<p style="text-align:justify"><span style="font-size:14px">Let us plot the vector diagrams for three phase voltages and currents as shown in figure 4.2.</span></p>

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160588211616058821.jpg" style="float:left; height:524px; width:800px" /></span></p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

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<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p><span style="font-size:14px">These are known as balanced voltages and currents. The electrical vector quantities are said to be balanced if:</span></p>

<ul>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">The magnitudes of all three phase quantities are equal.</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">The vectors are at an angle of 120<sup>0</sup>&nbsp;with each other.</span></p>
	</li>
	<li>
	<p style="text-align:justify"><span style="font-size:14px">The phase sequence is R-Y-B. This means that, when the vectors are rotated anticlockwise, vector Y will replace vector R and vector B will replace vector Y.</span></p>
	</li>
</ul>

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

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

<p style="text-align:justify"><span style="font-size:14px">However, when fault occurs, the voltages and currents are no more balanced. The unbalanced voltages and currents can be represented as shown in figure 4.3.</span></p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160588711616058871.jpg" style="float:left; height:524px; width:800px" /></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; I<sub>R</sub>&nbsp;= I<sub>R1</sub>&nbsp;+ I<sub>R2</sub>&nbsp;+ I<sub>R0</sub></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp;I<sub>Y</sub>&nbsp;= I<sub>Y1</sub>&nbsp;+ I<sub>Y2</sub>&nbsp;+ I<sub>Y0</sub></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp;I<sub>B</sub>&nbsp;= I<sub>B1</sub>&nbsp;+ I<sub>B2</sub>&nbsp;+ I<sub>B0</sub></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Positive sequence currents mean the balanced currents with phase sequence R-Y-B. Negative sequence currents are also balanced (vector sum I<sub>R2</sub>&nbsp;+ I<sub>Y2</sub>&nbsp;+ I<sub>B2</sub>&nbsp;= 0) but with phase sequence R-B-Y. Zero sequence currents are in phase with equal magnitudes as shown.</span></p>

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

<p><span style="font-size:14px">&nbsp;Similarly,</span></p>

<p><span style="font-size:14px">&nbsp;V<sub>R</sub>&nbsp;= V<sub>R1</sub>&nbsp;+ V<sub>R2</sub>&nbsp;+ V<sub>R0</sub></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">V<sub>Y</sub>&nbsp;= V<sub>Y1</sub>&nbsp;+ V<sub>Y2</sub>&nbsp;+ V<sub>Y0</sub></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">V<sub>B</sub>&nbsp;= V<sub>B1</sub>&nbsp;+ V<sub>B2</sub>&nbsp;+ V<sub>B0</sub></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Now, let us introduce vector operator &ldquo;a&rdquo; (figure 4.4).</span></p>

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160589031616058903.jpg" style="height:524px; width:800px" /></span></p>

<p><span style="font-size:14px">Vector operator a is a unit vector when operated on any other vector, rotates that vector through 120<sup>0</sup>&nbsp;in positive (anticlockwise) direction (obviously without changing the magnitude of the vector).</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Summarizing,</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(a=1\angle120^0\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(=-0.5+j\hspace{0.1cm}0.866\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(a^2=1\angle240^0\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp;= - 0.5- j 0.866</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(a^3=1\angle0^0\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(a^4=a=1\angle120^0\)</span></span></p>

<p><span style="font-size:14px">&nbsp;And so on.</span></p>

<p><span style="font-size:14px">&nbsp;1 + a + a<sup>2</sup>&nbsp;= 0</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Hence w. r. t. figure 4.3,</span></p>

<p><span style="font-size:14px">&nbsp;I<sub>R1</sub>&nbsp;= I<sub>1</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;I<sub>R2</sub>&nbsp;= I<sub>2</sub></span></p>

<p><span style="font-size:14px">I<sub>Y1</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>R1</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</sub>I<sub>Y2</sub>&nbsp;= a I<sub>2&nbsp;</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>

<p><span style="font-size:14px">I<sub>R0</sub>&nbsp;= I<sub>Y0</sub>&nbsp;= I<sub>B0</sub>&nbsp;= I<sub>0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</sub></span></p>

<p><span style="font-size:14px"><sub>&nbsp;</sub>I<sub>B1</sub>&nbsp;= a I<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I<sub>B2</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>2</sub></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Therefore,</span></p>

<p><span style="font-size:14px">&nbsp;I<sub>R</sub>&nbsp;= I<sub>1</sub>&nbsp;+ I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.3)</span></p>

<p><span style="font-size:14px">&nbsp; I<sub>Y</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>1</sub>&nbsp;+ a I<sub>2&nbsp;</sub>+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.4)</span></p>

<p><span style="font-size:14px">&nbsp;&nbsp;I<sub>B</sub>&nbsp;= a I<sub>1</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.5)</span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Similar expressions for voltages can be written.</span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">If we multiply equation 4.4 by a and equation 4.5 by a<sup>2</sup>,</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">&nbsp;I<sub>R</sub>&nbsp;= I<sub>1</sub>&nbsp;+ I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (Eq 4.6)</span></p>

<p><span style="font-size:14px">a I<sub>Y</sub>&nbsp;= I<sub>1</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>2</sub>&nbsp;+ a I<sub>0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</sub>(Eq 4.7)</span></p>

<p><span style="font-size:14px">a<sup>2</sup>&nbsp;I<sub>B</sub>&nbsp;= I<sub>1</sub>&nbsp;+ a I<sub>2</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>0</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.8)</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Therefore,</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(I_1=\frac{1}{3}\left(I_R+aI_Y+a^2I_B\right)\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.9)</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Similarly,</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(I_2=\frac{1}{3}\left(I_R+a^2I_Y+aI_B\right)\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.10)</span></p>

<p><span style="font-size:14px">And&nbsp;&nbsp;<span class="math-tex">\(I_0=\frac{1}{3}\left(I_R+I_Y+I_B\right)\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.11)</span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">All these expressions can be represented in matrix form also as follows:</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(\left[\matrix{I_R&amp;\cr I_Y&amp;\cr I_B}\right]=\left[\matrix{1 &amp;\cr 1&amp; \cr 1 }\matrix{1 &amp;\cr a^2 &amp;\cr a}\matrix{1&amp; \cr a &amp;\cr a^2 }\right]\left[\matrix{I_0 &amp;\cr I_1 &amp;\cr I_2 }\right]\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp; Or,&nbsp;&nbsp;I<sub>RYB</sub>&nbsp;= [A] I<sub>012</sub></span></p>

<p><span style="font-size:14px">Where,&nbsp;<span class="math-tex">\(\left[A\right]=\left[\matrix{1 &amp;\cr 1 &amp;\cr 1 }\matrix{1 &amp;\cr a^2 &amp;\cr a}\matrix{1 &amp;\cr a &amp;\cr a^2 }\right]\)</span></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px"><span class="math-tex">\(\left[\matrix{I_0 &amp;\cr I_1 &amp;\cr I_2 }\right]=\left[A\right]^{-1}\left[\matrix{I_R &amp;\cr I_Y &amp;\cr I_B }\right]\)</span>&nbsp; &nbsp; &nbsp;<span class="math-tex">\(Where\hspace{0.1cm}\left[A\right]^{-1}=\frac{1}{3}\left[\matrix{1 &amp;\cr 1&amp; \cr 1 }\matrix{1 &amp;\cr a &amp;\cr a^2}\matrix{1 &amp;\cr a^2 &amp;\cr a }\right]\)</span></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp; Or, I<sub>012</sub>&nbsp;= [A]<sup>-1</sup>&nbsp;I<sub>RYB</sub></span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Similar expressions for voltages can be written.</span></p>

<p style="text-align:justify"><span style="font-size:14px">Generator at the power plant generates only positive sequence emf E. When fault occurs the sequence voltages can be found out as follows:</span></p>

<p style="text-align:justify"><span style="font-size:14px">Figure 4.5 shows a star connected generator with neutral impedance Z<sub>n</sub>&nbsp;connected in neutral circuit. When fault occurs, there will be three sequence networks as shown in figures 4.6, 4.7 and 4.8. The positive sequence emf E is seen only in positive sequence network of figure 4.6 and not in figures 4.7 and 4.8. Currents I<sub>R1</sub>, I<sub>Y1</sub>&nbsp;and I<sub>B1</sub>&nbsp;pass only through phase windings (figure 4.5) and not through neutral as I<sub>R1</sub>&nbsp;+ I<sub>Y1</sub>&nbsp;+ I<sub>B1</sub>&nbsp;= 0.</span></p>

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

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160589991616058999.jpg" style="float:left; height:524px; width:800px" /></span></p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590491616059049.jpg" style="float:left; height:524px; width:800px" /><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590761616059076.jpg" style="height:524px; width:800px" /><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590951616059095.jpg" style="height:524px; width:800px" /></span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Hence, w. r. t. positive sequence network shown in figure 4.6,</span></p>

<p style="text-align:justify"><span style="font-size:14px">V<sub>1</sub>&nbsp;= E &ndash; I<sub>1</sub>&nbsp;Z<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.12)</span></p>

<p style="text-align:justify"><span style="font-size:14px">Where Z<sub>1</sub>&nbsp;= Positive Sequence Impedance of the Generator.</span></p>

<p style="text-align:justify"><span style="font-size:14px">As discussed earlier the value of Z<sub>1</sub>&nbsp;will be X<sub>d</sub>? (sub-transient direct axis reactance) if analysis of fault events is being done during one cycle following the inception of fault. The magnitude of fault current will be very large as X<sub>d</sub>? is usually very small. If analysis is being made for transient period (15 to 30 cycles following the instant of fault inception, time depending on X/R ratio) X<sub>d</sub>? has to be considered. The magnitude of the fault current, in this case will be less than that during sub-transient period, but yet quite large as&nbsp; transient reactance ( X<sub>d</sub>? ) of the large generator is of the order of 30 % (sub-transient reactance is of the order of 20 %). If one is calculating the fault current in a steady-state period, X<sub>d</sub>&nbsp;(or X<sub>s</sub>) is to be considered. This steady-state reactance, X<sub>s</sub>, is of the order of 200 %. Hence, steady-state fault current is less than the full load rated current of the generator.</span></p>

<p style="text-align:justify"><span style="font-size:14px">Refer negative sequence network shown in figure 4.7. It can be appreciated that the sum of the three negative sequence currents I<sub>R2</sub>, I<sub>Y2</sub>&nbsp;and I<sub>B2</sub>&nbsp;is zero, hence there is no current through neutral circuit (refer figure 4.5).</span></p>

<p style="text-align:justify"><span style="font-size:14px">Therefore, V<sub>2</sub>&nbsp;= - I<sub>2</sub>&nbsp;Z<sub>2</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.13)</span></p>

<p style="text-align:justify"><span style="font-size:14px">Where Z<sub>2</sub>&nbsp;= Negative Sequence Impedance of Generator.</span></p>

<p style="text-align:justify"><span style="font-size:14px">But, I<sub>R0</sub>&nbsp;+ I<sub>Y0</sub>&nbsp;+ I<sub>B0</sub>&nbsp;= 3 I<sub>0</sub>. Hence, referring to figure 4.5, the zero sequence currents passing through generator windings are I<sub>R0</sub>, I<sub>Y0</sub>&nbsp;and I<sub>B0</sub>&nbsp;respectively, but current will pass through neutral circuit and its magnitude will be 3I<sub>0</sub>&nbsp;causing a voltage drop I<sub>0</sub>&nbsp;(Z<sub>g0</sub>&nbsp;+ 3 Z<sub>n</sub>)</span></p>

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

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp; Z<sub>0</sub>&nbsp;= Z<sub>g0</sub>&nbsp;+ 3 Z<sub>n</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.14)</span></p>

<p style="text-align:justify"><span style="font-size:14px">Referring to zero sequence network shown in figure 4.8,</span></p>

<p style="text-align:justify"><span style="font-size:14px">V<sub>0</sub>&nbsp;= - I<sub>0</sub>&nbsp;Z<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.15)</span></p>

<p style="text-align:justify"><span style="font-size:14px">Now resuming back to calculation of fault current for a three-phase fault (L-L-L-g or L-L-L), let us write the conditions [figure 4.1(b)]:</span></p>

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

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160591671616059167.jpg" style="height:564px; width:800px" /></span></p>

<p><span style="font-size:14px">V<sub>R</sub>&nbsp;= V<sub>Y</sub>&nbsp;= V<sub>B</sub>&nbsp;= 0, at the point of fault.</span></p>

<p><span style="font-size:14px">I<sub>R</sub>&nbsp;+ I<sub>Y</sub>&nbsp;+ I<sub>B</sub>&nbsp;= 0</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045in}.}\hspace{.3in}I_0=\frac{1}{3}\left(I_R+I_Y+I_B\right)=0\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(V_1=\frac{1}{3}\left(V_R+aV_Y+a^2V_B\right)=0\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(V_2=\frac{1}{3}\left(V_R+a^2V_Y+aV_B\right)=0\)</span></span></p>

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

<p><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.045in}.}\hspace{.3in}V_0=-I_0Z_0=0\)</span></span></p>

<p><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; V<sub>2</sub>&nbsp;= - I<sub>2</sub>&nbsp;Z<sub>2</sub></span></p>

<p><span style="font-size:14px"><sub><span class="math-tex">\(\dot{.\hspace{.045in}.}\hspace{.3in}0=-I_2\hspace{0,1cm}Z_2\)</span></sub></span></p>

<p><span style="font-size:14px"><sub><span class="math-tex">\(\dot{.\hspace{.045in}.}\hspace{.3in}I_2=0\)</span></sub></span></p>

<p>&nbsp;</p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Hence negative and zero sequence networks do not exist, and only positive sequence network is there. Hence, only positive sequence currents are existing. Hence, this fault (L-L-L-g) is said to be symmetrical fault.</span></p>

<p style="text-align:justify"><span style="font-size:14px">Figure 4.9 is the positive sequence network (or reactance diagram) for L-L-L-g fault.</span></p>

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

<p><span style="font-size:14px"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160591971616059197.jpg" style="height:524px; width:800px" /></span></p>

<p style="text-align:justify"><span style="font-size:14px">V<sub>1</sub> = E &ndash; I<sub>1</sub> Z<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (refer figure 4.6)</span></p>

<p style="text-align:justify"><span style="font-size:14px"><sub>&nbsp;<sub><span class="math-tex">\(\dot{.\hspace{.025in}.}\hspace{.2in}0=E-I_1Z_1\)</span></sub></sub></span></p>

<p><span style="font-size:14px"><sub><sub><span class="math-tex">\(\dot{.\hspace{.025in}.}\hspace{.2in}I_1=\frac{E}{Z_1}\)</span></sub></sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.16)</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px"><span class="math-tex">\(I_R=I_1+I_2+I_0=\frac{E}{Z_1}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(I_Y=a^2I_1+aI_2+I_0=a^2\frac{E}{Z_1}\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(I_B=aI_1+a^2I_2+I_0=a\frac{E}{Z_1}\)</span></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">These are the three phase fault currents.</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(Fault\hspace{0.1cm} Current=\frac{V_{Base}}{Z_1}\)</span></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">If base voltage is considered as 1 pu&nbsp;<span class="math-tex">\((1\angle0^0)\)</span>&sbquo;</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(pu\hspace{0.1cm}Fault\hspace{0.1cm} Current=\frac{1}{pu Z_1}\)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(\sqrt3\times Base\hspace{0.1cm}Current\times pu\hspace{0.1cm} Fault\hspace{0.1cm}Current\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>

<p><span class="math-tex">\(=\frac{\sqrt3\times V_{Base}\times I_{Base}}{puZ_1}\)</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">As Base Voltage = 1 pu</span></p>

<p><span style="font-size:14px">And Base Current = 1 pu</span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px"><span class="math-tex">\(\dot{.\hspace{.025in}.}\hspace{.2in}Fault\hspace{0.1cm} Level \hspace{0.1cm}in \hspace{0.1cm}MVA=\frac{Base\hspace{0.1cm} MVA}{pu\hspace{0.1cm} Z_1}\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.17) &nbsp;</span></p>

<p>&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Fault current for three-phase fault can be calculated using Equation 4.16 or Equation 4.17. &nbsp; &nbsp; &nbsp;&nbsp;</span></p>