Help Needed: Cracking the Nut of Current in Field/Armature

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  • Thread starter nooby
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In summary, the power in the armature is equal to the power in the field + losses. To calculate the source current, you must know the input power.
  • #1
nooby
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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 can't connect the dots.
 
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  • #2
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
i got the supply current, but can't seem to understand the current and resistor relationship.
 

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  • #4
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 isn't there :(
 
  • #5
Find the excitation losses. Compare with excitation voltage drop.
 
  • #6
DC Shunt motor that means the field supply source is parallel with armature?
 
  • #7
Pel=Pmec+losses
 
  • #8
nooby said:
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
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
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
 

1. What is the purpose of cracking the nut of current in field/armature?

The purpose of cracking the nut of current in field/armature is to understand and control the flow of electricity in a circuit. This is important in many applications, such as electric motors, generators, and power systems, as it allows for efficient and safe operation.

2. Why is it difficult to crack the nut of current in field/armature?

Cracking the nut of current in field/armature can be difficult because it involves complex electrical concepts and calculations. The behavior of current in a circuit is affected by various factors, such as resistance, capacitance, and inductance, making it challenging to accurately predict and control.

3. What are some common techniques used to crack the nut of current in field/armature?

Some common techniques used to crack the nut of current in field/armature include Kirchhoff's laws, Ohm's law, and the use of circuit analysis tools such as nodal analysis and mesh analysis. These techniques allow for the calculation of current in different parts of a circuit and help in understanding its behavior.

4. How does cracking the nut of current in field/armature benefit society?

Cracking the nut of current in field/armature has numerous benefits for society. It allows for the development of more efficient and reliable electrical systems, which can lead to cost savings and improved performance. It also helps in the design and implementation of new technologies that rely on electricity, such as renewable energy sources and electric vehicles.

5. What are some potential challenges in cracking the nut of current in field/armature?

Some potential challenges in cracking the nut of current in field/armature include dealing with non-linear and time-varying circuits, as well as unexpected factors that can affect current flow, such as temperature and electromagnetic interference. Additionally, the complexity of modern electrical systems and the need for precise measurements and calculations can also pose challenges.

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