{"version":"1.0","provider_name":"Electrical Riddle","provider_url":"https:\/\/electrical-riddle.com\/en","author_name":"Electrical Riddle","author_url":"https:\/\/electrical-riddle.com\/en","title":"- Electrical Riddle","type":"rich","width":600,"height":338,"html":"<blockquote class=\"wp-embedded-content\" data-secret=\"bsE36dtrxx\"><a href=\"https:\/\/electrical-riddle.com\/en\/forums\/reply\/3169\/\">Reply To: Protection Riddle No.60 -Role of resistor in high impedance REF<\/a><\/blockquote><iframe sandbox=\"allow-scripts\" security=\"restricted\" src=\"https:\/\/electrical-riddle.com\/en\/forums\/reply\/3169\/embed\/#?secret=bsE36dtrxx\" width=\"600\" height=\"338\" title=\"&#8220;Reply To: Protection Riddle No.60 -Role of resistor in high impedance REF&#8221; &#8212; Electrical Riddle\" data-secret=\"bsE36dtrxx\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" class=\"wp-embedded-content\"><\/iframe><script>\n\/*! This file is auto-generated *\/\n!function(d,l){\"use strict\";l.querySelector&&d.addEventListener&&\"undefined\"!=typeof URL&&(d.wp=d.wp||{},d.wp.receiveEmbedMessage||(d.wp.receiveEmbedMessage=function(e){var t=e.data;if((t||t.secret||t.message||t.value)&&!\/[^a-zA-Z0-9]\/.test(t.secret)){for(var s,r,n,a=l.querySelectorAll('iframe[data-secret=\"'+t.secret+'\"]'),o=l.querySelectorAll('blockquote[data-secret=\"'+t.secret+'\"]'),c=new RegExp(\"^https?:$\",\"i\"),i=0;i<o.length;i++)o[i].style.display=\"none\";for(i=0;i<a.length;i++)s=a[i],e.source===s.contentWindow&&(s.removeAttribute(\"style\"),\"height\"===t.message?(1e3<(r=parseInt(t.value,10))?r=1e3:~~r<200&&(r=200),s.height=r):\"link\"===t.message&&(r=new URL(s.getAttribute(\"src\")),n=new URL(t.value),c.test(n.protocol))&&n.host===r.host&&l.activeElement===s&&(d.top.location.href=t.value))}},d.addEventListener(\"message\",d.wp.receiveEmbedMessage,!1),l.addEventListener(\"DOMContentLoaded\",function(){for(var e,t,s=l.querySelectorAll(\"iframe.wp-embedded-content\"),r=0;r<s.length;r++)(t=(e=s[r]).getAttribute(\"data-secret\"))||(t=Math.random().toString(36).substring(2,12),e.src+=\"#?secret=\"+t,e.setAttribute(\"data-secret\",t)),e.contentWindow.postMessage({message:\"ready\",secret:t},\"*\")},!1)))}(window,document);\n\/\/# sourceURL=https:\/\/electrical-riddle.com\/wp-includes\/js\/wp-embed.min.js\n<\/script>\n","description":"I am not able to follow the following para extracted from your riddle no 60The required earth fault setting current Iop is 250A (assume the maximum earth fault current is limited by the earthing resistor to 1000A). The chosen E\/F CT has an exciting current Ie of 1%, and hence using the equation: Iop = &hellip;  Read More &raquo;"}