Star to Delta Conversion MCQ Quiz - Objective Question with Answer for Star to Delta Conversion - Download Free PDF
Last updated on Jun 11, 2025
Latest Star to Delta Conversion MCQ Objective Questions
Star to Delta Conversion Question 1:
If each branch of a Delta circuit has impedance √3 Z, then each branch of the equivalent Y circuit has impedance:
Answer (Detailed Solution Below)
Star to Delta Conversion Question 1 Detailed Solution
Explanation:
Delta to Star (Y) Transformation
Definition: Delta to Star (or Y) transformation is a mathematical technique used in circuit analysis to simplify circuits with three-terminal networks. This conversion makes it easier to calculate the equivalent impedance and solve complex circuits by transforming a Delta (Δ) network into an equivalent Star (Y) network.
Problem Statement: If each branch of a Delta circuit has an impedance of √3Z, what is the impedance of each branch in the equivalent Star (Y) circuit?
Solution:
For a three-phase circuit, the relationship between the impedances of the Delta (Δ) and Star (Y) networks is given by the following formula:
Impedance of each branch in the Y-network:
ZY = (ZΔ × ZΔ) / (ZΔ + ZΔ + ZΔ)
Here, ZΔ is the impedance of each branch in the Delta network, and ZY is the impedance of each branch in the equivalent Star network.
Given:
- ZΔ = √3Z
Substitute the value of ZΔ into the formula:
ZY = (√3Z × √3Z) / (√3Z + √3Z + √3Z)
Simplify the numerator:
ZY = (3Z2) / (3√3Z)
Cancel out common terms:
ZY = Z / √3
Therefore, the impedance of each branch in the equivalent Star (Y) circuit is Z / √3.
Correct Option: Option 1: Z / √3
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 2: 3Z
This option is incorrect. The impedance of each branch in the Star network cannot exceed the impedance of the Delta network branches. The formula for Delta to Star conversion clearly shows that the impedance in the Star network is always lower than that in the Delta network.
Option 3: 3√3Z
This option is also incorrect. It vastly overestimates the equivalent impedance of the Star network. The correct Star network impedance is Z / √3, which is much smaller than the impedance values provided in this option.
Option 4: Z / 3
This option is incorrect because it underestimates the equivalent impedance. While the Star network impedance is smaller than the Delta network impedance, it is not as small as Z / 3. The correct value is Z / √3.
Conclusion:
Understanding the relationship between Delta and Star impedances is crucial for circuit analysis. The impedance of each branch in a Star network is always lower than that of the corresponding Delta network. The correct answer to the problem is Z / √3, as derived above. This transformation simplifies circuit calculations and is commonly used in power systems and network analysis.
Star to Delta Conversion Question 2:
The phase impedance of a balanced star connected load is ZY = 40 + j30 Ω. What will be the phase impedance of the equivalent Δ-connected load?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 2 Detailed Solution
Star to Delta Conversion
The delta equivalent of a star connected resistor is:
\(R_A={R_1R_2+R_2R_3+R_1R_3\over R_3}\)
\(R_B={R_1R_2+R_2R_3+R_1R_3\over R_2}\)
\(R_C={R_1R_2+R_2R_3+R_1R_3\over R_1}\)
If all resistances connected in the star are equal to 'R', then the delta equivalent resistance is:
\(R_A=R_B=R_C={3R}\)
Calculation
Given, ZY = 40 + j30 Ω
ZΔ = 3 × (40 + j30)
ZΔ = 120 + j90 Ω
Star to Delta Conversion Question 3:
In the star-delta transformation shown in the figure, if P = 150 Ω, Q = 180 Ω and R = 60 Ω, then A = ?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 3 Detailed Solution
Star to Delta Conversion
The delta equivalent of a star connected resistor is:
\(R_A={R_PR_Q+R_QR_R+R_RR_P\over R_R}\)
\(R_B={R_PR_Q+R_QR_R+R_RR_P\over R_Q}\)
\(R_C={R_PR_Q+R_QR_R+R_RR_P\over R_P}\)
Calculation
Given, P = 150 Ω, Q = 180 Ω and R = 60 Ω
\(R_A={150\times 180+180\times 60+60\times 150\over 60}\)
\(R_A=780\space {\Omega}\)
Star to Delta Conversion Question 4:
For the transformation of a delta-connected network to a wye-connected network, the resistance between any two terminals in the wye-connection is:
Answer (Detailed Solution Below)
Star to Delta Conversion Question 4 Detailed Solution
In electrical engineering, a transformation between Delta (Δ) and Wye (Y) networks is often used to simplify the analysis of three-phase circuits. The resistance between any two terminals in a Wye-connected (Y-connected) network can be derived from the resistances in a Delta-connected (Δ-connected) network.
Solution:
For converting a delta (Δ) network to a wye (Y) network, the resistance between any two terminals in the wye-connected network is given by the formula:
\( R_{Y} = \frac{R_{Δ1} \cdot R_{Δ2}}{R_{Δ1} + R_{Δ2} + R_{Δ3}} \)
Where RΔ1 ,RΔ2,RΔ3 are the resistances of delta network.
Star to Delta Conversion Question 5:
In a Delta network, each element has value of 'R', then the value of each element in equivalent star network will be -
Answer (Detailed Solution Below)
Star to Delta Conversion Question 5 Detailed Solution
Delta to Star Conversion:
The star equivalent of delta connected resistor is:
\(R_1={R_AR_B\over R_A+R_B+R_C}\)
\(R_2={R_AR_C\over R_A+R_B+R_C}\)
\(R_3={R_BR_C\over R_A+R_B+R_C}\)
If all resistances connected in the delta are 'R', then the star equivalent resistance is:
\(R_1=R_2=R_3={R\over 3}\)
Additional Information
The delta equivalent of a connected resistor is:
\(R_A={R_1R_2+R_2R_3+R_1R_3\over R_3}\)
\(R_B={R_1R_2+R_2R_3+R_1R_3\over R_2}\)
\(R_C={R_1R_2+R_2R_3+R_1R_3\over R_1}\)
If all resistances connected in the star are 'R', then the delta equivalent resistance is:
\(R_A=R_B=R_C={3R}\)
Top Star to Delta Conversion MCQ Objective Questions
Consider a Delta connection of resistors and its Star equivalent as shown below. If all the elements of the Delta connection are scaled by a factor k, with k > 0, the elements of the corresponding Star connection will be scaled by a factor of ____.
Answer (Detailed Solution Below)
Star to Delta Conversion Question 6 Detailed Solution
Download Solution PDFConcept:
For interconversions of delta and star, when all resistances are the same we can directly conclude two points
1. From Delta to Star resistance will be divided by 3
2. From Star to Delta resistance is multiplied by 3
Shortcuts for conversion:
Delta connection is shown below
Star connection of resistors is shown as
To convert Star from Delta
Branch resistance = product of connected resistance/sum of all resistances
For branch resistance PS, Ra and Rb are connected
\({R_A} = \frac{{{R_a} \times {R_b}}}{{{R_a} + {R_b} + {R_c}}}\)
For branch resistance QS, Ra and Rc are connected
\({R_B} = \frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}\)
For branch resistance, SR, Rb and Rc are connected
\({R_C} = \frac{{\left( {{R_c} \times {R_b}} \right)}}{{{R_a} + {R_b} + {R_c}}}\)
To convert Delta from Star
Branch resistance = ∑ connected resistances + \(\frac{{product\;of\;connected\;resistance}}{{resistance\;which\;is\;not\;connected}}\)
For branch resistance PQ, RA and RB are connected
\({R_a} = {R_A} + {R_B} + \frac{{{R_A} \times {R_B}}}{{{R_C}}}\)
For branch resistance PR, RB and RC are connected
\({R_b} = {R_C} + {R_B} + \frac{{{R_C} \times {R_B}}}{{{R_C}}}\)
For branch resistance QR, RA and RC are connected
\({R_c} = {R_C} + {R_A} + \frac{{{R_A} \times {R_C}}}{{{R_B}}}\)
Calculation:
Given that all resistances in Delta connections are scaled by ‘K’
From the formula let us calculate RA which is PS branch resistance,
\({R_{{A_1}}} = \frac{{k{R_a} \times k{R_b}}}{{k{R_a} + k{R_b} + k{R_c}}}\)
\( = \frac{{{k^2}\left( {{R_a} \times {R_b}} \right)}}{{k\left( {{R_a} + {R_b} + {R_c}} \right)}}\)
\( = k\frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}\)
\( = k{R_A}\)
the resistance in the Star network also scales by the same factor as in Delta network.
Find the resistance between terminals A and B in the electric circuit of below figure:
Answer (Detailed Solution Below)
Star to Delta Conversion Question 7 Detailed Solution
Download Solution PDFConcept:
Y(Wye) to Delta ‘or’ Delta to Y conversion technique is explained in the circuit below:
For a given Delta connection, the equivalent Wye connection will have resistances of values:
\({R_1} = \frac{R_bR_c}{{{R_a+R_b+R_c}}}\)
\({R_2} = \frac{R_aR_c}{{{R_a+R_b+R_c}}}\)
\({R_3} = \frac{R_aR_b}{{{R_a+R_b+R_c}}}\)
Application:
The given circuit is modified as shown below:
where R1, R2, and R3 are calculated using Wye-Delta transformation as:
\({R_1} = \frac{50\times 40}{{{50+40+10}}}=20~Ω\)
\({R_2} = \frac{10\times 50}{{{50+40+10}}}=5~Ω \)
\({R_3} = \frac{10\times 40}{{{50+40+10}}}=4~Ω \)
The above circuit is redrawn as:
The resistance between terminals A and B is 19 Ω
Find DELTA equivalent resistance R12, R23, and R31 from the given STAR configuration.
Answer (Detailed Solution Below)
Star to Delta Conversion Question 8 Detailed Solution
Download Solution PDFThe correct answer is option 3):(6 Ω, 18 Ω, 12 Ω)
Concept:
Star-to-delta conversion
RA = R1 + R2 + \(\frac{R_1\times R_2}{R_3}\)
RB = R1 + R3 + \(\frac{R_1\times R_3}{R_2}\)
RC = R2 + R3 + \(\frac{R_2\times R_3}{R_1}\)
Calculation:
Given
RA = R1 + R2 + \(\frac{R_1\times R_2}{R_3}\)
= 2 + 3 + \(\frac{2\times 3}{6}\) = 6
RB = R1 + R3 + \(\frac{R_1\times R_3}{R_2}\)
= 6 +2 + \(\frac{2\times 6 }{3}\)
= 6 +2 + 4 = 12
R3 = R2 + R3 + \(\frac{R_2\times R_3}{R_1}\)
= 3 + 6 + \(\frac{3 \times 6 }{2}\)
= 18
hence option 3 is correct
The resistor value in a Y network that is equivalent to a Δ containing three resistors of R Ω each is:
Answer (Detailed Solution Below)
Star to Delta Conversion Question 9 Detailed Solution
Download Solution PDFDelta to Star conversion:
The star equivalent of delta connected resistor is:
\(R_1={R_AR_B\over R_A+R_B+R_C}\)
\(R_2={R_AR_C\over R_A+R_B+R_C}\)
\(R_3={R_BR_C\over R_A+R_B+R_C}\)
If all resistances connected in the delta are 'R', then the star equivalent resistance is:
\(R_1=R_2=R_3={R\over 3}\)
For the networks shown below, what will be the resistance between the terminals A and B?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 10 Detailed Solution
Download Solution PDFConcept:
1) Delta to Star conversion:
\({R_A} = \frac{{{R_{AB}} × {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
\({R_B} = \frac{{{R_{AB}} × {R_{BC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
\({R_C} = \frac{{{R_{BC}} × {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
2) Star to Delta conversion:
\({R_{AB}} = \frac{{({R_A} × {R_C}) + ({R_A} × {R_B})+({R_B} × {R_C})}}{{{R_C}}}\)
\({R_{AC}} = \frac{{({R_A} × {R_C}) + ({R_A} × {R_B})+({R_B} × {R_C})}}{{{R_B}}}\)
\({R_{BC}} = \frac{{({R_A} × {R_C}) + ({R_A} × {R_B})+({R_B} × {R_C})}}{{{R_A}}}\)
Calculation:
In the given example RA = RB = RC = 3Ω
The equivalent circuit after star to delta conversion is:
\({R_{AB}} = {R_{BC}} = {R_{AC}} = \frac{{\left( {3 × 3} \right) + \left( {3 × 3} \right) + \left( {3 × 3} \right)}}{3}\)
= 9 Ω
The equivalent diagram can be drawn as
Now calculating the equivalent resistance between AC, AB, BC is
\(18||9 = \frac{{18 × 9}}{{18 + 9}} = {6\Omega }\)
The further circuit can be modified as
RAB = (12× 6) / 18
= 4 Ω
RA, RB, and RC are equivalent star-connected resistances of a delta circuit. Which one of the following is true?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 11 Detailed Solution
Download Solution PDFConcept:
Y(Wye) to Delta ‘or’ Delta to Y conversion technique is explained in the circuit below:
For a given Delta connection, the equivalent Wye connection will have resistances of values:
\({R_1} = \frac{R_bR_c}{{{R_a+R_b+R_c}}}\)
\({R_2} = \frac{R_aR_c}{{{R_a+R_b+R_c}}}\)
\({R_3} = \frac{R_aR_b}{{{R_a+R_b+R_c}}}\)
Calculation:
Putting on the respective values, we get:
\({R_A} = \frac{{\left( r \right)\left( {3r} \right)}}{{r + 2r + 3r}} = \frac{r}{2}\)
\({R_B} = \frac{{\left( r \right)\left( {2r} \right)}}{{r + 2r + 3r}} = \frac{r}{3}\)
\({R_C} = \frac{{\left( {3r} \right)\left( {2r} \right)}}{{r + 2r + 3r}} = r\)
Three resistors, whose values are 20 Ω, 30 Ω, and 50 Ω, are connected in the delta connection. If the delta to star conversion is done, what will be the equivalent resistance used in the star combination?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 12 Detailed Solution
Download Solution PDFConcept:
1) Delta to Star conversion:
\({R_A} = \frac{{{R_{AB}} \times {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
\({R_B} = \frac{{{R_{AB}} \times {R_{BC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
\({R_C} = \frac{{{R_{BC}} \times {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}\)
2) Star to Delta conversion:
\({R_{AB}} = \frac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_C}}}\)
\({R_{AC}} = \frac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_B}}}\)
\({R_{BC}} = \frac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_A}}}\)
Application:
Let, RAB = 20 Ω, RAC = 30 Ω, RBC = 50 Ω
According to the question, by converting the given Delta to star, we get
\({R_A} = \frac{{{20} \times {30}}}{{{20} + {30} + {50}}}\; = \frac{{{20} \times {30}}}{{100}}\;=6 \;\Omega \)
\({R_B} = \frac{{{20} \times {50}}}{{{20} + {30} + {50}}}\; =\frac{{{20} \times {50}}}{{100}}\;=10 \;\Omega \)
\({R_C} = \frac{{{30} \times {50}}}{{{20} + {30} + {50}}}\; =\frac{{{30} \times {50}}}{{100}}\;=15 \;\Omega \)
A balanced load of 5 + j4 is connected in delta. What is the impedance per phase of the equivalent star connection?
Answer (Detailed Solution Below)
Star to Delta Conversion Question 13 Detailed Solution
Download Solution PDFConcept:
The per phase impedance of balanced delta connected circuit = ZΔ
By using delta to star conversion,
The per phase impedance of balanced delta connected circuit = ZY = (1/3) ZΔ
Calculation:
Given delta connected load = 5 + j4
equivalent star connected load = 1/3 (5 + j4) = 1.66 + j1.33
In the given Data were conversion, Find the value of R1, R2 & R3.
Answer (Detailed Solution Below)
Star to Delta Conversion Question 14 Detailed Solution
Download Solution PDFConcept:
1) Delta to Star conversion:
\({R_A} = \dfrac{{{R_{AB}}\; \times\; {R_{AC}}}}{{{R_{AB}}\; + \;{R_{AC}}\; + \;{R_{BC}}}}\)
\({R_B} = \dfrac{{{R_{AB}} \;\times \;{R_{BC}}}}{{{R_{AB}}\; + \;{R_{AC}}\; + \;{R_{BC}}}}\)
\({R_C} = \dfrac{{{R_{BC}}\; \times\; {R_{AC}}}}{{{R_{AB}} \;+\; {R_{AC}}\; + \;{R_{BC}}}}\)
2) Star to Delta conversion:
\({R_{AB}} = \dfrac{{({R_A} \;×\; {R_C}) \;+\; ({R_A}\; × \;{R_B})\;+\;({R_B} \;×\; {R_C})}}{{{R_C}}}\)
\({R_{AC}} = \dfrac{{({R_A}\; × \;{R_C})\; +\; ({R_A} \;×\; {R_B})\;+\;({R_B} \;× \;{R_C})}}{{{R_B}}}\)
\({R_{BC}} = \dfrac{{({R_A}\; ×\; {R_C})\; +\; ({R_A} \;×\; {R_B})\;+\;({R_B}\; × \;{R_C})}}{{{R_A}}}\)
Calculation:
RAB = 4 Ω , RBC = 3 Ω , RAC = 2 Ω (Given)
\({R_1} = {R_A} = \frac{{4 \times 2}}{9} = \frac{8}{9}\;\)
\({R_2} = {R_C} = \frac{{3 \times 2}}{9} = \frac{6}{9}\;\)
\({R_3} = {R_B} = \frac{{4 \times 3}}{9} = \frac{{12}}{9}\;\)
Determine the value of equivalent resistance across nodes a and b.
Answer (Detailed Solution Below)
Star to Delta Conversion Question 15 Detailed Solution
Download Solution PDFThe correct answer is option 1): 1.5 Ω
Concept:
When two resistance R,R are connected in parallel the equivalent resistance is (\(1\over R\)+ \(1\over R\))-1
Delta (Δ) to Y(Wye) conversion is shown in the circuit below:
\({R_1} = \frac{R_bR_c}{{{R_a+R_b+R_c}}}\)
\({R_2} = \frac{R_aR_c}{{{R_a+R_b+R_c}}}\)
\({R_3} = \frac{R_aR_b}{{{R_a+R_b+R_c}}}\)
Calculation:
The given circuit is
here 2 Ω || 2 Ω
Total resistance = ( \(1 \over 2\) + \(1 \over 2\))-1
So the total resistance is 1 Ω ,
here 4 Ω || 4 Ω
Total resistance = ( \(1 \over 4\) + \(1 \over 4\))-1
So the total resistance is 2 Ω,
The circuit is redrawn as
As we see that the above figure is a balanced wheat stone bridge and no current would flow from 3 ohms resistor.
So, the resistance across "a-b" would be equal to
R = (1 + 2) II (1 + 2)
= 1.5 ohms