Star to Delta Conversion MCQ Quiz - Objective Question with Answer for Star to Delta Conversion - Download Free PDF

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:

Z_Y = (Z_Δ × Z_Δ) / (Z_Δ + Z_Δ + Z_Δ)

Here, Z_Δ is the impedance of each branch in the Delta network, and Z_Y is the impedance of each branch in the equivalent Star network.

Given:

• Z_Δ = √3Z

Substitute the value of Z_Δ into the formula:

Z_Y = (√3Z × √3Z) / (√3Z + √3Z + √3Z)

Simplify the numerator:

Z_Y = (3Z²) / (3√3Z)

Cancel out common terms:

Z_Y = 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

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

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

\(R_A={150\times 180+180\times 60+60\times 150\over 60}\)

\(R_A=780\space {\Omega}\)

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:

\( R_{Y} = \frac{R_{Δ1} \cdot R_{Δ2}}{R_{Δ1} + R_{Δ2} + R_{Δ3}} \)

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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:

\(R_A=R_B=R_C={3R}\)

Concept: 

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, R_a and R_b are connected

\({R_A} = \frac{{{R_a} \times {R_b}}}{{{R_a} + {R_b} + {R_c}}}\)

For branch resistance QS, R_a and R_c are connected

\({R_B} = \frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}\)

For branch resistance, SR, R_b and R_c 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, R_A and R_B are connected

\({R_a} = {R_A} + {R_B} + \frac{{{R_A} \times {R_B}}}{{{R_C}}}\)

For branch resistance PR, R_B and R_C are connected

\({R_b} = {R_C} + {R_B} + \frac{{{R_C} \times {R_B}}}{{{R_C}}}\)

For branch resistance QR, R_A and R_C 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 R_A 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.

Concept:

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, R_2, 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

The correct answer is option 3):(6 Ω, 18 Ω, 12 Ω)

Concept:

Star-to-delta conversion

R_A = R₁ + R₂ + \(\frac{R_1\times R_2}{R_3}\)

R_B = R1 + R₃ + \(\frac{R_1\times R_3}{R_2}\)

R_C = 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

 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:

\(R_1=R_2=R_3={R\over 3}\)

Concept:

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 R_A = R_B = R_C = 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

R_AB = (12× 6) / 18

= 4 Ω 

Concept:

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:

\({R_A} = \frac{{\left( r \right)\left( {3r} \right)}}{{r + 2r + 3r}} = \frac{r}{2}\)

\({R_B} = \frac{{\left( r \right)\left( {2r} \right)}}{{r + 2r + 3r}} = \frac{r}{3}\)

\({R_C} = \frac{{\left( {3r} \right)\left( {2r} \right)}}{{r + 2r + 3r}} = r\)

Concept:

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, R_AB = 20 Ω, R_AC = 30 Ω, R_BC = 50 Ω

According to the question, by converting the given Delta to star, we get

\({R_A} = \frac{{{20} \times {30}}}{{{20} + {30} + {50}}}\; = \frac{{{20} \times {30}}}{{100}}\;=6 \;\Omega \)

\({R_B} = \frac{{{20} \times {50}}}{{{20} + {30} + {50}}}\; =\frac{{{20} \times {50}}}{{100}}\;=10 \;\Omega \)

\({R_C} = \frac{{{30} \times {50}}}{{{20} + {30} + {50}}}\; =\frac{{{30} \times {50}}}{{100}}\;=15 \;\Omega \)

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

Concept:

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:

R_AB = 4 Ω , RBC =  3 Ω , RAC = 2 Ω (Given)

\({R_1} = {R_A} = \frac{{4 \times 2}}{9} = \frac{8}{9}\;\)

\({R_2} = {R_C} = \frac{{3 \times 2}}{9} = \frac{6}{9}\;\)

\({R_3} = {R_B} = \frac{{4 \times 3}}{9} = \frac{{12}}{9}\;\)

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:

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

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

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

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