<p style="text-align:justify"><img alt="" src="https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16259077471625907747.jpg" style="height:374px; width:800px" /></p>

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

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

<p style="text-align:justify"><span style="font-size:14px"><strong>Example 6.2:</strong> Referring to figure 6.12,</span></p>

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

<ul>
	<li style="text-align:justify">
	<p><span style="font-size:14px">R is a standard normal inverse IDMT overcurrent relay with normal inverse characteristic (setting range 50-200 % of C. T. secondary rating in seven equal steps).</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">P. S. of relay R = 75 %, TMS of relay R = 0.8.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">In-built high-set instantaneous unit provided with relay R is set at 600 %.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">C. T. ratio = 500/1 amp.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Fault current = 1500 amp.</span></p>
	</li>
</ul>

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

<p style="text-align:justify"><span style="font-size:14px">Find out the time of operation of the relay. If the fault current is 3500 amp, what will be the time of operation of the relay?</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><strong>Solution:</strong> Plug-setting in C. T. primary terms <em>= 500&nbsp;</em><em>&times; 0.75 =375 amp</em></span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><em><span class="math-tex">\(PSM=\frac{I_f}{P.S.}\)</span></em></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<em><span class="math-tex">\(=\frac{1500}{375}\)</span></em></span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 4</span></p>

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

<p style="text-align:justify"><span style="font-size:14px"><em>Time of Operation of the Relay</em></span></p>

<p style="text-align:justify"><span style="font-size:14px"><em><span class="math-tex">\(=\frac{3}{log_{10}PSM}\times TMS\)</span></em></span></p>

<p style="text-align:justify"><span style="font-size:14px"><em><span class="math-tex">\(=\frac{3}{log_{10}4}\times 0.8\)</span></em></span></p>

<p style="text-align:justify"><span style="font-size:14px"><em>=3.986 sec.</em></span></p>

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

<p style="text-align:justify"><span style="font-size:14px">Fault current 3500 amp is (3500/500 = 7) 700 % of C. T. primary rating (or secondary, if fault current is transformed to secondary). This is more than H. S. setting, 600 %. Therefore, the relay will operate within 3 cycles (60 milliseconds).</span></p>