# Circuits II Polyphase System Help

• Engineering
• md5fungi
In summary, the conversation discusses how to determine the value of x in order to achieve a power factor of unity at the source. The circuit is represented as a 240angle0 source in series with the load and an equivalent capacitor (-jx/3) in parallel. It is important to consider the balanced 3 phase system in order to solve for the current going into the loads. The addition of a capacitor in parallel can help achieve a unity power factor, with the impedance of the capacitor being (240+j0)/(+j10). The capacitance can then be calculated using the impedance, and the answer can be checked by calculating the impedance of the capacitor and load.
md5fungi

## Homework Statement

"Determine x for which the power factor at the source is in unity."

The data given is in the diagram; The impedance of the capacitors is the only thing we don't know. The load and sources are balanced.

## Homework Equations

Just the basics: Kirchoff's Current Law, possibly some Loop/Mesh equations?, basic Polyphase system properties.

## The Attempt at a Solution

Basically I was taught that if I needed to find the impedance of the capacitor in this particular setup, I could represent the circuit as a 240angle0 source in series with the load and equivalent capacitor (-jx/3) in parallel with each other. The impedances resistor and additional load in this example can be combined to make it easier. But the only way I can solve for x as far as I understand is form an equation involving the current through the combined load and current through the capacitor, so that they add up to the current through the source. However, since we aren't given the current through the source, I am very confused.

Any ideas? (This isn't homework really, just a practice problem from an old test that is supposedly going to help me with an exam I have this Friday)

Last edited:
First find the current going into the loads. Luckily you have a balanced 3 phase system otherwise it would be difficuilt to solve it. Consider each phase individually.

So if the current is 5-j10 (I am just using some number here calc it by V/Z) and it should be a lagging current due to an inductive load, you need to put a capacitor in parallel to get a unity pf.

So previously the source provided a current had a reactive component of -j10. With the addition of the cap, you want the cap to absorb a current of the same magnitude but opposite phase to cancel it out and get unity pf. So the cap current must be +j10. Hence the impedance of the cap must be (240+j0)/(+j10). Once you have the impedance use that to find the capacitance 1/(2*pi*f*C)= |Impedance|

To check your answer calculate the impedance of the cap and the load (75 angle 25 || 10). There should be no reactive component (or very little due to rounding off errors)

C= 14.94uF from my calcs.

Thanks; I actually finally realized how the fact the power factor was in unity could help me this morning and managed to solve it. Thanks for your help and time, though!

md5fungi said:
Thanks; I actually finally realized how the fact the power factor was in unity could help me this morning and managed to solve it. Thanks for your help and time, though!

Did you get the same answer as I did?

## 1. What is a polyphase system?

A polyphase system is an electrical power distribution system that uses multiple phases to transmit power. It is commonly used in industrial and commercial applications to deliver higher amounts of power over longer distances.

## 2. How does a polyphase system differ from a single-phase system?

In a single-phase system, only one voltage waveform is used to transmit power. In a polyphase system, multiple voltage waveforms with different phases are used, resulting in a more efficient and powerful distribution of electrical energy.

## 3. What are the advantages of using a polyphase system?

One of the main advantages of a polyphase system is its ability to transmit higher amounts of power over longer distances without significant voltage drop. It also allows for more efficient use of power and better control of power flow.

## 4. What are some common applications of polyphase systems?

Polyphase systems are commonly used in industrial and commercial settings, such as factories, power plants, and large buildings. They are also used in electrical grids to transmit power from power plants to homes and businesses.

## 5. How do you analyze a polyphase system?

To analyze a polyphase system, you can use techniques such as phasor diagrams, power triangle, and Kirchhoff's laws. These tools can help you understand the behavior and performance of the system, and make calculations for power, voltage, and current.

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