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Question:
Published on: 22 June, 2022

Explain

  • Star – delta conversion
  • Delta – star conversion with the help of a purely resistive circuit
Answer:

Fig. 24(a)

Fig. 24(b)

In this transformation any resistance in delta network can be found out if the resistances of corresponding star network are known.

Here the branch resistances in the star connections are

\({R_1}^\prime=\frac{R_1R_2}{(R_1+R_2+R_3)}\) … (1)

\({R_2}^\prime=\frac{R_2R_3}{(R_1+R_2+R_3)}\) …(2)

\({R_3}^\prime=\frac{R_1R_3}{(R_1+R_2+R_3)}\) … (3)

Starting three equations (1), (2) and (3), we have

\({R_1}^\prime{R_2}^\prime+{R_2}^\prime{R_3}^\prime+{R_3}^\prime{R_1}^\prime=\frac{R_1{R_2}^2R_3+R_1R_2{R_3}^2+{R_1}^1R_2R_3}{{(R_1+R_2+R_3)}^2}\)

\({R_1}^\prime{R_2}^\prime+{R_2}^\prime{R_3}^\prime+{R_3}^\prime{R_1}^\prime=\frac{R_1R_2R_3(R_1+R_2+R_3)}{{(R_1+R_2+R_3)}^2}=\frac{R_1R_2R_3}{(R_1+R_2+R_3)}\) … (4)

Dividing equation (4) alternatively by \({R_1}^\prime\), \({R_2}^\prime\) and \({R_3}^\prime\) and using equation (1), (2) and (3), we have

\(R_1=\frac{{R_1}^\prime{R_2}^\prime+{R_2}^\prime{R_3}^\prime+{R_3}^\prime{R_1}^\prime}{{R_2}^\prime}\) … (5)

\(R_2=\frac{{R_1}^\prime{R_2}^\prime+{R_2}^\prime{R_3}^\prime+{R_3}^\prime{R_1}^\prime}{{R_3}^\prime}\) … (6)

\(R_3=\frac{{R_1}^\prime{R_2}^\prime+{R_2}^\prime{R_3}^\prime+{R_3}^\prime{R_1}^\prime}{{R_1}^\prime}\) … (7)

Thus the above equations provide the equivalent resistance in delta network in terms of the resistances of star network.

 

In this transformation any resistance in star network can be expressed in terms of the resistances of delta network.

The resistance between the nodes A and C of delta network \(R_{AC}\) is the equivalent resistance of \(R_2\) in series with a parallel connection of the resistance \((R_1+R_3)\)

Here

\(R_{AC}=\frac{R_1R_2}{(R_1+R_2+R_3)}\)

Again the total resistance looking at the terminals A and C of star network is \(R_{AC}=({R_1}^\prime+{R_2}^\prime)\)

Thus

\(\left({R_1}^\prime+{R_2}^\prime\right)=\frac{R_1R_2+R_2R_3}{(R_1+R_2+R_3)}\) … (8)

Similarly equating the resistance between the nodes B and C in delta network with that in star network one obtains

\(\left({R_1}^\prime+{R_3}^\prime\right)=\frac{R_1R_3+R_2R_3}{(R_1+R_2+R_3)}\) … (9)

Considering the terminals A and B of delta and star network, a similar equation is obtained

\(\left({R_1}^\prime+{R_3}^\prime\right)=\frac{R_1R_2+R_2R_3}{(R_1+R_2+R_3)}\) … (10)

From equation (8) and (9), we have

\(\left({R_1}^\prime-{R_3}^\prime\right)=\frac{R_1R_2-R_1R_3}{(R_1+R_2+R_3)}\) … (11)

Solving \({R_1}^\prime\) and \({R_3}^\prime\) from equation (10) and (11), we have

\({R_1}^\prime=\frac{R_1R_2}{(R_1+R_2+R_3)}\) … (12)

\({R_3}^\prime=\frac{R_1R_3}{(R_1+R_2+R_3)}\) … (13)

Substituting the value of \({R_3}^\prime\) from equation (8) and (13), we have

\({R_2}^\prime=\frac{R_2R_3}{(R_1+R_2+R_3)}\) … (14)

Thus the equivalent resistance connected to a node in star network can be obtained from equation (12) to (14) if the resistances in delta network are known.

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