Star To Delta Transformation Or Star To Delta Conversion And Formulas

STAR TO DELTA TRANSFORMATION OR STAR TO DELTA CONVERSION AND FORMULAS

Fig:-

Suppose, R_A  , R_B and R_C is three resistances connecting to each other such that they forms a star connections or star networks.

 Now, we have to convert this circuit in delta for solving or simplifying given star connected circuits. Because if you are not convert may not get equivalent resistances of that given circuit.

Fig:-

So, we make a delta connection. Connecting pt.1 to pt.2 and pt.3 and also connecting pt.2 to pt.3. It will form a new network as ‘delta connection’ as I shown through dotted diagram and hence forms a new ‘Delta circuit’.

For new circuits containing new resistances is R₁₂  , R₁₃ and R₂₃ having new value’s. For these new resistances we have direct formula’s I am going to give you because no need of derivation, kindly used direct formulas as given below:-

Fig:-

How to remember?

“The equivalent delta resistance between any two terminals is given by the sum of star resistances between those terminals plus the product of these two star resistances divided by the third star resistance”.

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Ques.) Find the value of R_L for the given network below that the power is maximum? And also find the Max Power through load-resistance R_L by using maximum power transfer theorem?

Solution: -

For the above network, we are going to find-out the value of unknown resistance called “R_L”. In previous post, I already show that when power is maximum through load-resistance is equals to the equivalent resistance between two ends of load-resistance after removing.

So, for finding load-resistance R_L. We have to find-out the equivalent resistance like that for this circuit:-

Now, For finding Maximum Power through load-resistance we have to find-out the value of V_o.c.

Here, V_o.c is known as voltage between open circuits.

So, steps are:-

For this circuit using Mesh-analysis. We get

Applying Kvl in loop 1^st:-

6-6I₁-8I₁+8I₂=0

-14I₁+8I₂=-6         ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (1)

Again, Applying Kvl in loop 2^nd:-

-8I₂-5I₂-12I₂+8I₁=0

8I₁-25I₂=0             ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (2)

On solving,eqn (1) & eqn (2), We get

I₁ = 0.524 A

I₂ = 0.167 A

Now, From the circuit V_o.c is

V_A-5I₂- V_B = 0

V_o.c/ V_AB= 5I₂ = 5X0.167 = 0.835v

So, the maximum power through the R_L is given by:-