the current in the field. I got a feeling i'm chasing something that isnt there :(
 
  • #5
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Find the excitation losses. Compare with excitation voltage drop.
 
  • #6
471
126
DC Shunt motor that means the field supply source is parallel with armature?
 
  • #8
cnh1995
Homework Helper
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i got the supply current
The power you used in your equation is the 'output' power. To calculate the source current, you must know the 'input' power.
Output power + losses gives you the input power.
So identify where the losses are taking place and recall/look up their corresponding formulae.
Further, have you studied the concept of back-emf?
 
  • #9
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I had a closer read of the question, it states "Note – the rated power of a motor is the output power under full-load conditions."
so by using the efficiency formula to work out Power input (1680.9W).Which means i can work out the current the total current. This then allows me to work out the current in the armature which. Then i simple use KCL to work out the field current :)

hope that makes sense :)

Thanks you all for your assistance
 
  • #10
471
126
You may use the equation Is=Iarm+If and RA.Iarm^2+Rf.If^2=losses

Rf.If^2=Vs.If

and you'll get an equation like: a.Iarm^2+b.Iarm+c=0
 

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