<p style="text-align:justify"><span style="font-size:14px">Let us consider Y-B fault. Referring to figure 4.34, </span></p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161538841616153884.jpg" style="height:545px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px">the conditions for Y-B fault are:</span></p>

<p style="text-align:justify"><span style="font-size:14px">I<sub>R</sub> = 0, I<sub>Y</sub> = -<sub> </sub>I<sub>B</sub>, V<sub>Y</sub> = V<sub>B</sub> + I<sub>Y</sub> Z<sub>f</sub>.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_0=\frac{1}{3}(I_R+I_Y+I_B)=0\)</span></span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">V<sub>0</sub> = - I<sub>0</sub> Z<sub>0</sub> = 0</span></p>

<p style="text-align:justify"><span style="font-size:14px">Thus, zero sequence current and voltage are zero.</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp; I<sub>Y</sub> = - I<sub>B</sub> &nbsp;&nbsp;</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(a^2I_1+aI_2=-(aI_1+a^2I_2)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{0.045in}.}\hspace{0.3cm}I_1(a^2+a)=-I_2(a^2+a)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{0.045in}.}\hspace{0.3cm}I_1=-I_2\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (EQ 4.21)</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp; V<sub>Y</sub> = V<sub>B</sub> + I<sub>Y</sub> Z<sub>f</sub> </span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(a^2\hspace{0.1cm}V_1+a\hspace{0.1cm}V_2=a\hspace{0.1cm}V_1+a^2\hspace{0.1cm}V_2+(a^2\hspace{0.1cm}I_1+a\hspace{0.1cm}I_2)Z_f\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(V_1(a^2-a)=V_2(a^2-a)+(a^2I_1-aI_1)Z_f\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">V<sub>1</sub> = V<sub>2</sub> + I<sub>1</sub> Z<sub>f</sub>&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.22)</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">For satisfying equations 4.21 and 4.22, sequence networks are to be connected as shown in figure 4.35.</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161539111616153911.jpg" style="height:545px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px">From figure 4.35,</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_1=\frac{E}{Z_1+Z_2+Z_f}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Z<sub>f</sub> is the arc resistance due to flashover between two conductors of different phases. Z<sub>f</sub> is resistive, hence we can write R<sub>f</sub> also.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_Y=-I_B=a^2I_1+aI_2=I_1(a^2-a)\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class="math-tex">\(=\frac{-j\sqrt{3E}}{Z_1+Z_2+Z_f}\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">Z<sub>f</sub> (or R<sub>f</sub>) is quite small as only arc (flashover) resistance is involved. It can be 2 to 4 ohms.</span></p>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify">&nbsp;</p>