Saturday , October 23 2021

Wye Delta conversion

Wye Delta conversion

 

Question: For a balanced wye-delta prove that {R_Y=\frac{R_\triangle}3\;or\;R_\triangle=3R_Y}

 

Solution

To obtain the equivalent resistances in the wye network, we compare the two networks and make sure that the resistance between each pair of nodes in the Δ network is the same as the resistance between the same pair of nodes in the Y network.

Figure:1

For terminals 1 and 2 in  figure 1(a) and 1(b), we obtain

{\begin{array}{l}R_{12}(Y)=R_1+R_2\\R_{12}(\triangle)=R_b\parallel(R_a+R_b)\\\\Setting\;R_{12}(Y)=R_{12}(\triangle)\;gives\\\\R_{12}=R_1+R_3=\frac{R_b(R_a+R_c)}{R_a+R_b+R_c}...................(i)\\\\Similarly\\\\R_{13}=R_1+R_2=\frac{R_c(R_a+R_b)}{R_a+R_b+R_c}...................(ii)\\\\R_{34}=R_2+R_3=\frac{R_a(R_b+R_c)}{R_a+R_b+R_c}...................(i)\\\end{array}}

 

Subtracting equation (iii) from equation (i) we get

{\begin{array}{l}\\R_1+R_3-R_2-R_3=\frac{R_b(R_a+R_c)}{R_a+R_b+R_c}-\frac{R_a(R_b+R_c)}{R_a+R_b+R_c}\\\\\boldsymbol\Rightarrow R_1-R_2=\frac{R_{\mathbf b}R_{\boldsymbol a}+R_{\boldsymbol b}R_{\mathbf c}-R_{\mathbf a}R_{\mathbf b}-R_{\boldsymbol a}R_{\mathbf c}}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}\\\\\boldsymbol\Rightarrow R_{\mathbf1}-R_{\mathbf2}=\frac{R_{\boldsymbol c}(R_{\mathbf b}-R_{\boldsymbol a})}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}..............(iv)\\\end{array}}

 

Adding equations (ii) and (iv) we get

{\begin{array}{l}\\R_1+R_2+R_1-R_2=\frac{R_c(R_a+R_b)}{R_a+R_b+R_c}+\frac{R_c(R_b-R_a)}{R_a+R_b+R_c}\\\\\boldsymbol\Rightarrow\boldsymbol\;\boldsymbol\;\boldsymbol\;\mathbf2R_1=\frac{R_cR_{\boldsymbol a}+R_{\boldsymbol b}R_{\mathbf c}+R_{\mathbf c}R_{\mathbf b}-R_cR_{\mathbf a}}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}\\\\\boldsymbol\Rightarrow\boldsymbol\;\boldsymbol\;\boldsymbol\;R_{\mathbf1}=\frac{R_bR_c}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}..............(v)\\\end{array}}

 

Subtracting equation (iv) from equation (ii) we get

{\begin{array}{l}\\R_1+R_2-R_1+R_2=\frac{R_c(R_a+R_b)}{R_a+R_b+R_c}-\frac{R_c(R_b-R_a)}{R_a+R_b+R_c}\\\\\boldsymbol\Rightarrow\boldsymbol\;\boldsymbol\;\boldsymbol\;\mathbf2R_2=\frac{R_cR_{\boldsymbol a}+R_cR_{\mathbf b}-R_{\mathbf c}R_{\mathbf b}+R_cR_{\mathbf a}}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}\\\\\boldsymbol\Rightarrow\boldsymbol\;\boldsymbol\;\boldsymbol\;R_2=\frac{R_cR_a}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}\\\end{array}}

 

Subtracting equation (v) from equation (i) we get

{\begin{array}{l}\\R_1+R_3-R_1=\frac{R_b(R_a+R_c)}{R_a+R_b+R_c}-\frac{R_bR_c}{R_a+R_b+R_c}\\\\\;\;\;\;\;\;\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;R_2=\frac{R_cR_a}{R_{\boldsymbol a}+R_{\mathbf b}+R_{\boldsymbol c}}\\\end{array}}

When the Y and Δ networks are balanced then {R_1=R_2=R_3=R_{Y\;}\;\;\;\;\;R_a=R_b=R_c=R_\triangle}

Under this conditions, we get from equation (v)

{\begin{array}{rcl}\;\;\;R_{\mathbf Y\boldsymbol\;}&=&\frac{R_\triangle\;R_\triangle}{R_\triangle+R_\triangle+R_\triangle}\\&&\\R_{\mathrm Y\boldsymbol\;}&=&\frac{R_\triangle\;R_\triangle}{3R_\triangle}\\&&\\R_{\mathrm Y\;}&=&\frac{\;R_\triangle}3\;(Showed)\\&&\\&&or\\&&\\R_\triangle&=&3R_Y\;(Shoewd)\\&&\\&&\\&&\end{array}}

Wye Delta conversion

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