A Delta Wye Transformation according to Sadaghar M. and McAllister W. (2018) is a very useful solution to simplify complex resistor networks, with proper resistor values for each configuration, they are proven to be behaving the same. It is an extra technique for transforming certain resistor combinations that cannot be handled by the series and parallel equations. This is also referred to as a Pi – T transformation. It will make us be able to convert impedances connected together in a 3-phase configuration from one type of connection to another.

Lastly, the origin of the name is based from the shapes of the circuit diagrams, which is the Y and the Greek capital letter ?.

Table 1 showed us the transformation that happened to Delta going to the Wye Transformation, and it will show you the determined values of Resistors Currents and voltages that happened after doing such transformation. First step that we did is we connect the power supply, measured the voltage readings together with currents and after all knowing the values, we recorded it on the Delta Measured that you can see below.

For determining the values in Wye Measured, we recorded the resistances using a potentiometer and applied the formulas in determining the RA, RB and RC. After all the computations, we can see that in the Wye measured it is somehow near to the calculated values and that gave us a small percentage of error.

### Delta R1 R2 R3 RL I1 I2 VA VB VC

Measured 148 1977 327 980 16mA 3.3mA 1.87 4.97 3.09

### Wye RA RB RC RL I1 I2 VA VB¬ VB

Measured 120 270 20 983 15.7mA 3.24mA 1.79 4.95 3.07

Calculated 119.31 263.93 19.76 978 15.24mA 3.18mA 1.88 5 3.12

### TABLE 1. DELTA WYE TRANSFORMATION

### Wye RA RB RC RL I1 I2 VA VB VC

Measured 148.7 1977 327.3 979 14mA 9.5mA 4.6 15 10.27

Delta R1 R2 R3 RL I1 I2 VA VB¬ VB

Measured 492 3017 6649 984 13.2mA 9.8mA 4.1 13.9 10.35

Calculated 500 3023 6655 873 13.7mA 9.2mA 4.3 14.6 10.42

### TABLE 2. WYE DELTA TRANSFORMATION

For the Table 2, it showed us the changed in values after it undergoes a Wye to Delta transformation. We can see below that it determined the values of currents, resistors and voltages after the said transformation. First step that we did is we connect the power supply and ensure that the output voltage is set to 15V dc, next is we list down the Voltage and Current values that weve got using VOM to the Wye measure to Table 2. In getting the Delta Equivalent, we used a potentiometer in getting the resistances and listed it down to the Table 2, after getting all the values we used the formulas in finding R1, R2 and R3. From the table below we can see that Delta Measured is somehow near to the values of Calculated which means that weve attained a small amount of percentage error.

## CONCLUSION

To identify the delta connection of resistances and the wye connection in complicated network circuits.

One of the aims given to us is to know the delta connection of the resistances and the wye connection in a certain network circuit. These delta-wye transformations enable us to substitute the specified resistor in ? and y configuration. With this experiment we get to identify a certain connection and also to know what the conditions for it are to be a delta connection or a wye connection, Delta (?) connection of resistances refers to the three independent single-phase units whiles a neutral wire is considered to be a wye connection. and lastly, with this experiment we learned and demonstrated the transformation principles involved in converting the delta connection of resistors to the wye connection or vice versa.

To demonstrate and verify the corresponding responses between delta connected resistors and its equivalent wye connected resistors.

Next objective that was given to us is to illustrate the corresponding responses between delta and wye connection and Experiment 4 talks with delta-wye transformation which is a known technique use in simplifying resistors combinations that cant be determined with parallel series equations. With this experiment we noticed that a delta has three nodes while a wye has four nodes which is in the center. The circuits can be redrawn to square and even triangular. In getting the Rwye the product of adjacent resistances in delta divided by sum of resistances in delta. While in getting Rdelta, the sum of products adjacent resistances in wye divided by the opposite resistance in wye. The transformation formula is based on the concept that the two connections are equal to the resistances seen across the pair of terminals.

To learn and demonstrate the transformation principles involved in converting the delta connection of resistors to the wye connection or vice versa.

Experiment 4 demonstrated the transformation involved in converting the connection of resistors and with this experiment we get to know the different equations in delta to wye or even wye-delta transformations. Given three resistors R1,R2 and R3 we will be representing the solved values with RA, RB ¬and Rc which will be used for the actual equations. After these transformations we now have the values for series and parallel resistors, simplify the values until it gets down to single resistors and lastly redraw the diagram to a simpler and more familiar diagram. Lastly, after observing the output we conclude that that RAB (delta) is equal to RAB (wye), RBC (delta) is equal to RBC (wye), and RAC (delta) is equal to RAC (wye).

### QUESTIONS AND PROBLEMS:

When is the delta connection of resistors equivalent to the wye connected resistors?

Wye Delta Transformation is an additional technique for transforming certain resistors combination that can be solved or determined by series or even parallel equation or formulas. We can rely to these formulas in converting values from wye to delta and even delta to wye:

What are the practical applications of the technique delta wye transformation? Discuss briefly the different practical applications.

This method of transforming diagram in a simpler circuit can be very useful for example in Delta it can be very beneficial in determining the line current and phase current in given 3 phase circuit. Delta also is helpful in creating connections between 3-phases supply machines. These transformations are not just a useful application for a circuit with three resistors, it can also be useful for Mathematics, in where it plays a vital role in the theory of circular planar graphs.

Determine the total resistance across the terminals from the figure below.

## Solution:

Transform the delta (upper loop) to wye

R_1=((12k?)(16k?))/(12k?+16k?+14k?)=6k?

R_2=((12k?)(4k?))/(12k?+16k?+14k?)=1.5k?

R_3=((16k?)(4k?))/(12k?+16k?+14k?)=2k?

### Find the total resistance

R_T=6k?+(1/(2k?+3k?)+1/(1.5k?+6k?))^(-1)=9k?

### Answer:The total resistance is 9k?

Determine the total resistance ce across the terminal from the figure below

## Solution:

Transform Loop A and B (which are in delta) to wye.

### R_1=(6?(1?))/(6?+3?+1?)=0.6?

### R_2=(3?(1?))/(6?+3?+1?)=0.3?

### R_3=(6?(3?))/(6?+3?+1?)=1.8?

### R_4=(5?(1?))/(5?+1?+4?)=0.5?

### R_5=(1?(4?))/(5?+1?+4?)=0.4?

### R_6=(4?(5?))/(5?+1?+4?)=2?

### b. Find total resistance.

R_T={[0.6?+2?+(1/((0.3?+0.6?+0.4?)+1/(1.8?+0.5?+2?))^(-1) ]^(-1)+[0.6?]^(-1) }^(-1)= 2.2493 ?

Answer: The total resistance of the circuit is 2.2493 ?.

Determine the total resistance across the terminals from the figure below.

## Solution:

### Combine R75? and R25? : Ra

## Ra = 75?+25? = 100?

### Combine R15? and R35? : Rb

## Rb = 15?+35? = 50?

Transform ? to Y (upper: R100?, R20?, R30?; and lower R30?, R20?, R50?)

### R_1=((100?)(20?))/(100?+20?+80?)=10?

### R_2=((20?)(80?))/(100?+20?+80?)=8?

### R_3=((100?)(80?))/(100?+20?+80?)=40?

### R_4=((30?)(20?))/(20?+30?+50?)= 6?

### R_5=((50?)(20?))/(20?+30?+50?)=10?

### R_6=((30?)(50?))/(20?+30?+50?)=15?

### Combine the resistors in parallel

Let: Sum of resistors on the right in series be Rc

### Rc = 40? + 10 ? + 10 ? = 60 ?

Let: Sum of resistors on the right in series be Rd

### Rd = 26? + 8 ? + 6 ? = 40 ?

## Rd||Rc: Re

### R_e=((60?)(40?))/(40?+60?)=24?

## Find RT

### RT = 24 ?+1 ?+10 ?+15 ? = 50 ?

Answer: The total resistance of the given circuit is 50 ?.

Determine the I0 from the given circuit figure below.

## Solution:

### Transform the upper loop into wye.

### R_1=30(20)/(20+50+30)=6?

### R_2=50(20)/(20+50+30)=10?

### R_3=(30(50))/(20+50+30)=15?

## Find total resistance

R_T=((10?+46?)(15?+9?))/(10?+46?+15?+9?))+6?+2.2?=25?

## Find total current

## I=V/R=500V/25=20A

## Find I1 and I2

### By Current Divider Principle

### I_1=20A(24?/(56+24?))=6A

### I_2=20A(56?/(56+24?))=14A

## Find io

Using Kirchhoffs Voltage Law (Lower Triangular Loop]

### from the original Figure) clockwise

(46 ?)( 6A) (50 ?)(i¬o)-(9 ?)(14A) = 0

## Io = 3 A

### Answer: Current io is 3 A.

## REFERENCES

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