Balanced Three Phase Load System

In summary: Omega}{754 \Omega}}\tan{\theta} = -3R/754\Omega\tan{\theta} = -0.75{\theta} = \arctan{-0.75}{\theta} = -37In summary, the conversation discusses finding the total impedance in a phase, given the frequency, power factor, and equations for calculating impedance. The steps for solving the problem are outlined, including converting resistors to an equivalent Delta connection and using an arbitrary value for voltage. The final result is a total impedance of 754j ohm, with a current angle of -37 degrees.
  • #1
JosefMTD
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0

Homework Statement



b17d9a210e.png

f = 60 Hz
pf = 0.8 (lagging)

Homework Equations



Z(L) = 2 (pi) * f * L
Given z = x + iy
tan(angle) = y/x
pf = cos(angle)

The Attempt at a Solution



I changed the Wye-connected resistors to an equivalent Delta-connected resistors:

36f2d8c63b.png


The new resistors become 3R in the equivalent Delta-connected resistors

Next step is calculating Z(L)
Z(L) = j*2 (pi) * 60 Hz * 2 H = 754j ohm

1/Zphase = 1/Z(L) + 1/Z(R)
1/Zphase = 1/754j ohm + 1/3R ohm = -j/754 ohm + 1/3R ohm
1/Zphase = (754 ohm - 3Rj)/754 ohm*3R ohm

Since V is not given value in this problem, I choose to use an arbitrary value of x with angle of 0 degree, thus making it a constant.

Iphase = Vphase/Zphase = x*(754 ohm -3Rj)/754 ohm*3R ohm

It's known that power factor is 0.8

cos(angle) = 0.8
angle = arc cos(0.8) = +/- 37

Since known that the current is lagging, the angle is lower than 0 degree, thus making the appropriate angle is -37 degree.

Since I know the angle of phase I, I put it in the formula to find the angle of a complex number:
tan(angle) = Im(complex variable)/Re(complex variable)
tan(-37 degree) = [(x/754 ohm*3R)(-3Rj)]/[(x/754 ohm*3R)(754)] = -3R / 754
- 0.75 = -3R / 754
R = 754/4 Ohm = 188.5 ohm

Am I doing it right?
Thank you in advance for helping me to solve this!
 

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  • #2
Sorry for not using the latex:

Calculating total impedance in a phase
[tex]{\frac{1}{Z_p}} = {\frac{1}{Z_L}} + {\frac{1}{Z_R}}[/tex]
[tex]{\frac{1}{Z_p}} = {\frac{1}{754 j\Omega}} + {\frac{1}{3R}}[/tex]
[tex]{\frac{1}{Z_p}} = {\frac{-j}{754 \Omega}} + {\frac{1}{3R}}[/tex]
[tex]{\frac{1}{Z_p}} = {\frac{754\Omega - 3Rj}{754 j\Omega * 3R}}[/tex]

Making the i = V/Z
[tex]I = V*{\frac{754\Omega - 3R}{754 j\Omega * 3R}}[/tex]
[tex]I = {\frac{V}{754 j\Omega * 3R}}*(754\Omega - 3Rj)[/tex]

Finding the angle of I:
[tex]\tan{\theta} = ({\frac{V}{754 j\Omega * 3R}}*-3R)/({\frac{V}{754 j\Omega * 3R}}*754\Omega)[/tex]
[tex]\tan{\theta} = {\frac{-3R}{754 \Omega}}[/tex]
 
Last edited:

1. What is a balanced three phase load system?

A balanced three phase load system is a type of electrical power distribution system that consists of three phases, each carrying equal currents and voltage, resulting in a balanced load across all three phases. This is commonly used in industrial and commercial settings.

2. How does a balanced three phase load system work?

In a balanced three phase load system, the power is distributed evenly across all three phases, with each phase carrying one-third of the total load. This is achieved by using three separate conductors, each connected to a different phase, to distribute the load.

3. What are the advantages of a balanced three phase load system?

The main advantage of a balanced three phase load system is its efficiency. By distributing the load evenly across all three phases, it reduces the amount of stress on each individual phase, resulting in a more stable and reliable power supply. It also allows for the use of smaller and more cost-effective equipment.

4. How is a balanced three phase load system different from a single phase load system?

A single phase load system only has one phase, carrying the entire load. This can result in unbalanced loads and can lead to power quality issues. In contrast, a balanced three phase load system distributes the load evenly across all three phases, resulting in a more stable and efficient power supply.

5. Is a balanced three phase load system necessary for all types of electrical systems?

No, a balanced three phase load system is not necessary for all types of electrical systems. It is most commonly used in industrial and commercial settings that require large amounts of power. Residential buildings typically use a single phase load system, as the power demand is lower and does not require the use of three phases.

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