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:  

  1. Z/√3
  2. 3Z
  3. 3√3Z 
  4. Z/3  

Answer (Detailed Solution Below)

Option 1 : Z/√3

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?

  1. ZΔ = 40 + j30 Ω
  2. ZΔ = 120 + j90 Ω
  3. ZΔ = (40+j30) × √3 Ω
  4. ZΔ = (120 + j90)/3 Ω

Answer (Detailed Solution Below)

Option 2 : ZΔ = 120 + j90 Ω

Star to Delta Conversion Question 2 Detailed Solution

Star to Delta Conversion

qImage535

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:

qImage67b2eb09914eb771f62f544e

In the star-delta transformation shown in the figure, if P = 150 Ω, Q = 180 Ω and R = 60 Ω, then A = ?

  1. 260 Ω
  2. 470 Ω
  3. 780 Ω
  4. 312 Ω

Answer (Detailed Solution Below)

Option 3 : 780 Ω

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: 

  1. The resistance is equal to the sum of the two delta resistances connected to the common terminal.
  2. The resistance is equal to the sum of the two delta resistances connected to the common terminal, divided by the total number of resistances in the delta network. 
  3. The resistance is equal to the difference between the highest and lowest delta resistances.
  4. The resistance is equal to the product of the two delta resistances connected to the common terminal, divided by the sum of all three delta resistances.

Answer (Detailed Solution Below)

Option 4 : The resistance is equal to the product of the two delta resistances connected to the common terminal, divided by the sum of all three delta resistances.

Star to Delta Conversion Question 4 Detailed Solution

Concept:

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.RΔ1" id="MathJax-Element-315-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ1" id="MathJax-Element-333-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ1" id="MathJax-Element-346-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ1
RΔ1" id="MathJax-Element-315-Frame" role="presentation" style=" word-spacing: 0px; position: relative;" tabindex="0">RΔ1" id="MathJax-Element-333-Frame" role="presentation" style=" position: relative;" tabindex="0">RΔ1Unknown node type: span" id="MathJax-Element-347-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ1Unknown node type: span RΔ2" id="MathJax-Element-316-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ2" id="MathJax-Element-334-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ2Unknown node type: span" id="MathJax-Element-348-Frame" role="presentation" style="position: relative;" tabindex="0">RΔ2Unknown node type: span This formula indicates that the resistance between any two terminals in the wye-connected network is equal to the product of the two delta resistances connected to the common terminal, divided by the sum of all three delta resistances.

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 -

  1. R/6
  2. R/4
  3. R/2
  4. R/3

Answer (Detailed Solution Below)

Option 4 : R/3

Star to Delta Conversion Question 5 Detailed Solution

Delta to Star Conversion:

qImage535

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 ____.

qImage66923fc53503e7f37472c538

qImage66923fc53503e7f37472c539

  1. k2
  2. k
  3. 1/k
  4. \(\sqrt k \)

Answer (Detailed Solution Below)

Option 2 : k

Star to Delta Conversion Question 6 Detailed Solution

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Concept: 30July1

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:

F1 S.B Deepak 04.03.2020 D1

  1. 30 Ω 
  2. 19 Ω 
  3. 50 Ω 
  4. 100 Ω 

Answer (Detailed Solution Below)

Option 2 : 19 Ω 

Star to Delta Conversion Question 7 Detailed Solution

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Concept:

Y(Wye) to Delta ‘or’ Delta to Y conversion technique is explained in the circuit below:

F1 S.B Madhu 24.12.19 D 2

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:

F1 S.B Madhu 09.03.20 D1

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:

F1 S.B Madhu 09.03.20 D2

The resistance between terminals A and B is 19 Ω

Find DELTA equivalent resistance R12, R23, and R31 from the given STAR configuration.

F3 Madhuri Engineering 10.01.2023 D2

  1. Ω, 24 Ω, 12 Ω 
  2. 18 Ω, 24 Ω, 9 Ω
  3. Ω, 18 Ω, 12 Ω
  4. 18 Ω, 12 Ω, 24 Ω

Answer (Detailed Solution Below)

Option 3 : 6 Ω, 18 Ω, 12 Ω

Star to Delta Conversion Question 8 Detailed Solution

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The correct answer is option 3):(Ω, 18 Ω, 12 Ω)

Concept:

Star-to-delta conversion

 

F1 Vinanti Engineering 18-11-22 D28

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

F3 Madhuri Engineering 10.01.2023 D2

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:

  1. \(\rm\frac{R}{3}\) Ω each
  2. 3 R2 Ω each
  3. \(\rm​\frac{R^2}{3}\) Ω each
  4. 3 R Ω each

Answer (Detailed Solution Below)

Option 1 : \(\rm\frac{R}{3}\) Ω each

Star to Delta Conversion Question 9 Detailed Solution

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 Delta to Star conversion:

qImage535

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?

F29 Shubham B 19-4-2021 Swati D3

  1. 4 Ω
  2. 6 Ω
  3. 9 Ω
  4. 3 Ω

Answer (Detailed Solution Below)

Option 1 : 4 Ω

Star to Delta Conversion Question 10 Detailed Solution

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Concept:

SSC JE Electrical 80 D2

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 

F1 Ravi Ranjan EE 21-5-2021 Swati D7

Now calculating the equivalent resistance between AC, AB, BC is

 \(18||9 = \frac{{18 × 9}}{{18 + 9}} = {6\Omega }\)

F1 Ravi Ranjan EE 21-5-2021 Swati D8

The further circuit can be modified as

F1 Ravi Ranjan EE 21-5-2021 Swati D9

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?

F1 U.B. N.J 22.08.2019 D 9

  1. RA = r/2, RB = r/3, RC = r
  2. RA = r/3, RB = r/3, RC = r/3
  3. RA = r/3, RB = r/2, RC = r
  4. \({R_A} = \frac{3}{4}r,\;{R_B} = \frac{2}{3}r,\;{R_C} = \frac{6}{5}r\)

Answer (Detailed Solution Below)

Option 1 : RA = r/2, RB = r/3, RC = r

Star to Delta Conversion Question 11 Detailed Solution

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Concept:

Y(Wye) to Delta ‘or’ Delta to Y conversion technique is explained in the circuit below:

F1 S.B Madhu 24.12.19 D 2

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?

  1. 6Ω, 10Ω, 10Ω
  2. 6Ω, 10Ω, 15Ω
  3. 15Ω, 15Ω, 10Ω
  4. 6Ω, 15Ω, 15Ω

Answer (Detailed Solution Below)

Option 2 : 6Ω, 10Ω, 15Ω

Star to Delta Conversion Question 12 Detailed Solution

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Concept:

SSC JE Electrical 80 D2

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:

F1 Shraddha Uday 24.12.2020 D8

Let, RA= 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?

  1. 5 + j4
  2. 1.66 + j1.33
  3. 15 + j12
  4. 2.5 + j2

Answer (Detailed Solution Below)

Option 2 : 1.66 + j1.33

Star to Delta Conversion Question 13 Detailed Solution

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Concept:

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.

F1 Shubham Madhuri 11.05.2021 D9

  1. 9/12, 9/6, 9/8
  2. 6/9, 12/9, 8/9
  3. 8/9, 6/9, 12/9
  4. 9/6, 9/12, 9/8

Answer (Detailed Solution Below)

Option 3 : 8/9, 6/9, 12/9

Star to Delta Conversion Question 14 Detailed Solution

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Concept:

SSC JE Electrical 80 D2

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:

F1 Shubham Madhuri 11.05.2021 D9

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.

F1 Vinanti Engineering 02.12.22 D3

  1. 1.5 Ω
  2. 2.5 Ω
  3. 3 Ω
  4. 4.5 Ω

Answer (Detailed Solution Below)

Option 1 : 1.5 Ω

Star to Delta Conversion Question 15 Detailed Solution

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The 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:

F1 Raviranjan 17-09-21 Savita D6

 

\({R_1} = \frac{R_bR_c}{{{R_a+R_b+R_c}}}\)

\({R_2} = \frac{R_aR_c}{{{R_a+R_b+R_c}}}\)

\({R_3} = \frac{R_aR_b}{{{R_a+R_b+R_c}}}\)

Calculation:

The given circuit is 

F1 Vinanti Engineering 02.12.22 D3

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 

F1 Vinanti Engineering 02.12.22 D4

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

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