**Star to delta or Delta to star Transformation** is used to simplify the circuit when we cant not able to solve the circuit by using series or parallel combination. **Star and delta** can be transformed from one form to another form easily with the help of driven parameters.

A **star** connection consists of three branches of a network often connected to a mutual point in Y -model.

A **delta **connection consists of the three branches of a network is connected in a closed loop in the form of a delta model.

## Delta(Δ) To Wye (Y) Or Star Conversion

It consist of three three- terminal network of three resistance R_{AB},R_{BC},R_{CA} are connected in delta. We need to** convert** this **Delta to star** of resistance R_{A},R_{B},R_{C} in wye or star network.

Lets solve the circuit and calculate the equivalent resistance between the two terminals

Equivalent resistance between A and C

R_A+R_C=\frac { R_{CA}(R_{BC}+R_{AB}) }{ R_{AB}+R_{BC}+R_{CA} }

Equivalent resistance between B and A

R_A+R_B=\frac { R_{AB}(R_{BC}+R_{CA}) }{ R_{AB}+R_{BC}+R_{CA} }

Equivalent resistance between C and B

R_B+R_C=\frac { R_{BA}(R_{CA}+R_{AB}) }{ R_{AB}+R_{BC}+R_{CA} }

Combining all the equations then

R_A+R_B+R_C=\frac { R_{AB}R_{BC}+R_{BC}R_{CA}+R_{CA}R_{AB} }{ R_{AB}+R_{BC}+R_{CA} }

By solving above equations we can get the the unknown resistance of** why or star** circuit

R_A=\frac { R_{CA}R_{AB} }{ R_{AB}+R_{BC}+R_{CA} }

R_B=\frac { R_{AB}R_{BC} }{ R_{AB}+R_{BC}+R_{CA} }

R_C=\frac {R_{BC}R_{CA} }{ R_{AB}+R_{BC}+R_{CA} }

## Wye (Y) Or Star To Delta(Δ) Conversion

Lets us consider a three- terminal network of three resistance R_{AB},R_{BC},R_{CA} are connected in star or wye . We need to **convert** this **star to delta** of resistance R_{A},R_{B},R_{C} in network.

Current equation through the R_A can be written as

I_A=\frac { (V_A-V_N) }{ R_A }

By apply KCL at node N for star connection

\frac { (V_A-V_N) }{ R_A } +\frac { (V_B-V_N) }{ R_B } +\frac { (V_C-V_N) }{ R_C } =0

By solving this equation we get V_N=\frac { \left( \frac { V_A }{ R_A } +\frac { V_B }{ R_B } +\frac { V_C }{ R_C } \right) }{ \left( \frac { 1 }{ R_A } +\frac { 1 }{ R_B } +\frac { 1 }{ R_C } \right) }

As we know that current entering from one node= current leaving to that node

I_A=\frac { V_{AB} }{ R_{AB} }+\frac { V_{AC} }{ R_{AC} } (for delta network)

From above two current equation we can compare both then

\frac { V_{AB} }{ R_{AB} }+\frac { V_{AC} }{ R_{AC} }=\frac { (V_A-V_N) }{ R_A }

Put the value of V_N in the above equation

\frac { V_{AB} }{ R_{AB} }+\frac { V_{AC} }{ R_{AC} }=\frac { V_A-\left( \frac { \frac { V_A }{ R_A } +\frac { V_B }{ R_B } +\frac { V_C }{ R_C } }{ \frac { 1 }{ R_A } +\frac { 1 }{ R_B } +\frac { 1 }{ R_C } } \right) }{ R_A }

By solving this equation and equating coefficient of V_{AB} and V_{AC} both side we get

\frac { 1 }{ R_{AB} } =\frac { 1 }{ R_AR_B\left( \frac { 1 }{ R_A } +\frac { 1 }{ R_B } +\frac { 1 }{ R_B } \right) }

R_{AB}=R_A+R_B+\frac { R_AR_B }{ R_C }

\frac { 1 }{ R_{AC} } =\frac { 1 }{ R_AR_C\left( \frac { 1 }{ R_A } +\frac { 1 }{ R_B } +\frac { 1 }{ R_C } \right) }

R_{AC}=R_A+R_C+\frac { R_AR_C }{ R_B }

similary we can drive the resistance for R_{BC}

steps for calculating R_{BC}

- Calculate the current I_B for star circuit as we find above
- Calculate the current I_B for delta circuit as we find above
- Equating the above two equations and using the value of V_N

Then we get

\frac { 1 }{ R_{BC} } =\frac { 1 }{ R_BR_C\left( \frac { 1 }{ R_A } +\frac { 1 }{ R_B } +\frac { 1 }{ R_C } \right) }

R_{BC}=R_B+R_C+\frac { R_BR_C }{ R_A }

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