# A balanced wye delta connection

• mingming
In summary, the given problem involves a balanced delta load with 18ohm resistance and 24 capacitive reactance in each phase, supplied through lines with 1 ohm resistance and 2 ohm inductive reactance. By converting the delta load to a wye load and calculating the equivalent impedance, the phase current and line current can be determined to be 4.04 amps at an angle of -40.60 degrees and 7.0 amps at an angle of -40.60 degrees, respectively.
mingming

## Homework Statement

A balanced delta load having 18ohm resistance and 24 capacitive reactance in series in each phase is supplied through lines each having 1 ohm resistance and 2 ohm inductive reactance. If the line to line voltage at the sending end is 250 V, find the phase current and line currents.

Zy = 1/3 Z delta

## The Attempt at a Solution

i first convert the delta load to wye so the load now becomes Zy= (6-j8) and then i add the line impedance and load impedance...Zeq= (6-j8)+ (1+j2) = 7-j6 and then i solved for the line currents and phase currents which is equal for wye Ip=Il= 144.34/ 9.22 angle of -40.60
it's answer is 15.66 angle of 40.60 degrees...THIS IS WHERE I AM CONFUSED HOW DO I SOLVE THE PHASE CURRENTS AND LINE CURRENTS WHEN THE LOAD IS IN DELTA...it is not supposed to be equal since the load is in delta...please help ...

Hi there,

Thank you for sharing your attempt at solving this problem. It's great to see you working through the steps and trying to understand the concept.

Firstly, let's start by clarifying the problem statement. The given values of 18ohm resistance and 24 capacitive reactance are for each phase in a balanced delta load. This means that each phase has a combination of resistance and reactance, and the total load is balanced across the three phases.

Now, when converting the delta load to a wye load, you correctly calculated the equivalent impedance as (7-j6). However, this is the total equivalent impedance seen by each line in the wye configuration.

To find the individual phase currents, we need to divide this equivalent impedance by sqrt(3). This is because in a balanced delta load, the line current is sqrt(3) times the phase current.

So, the phase current (Ip) and line current (Il) can be calculated as:
Ip = (7-j6)/ sqrt(3) = 4.04 - j3.5
Il = sqrt(3) * Ip = 7.0 - j6.06

Therefore, the phase current is 4.04 amps at an angle of -40.60 degrees, and the line current is 7.0 amps at an angle of -40.60 degrees.

I hope this helps clarify the solution for you. Keep up the good work!

## 1. What is a balanced wye delta connection?

A balanced wye delta connection is a type of electrical connection used in three-phase power systems. It consists of three single-phase loads connected in a triangular or delta configuration, and three single-phase loads connected in a Y or wye configuration. This type of connection is used to supply power to large motors, generators, and other high-power equipment.

## 2. How does a balanced wye delta connection work?

In a balanced wye delta connection, the three phases of the power system are connected to the three single-phase loads in a triangular or delta configuration. This creates a closed loop, with each phase connected to the next in sequence. The other ends of the single-phase loads are then connected to a common point, forming a Y or wye configuration. This allows for balanced distribution of power across the three phases, with each phase carrying the same amount of current.

## 3. What are the advantages of a balanced wye delta connection?

One of the main advantages of a balanced wye delta connection is that it provides a way to supply power to large motors and other high-power equipment without the need for a neutral wire. This reduces the complexity and cost of the electrical system. Additionally, this type of connection allows for balanced distribution of power, which can help prevent overloading of one phase and improve the overall efficiency of the system.

## 4. What are the potential issues with a balanced wye delta connection?

One potential issue with a balanced wye delta connection is that it can be difficult to troubleshoot and repair if one of the phases or loads fails. This is because the loads are connected in a closed loop, making it difficult to isolate the specific problem. Additionally, if the loads are not balanced, it can lead to unequal distribution of power and potential overloading of one phase.

## 5. How is a balanced wye delta connection different from an unbalanced connection?

A balanced wye delta connection has equal currents and voltages across all three phases, whereas an unbalanced connection has unequal currents and voltages. In an unbalanced connection, one or more of the single-phase loads may be larger or smaller than the others, leading to unequal distribution of power. This can result in overloading of one phase and potential damage to the equipment. Balanced connections are typically preferred for high-power applications to prevent these issues.

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