North South University

Department of Electrical & Computer Engineering

LAB REPORT
Course Code: EEE141L.2
Course Title: EEE141L Electrical Circuits Lab
Course Instructor: Dr. Mohammad Abdul Matin
Experiment Number: 02 04
Experiment Name:
Delta-Wye Conversion.

Experiment Date: 21/11/2021

Date of Submission: 26/11/2021
Section: 02
Group Number: None
Submitted To: Tabia Hossain
Submitted By Score

Student Name and ID:

Name: Md. Misbah Khan
ID: 2132089643
Objectives:
• To perform Delta-Wye Conversion
• To verify the results with measured data.
• Solve a complex circuit using Delta-Wye Conversion.

List of Components:
• Trainer Board.
• DMM.
• 5 x 15kΩ resistor.
• 3 x 5 kΩ resistor.

Theory:
During simplifying a resistor network, you get stuck. Some resistor networks cannot
be simplified using the usual series and parallel combinations. This situation can often be
handled by trying the Delta-Wye transformation, or 'Delta-Wye' transformation. The
names Delta and Wye come from the shape of the schematics, which resemble letters.
The transformation allows you to replace three resistors in a Δ configuration by three
resistors in a Y configuration, and the other
way around.

Figure-1 Figure-2

Schematics like above are neither series network, nor parallel. We can solve Figure 1 like
bellow:
We’ll derive the Y equvalent using the Δ-Y transformation.

Similarly for Figure 2 we can do Y- Δ transformation like this:

Circuit diagram:
A 1
A

R1
R2 R3
2 5.0kΩ
15.0kΩ 15.0kΩ
R1
V 3 4 R2 R3
2
10V B 15.0kΩ C V
10V 5.0kΩ 5.0kΩ
B 3 4C
R5 R4
0
R5 R4
15.0kΩ D 15.0kΩ 0
15.0kΩ D 15.0kΩ

Circuit 1 Circuit 2
Data:
Table 1:
Readings Circuit 1 Circuit 2 % Error
10 10 0%
5 5 0%
5 5 0%
5 5 0%
0 0 0%
5 5 0%

Circuit 1 with DMM

Circuit 2 with DMM

Answer of question 1: The resistors in Circuit 1 are neither in series network, nor
parallel.

Answer of question 2: We will use 'Delta-Wye' transformation to find the equivalent

resistance.

Answer of question 3: In circuit 1, the upper portion (ΔABC) is a Delta network.

So in order to calculate the equivalent resistance we need to
transform it to a Y network, like so:

Figure-3

Now we’ll derive the Y equvalent using the Δ-Y transformation.

𝑅𝑎 𝑅𝑏 15×15 225
𝑅1 = = = = 5kΩ
𝑅𝑎 +𝑅𝑏 +𝑅𝑐 15+15+15 45

𝑅𝑎 𝑅𝑐 15×15 225
𝑅2 = = = = 5kΩ
𝑅𝑎 +𝑅𝑏 +𝑅𝑐 15+15+15 45

𝑅𝑏 𝑅𝑐 15×15 225
𝑅3 = = = = 5kΩ
𝑅𝑎 +𝑅𝑏 +𝑅𝑐 15+15+15 45
Answer of question 4: Redrawn equivalent circuit after applying Delta-Wye
conversion will be like this:

Figure-4
It’s clear that this circuit is same as circuit 2.

Answer of question 5: In figure 4 (which is actually circuit 2), we can see that
R2 & R5 are in series connection, we’ll denote their
equivalent resistance as Req1, and R3 & R6 are in series
connection,like before we’ll denote their equivalent
resistance as Req2. Now Req1 & Req2 are in parallel
network. We’ll denote their equivalent resistance as
Req3. Now Req3 & R1 are in series connection. So if total
equivalent resistance of the entire circuit is Req, then :

Req1 = (R2 + R5) = (5+15) kΩ =20 kΩ

Req2 = (R3 + R6) = (5+15) kΩ = 20 kΩ
20×20
So, Req3 = (Req1 ∥ Req2) = kΩ = 10 kΩ
20+20

Now, Req = Req3 + R1 = (10+5) kΩ = 15 kΩ

So, Req is 15 kΩ.

Answer of question 6: From the answer of question 5, we got that,
Req = 15 kΩ
Given that, V = 10 v

𝑉 10
Is = = mA = 0.667 mA
𝑅𝑒𝑞 15
IR1 = Is (as R1 is in series with the power source, so it remains
unchanged)
And IR2 = IR3 ( as 𝑅𝑒𝑞1 = 𝑅𝑒𝑞2 )
𝑅𝑒𝑞1 20 kΩ
Now, IR2 = IR3 = × Is = × 0.662 mA = 0.331 mA
𝑅𝑒𝑞1 +𝑅𝑒𝑞2 20 kΩ +20 kΩ

Now, if voltage through the R1, R2 and R3 resistor are

respectively VR1, VR2, VR3 then:

VR1 = IR1R1 = 0.667 mA × 5 kΩ = 3.335 v

VR2 = IR2R2 = 0.331 mA × 5 kΩ = 1.655 v
VR3 = IR3R3 = 0.331 mA × 5 kΩ = 1.655 v

Answer of question 7: In circuit 2,

VAB = VR1 + VR2 = (3.335 + 1.655) v = 4.99 v
VAC = VR1 + VR3 = (3.335 + 1.655) v = 4.99 v
VBC = VAB - VAC = (4.99 - 4.99) v = 0 v
Now, IR2 = IR5 = 0.331 mA (as R2 & R5 are in seies network)
And, IR3 = IR6 = 0.331 mA (as R3 & R6 are in seies network)

So, VBD = VR5 = IR5 × R5 = 0.331 mA × 15 kΩ = 4.97 v

VCD = VR6 = IR6 × R6 = 0.331 mA × 15 kΩ = 4.97 v
In order to mesure, VAD we need to mesure VR5 (as VR2 & VR5 are in parallel with VR3
& VR6. And in a parallel network, voltage remains same.)
So, VR5 = IR5 × R5 = 0.331 mA × 15 kΩ = 4.965 v
Now, VAD = VR1 + VR2 + VR5 = ( 3.335+1.655 +4.965) v = 9.955 v
Error:
|5−4.99|
Error for VAB = % = 0.002%
5
|5−4.99|
Error for VAC = % = 0.002%
5
 |0−0|
Error for VBC = % = 0%
0
|5−4.97|
Error for VBD = % = 0.2%
5
|5−4.97|
Error for VCD = % = 0.2%
5
|10−9.955|
Error for VAD = % = 0.0045%
10

Answer of question 8: In table 2 we can see that every value for circuit 1 is same to
circuit 2. So we can say that they are equivalent. And as circuit 2 is basically circuit 1
after applying Delta-Wye conversion, we can say that Detla-Wye conversion is successful
in our experiment.

Discussion: In this experiment, we learned about Delta-Wye Conversion, how to

apply then in a non series-parallel schematic. We learned about solve a complex circuit
using Delta-Wye Conversion. Sometimes we saw some circuit that are unusal, they are
neither parallel network, nor series. We can solve them using Delta=Wye conversion. By
our experiment we observed that Delta-Wye transformation doesn’t change the circuit,
we used it just to simplify the circuit.
Experiment are done in NI multisim 14.0 software. We find a slight difference in
experimental and theoretical value which can be caused for using different method and
for calculation with various floating number.