# How to Analyze a Balanced Three Phase Star Load Diagram?

• DunceKirchhoff
In summary, the conversation discusses the correct angles to use for voltage calculations, with the main focus on using VAB as the reference phasor. The question asks for a defined angle of zero degrees for VAB, leading to the other voltages being shifted by -30 degrees. The conversation also mentions the importance of using the correct angles for accurate calculations.
Anyone??

It's a bit of a pain having to follow links to question bodies. Consider loading your pics to PF rather than some other image hosting site, and including them in the post body (via the "Go Advanced" editing frame where icons on the frame give you the ability).

The only issue I'd have with your results as shown is that the question asks you to use VAB as the reference phasor for the calculations, which should, I think, yield a "defined" angle for VAB of zero degrees rather than 30°. The other voltages would then similarly be shifted by -30° for part (a), and would carry through to the voltages calculated for part (b) and currents for part (c).

Hi Neil

Sorry will do that in the future.

Yeah usually the ref is 0 degrees agreed.
But in this question the line voltage is not given but the phase voltage Vao.
So given that information I have tried to work it out.
Also the line voltage leads the phase voltage by 30 degrees.

Could be wrong

I am wrong sorry, yes it has asked me to use vab as a reference.
So vab=0 degrees
Vbc -120 degrees
Vca 120 degrees
Whats the phase voltage?

DunceKirchhoff said:
I am wrong sorry, yes it has asked me to use vab as a reference.
So vab=0 degrees
Vbc -120 degrees
Vca 120 degrees
Whats the phase voltage?

For VAO', VBO' and VCO' you can just shift their normal values by the same -30°. Confirm by re-calculating VAB etc., from them to see that the resulting angles match what you got for part (a).

So basically I have put vab, vbc and vca where vao, vbo and vco should be? And the phase angle is -30 instead of 30?

DunceKirchhoff said:
So basically I have put vab, vbc and vca where vao, vbo and vco should be? And the phase angle is -30 instead of 30?

##V_{AO'} = 100\ ∠\ 0°##
##V_{BO'} = 100\ ∠ -120°##
##V_{CO'} = 100\ ∠ \ 120°##

which I believe is a "standard" set of angles for the lines of a star configuration. Now you want to have the voltages VAB, etc., come out having the desired set of angles where VAB has an angle of zero degrees. Those angles were determined in part (a).

Since all the various relative angles are fixed on a phasor diagram, if you had to shift VAB by -30° to make it the zero degree reference then ALL the angles will shift by the same amount. It's equivalent to rotating the whole phasor diagram by 30°. So if you were to subtract 30° from each of the above you should find that if you recalculate VAB etc. from VAO' VBO' and VCO', the angles will turn out to match the ones from part (a).

Ok thanks I got you. Well explained.

## 1. What is a balanced three phase star load?

A balanced three phase star load is a type of electrical load that is connected in a star or Y configuration, where three equal impedances are connected at each phase. It is used to distribute power evenly and efficiently in three-phase electrical systems.

## 2. How does a balanced three phase star load differ from other types of loads?

A balanced three phase star load differs from other types of loads, such as delta loads, in that it allows for a neutral connection in the system and the current is evenly divided among the three phases. This results in a more stable and efficient distribution of power.

## 3. What are the advantages of using a balanced three phase star load?

The advantages of using a balanced three phase star load include a more balanced distribution of power, reduced current in each phase, and the ability to use a neutral connection. This can result in a more reliable and efficient electrical system.

## 4. How is a balanced three phase star load calculated?

To calculate a balanced three phase star load, you can use the formula: P = √3 V I cos θ, where P is the total power, V is the line voltage, I is the line current, and cos θ is the power factor. This formula takes into account the three phases and the power factor to determine the total power of the load.

## 5. Can a balanced three phase star load be converted to a delta load?

Yes, a balanced three phase star load can be converted to a delta load by rearranging the connections of the impedances. However, this may require additional equipment and careful consideration of the load requirements, as the voltage and current will be affected by the conversion.

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