{"id":35,"date":"2019-07-04T10:56:51","date_gmt":"2019-07-04T14:56:51","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/singlephasetransformers\/?post_type=chapter&p=35"},"modified":"2022-04-26T13:45:32","modified_gmt":"2022-04-26T17:45:32","slug":"percent-impedance","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/singlephasetransformers\/chapter\/percent-impedance\/","title":{"raw":"Percent Impedance","rendered":"Percent Impedance"},"content":{"raw":"

The\u00a0percent impedance<\/em>\u00a0(%Z) is the percent of the rated load impedance possessed by a transformer. The percent impedance is important in that it allows us to:<\/p>\r\n\r\n

    \r\n \t
  1. Calculate available fault currents (both individual and bank).<\/li>\r\n \t
  2. Determine whether two transformers are suitable for paralleling.<\/li>\r\n<\/ol>\r\nIn the last chapter, we learned how to determine the short-circuit voltage of a transformer. We now can use that voltage to determine the percent impedance:\r\n

    Eshort circuit<\/sub>\/ERated<\/sub>\u00a0= %Z<\/strong><\/p>\r\n\r\n

    Video Alert!<\/strong><\/h2>\r\n[embed]https:\/\/www.youtube.com\/watch?v=bFQ57hONOWg[\/embed]\r\n

    Fault-Current Calculations<\/strong><\/h2>\r\n

    To calculate the fault current available from a transformer if a dead short<\/i>\u00a0occurs across the secondary terminals, use the formula:<\/p>\r\n

    Irated (Secondary)<\/sub>\/%Z = Fault current<\/strong><\/p>\r\n

    Remember to use the percentage as a decimal not the full number. For example, 2.5% is actually .025.<\/p>\r\n\r\n

    Video Alert!<\/strong><\/h2>\r\nHow to calculate fault current using %Z:\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=teOCt9G2amU&t=32s[\/embed]\r\n
    \r\n

    Example #1<\/strong><\/span><\/h2>\r\n<\/header>\r\n
    \r\n\r\nWhat is the available fault current of a step-down transformer rated at 50 kVA, 1200 V \u2013 120 V and 2.75% percent impedance?\r\n\r\nI(full load) = [latex]\/frac{50000}{120}[\/latex]\r\n\r\nI(full load) = 417 Amps\r\n\r\nI(short circuit) = [latex]\\frac{417}{.0275}[\/latex]\r\n\r\nI(short circuit) = 15,151 Amps\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Exampe #2<\/strong><\/span><\/h2>\r\n<\/header>\r\n
    \r\n\r\nA transformer is rated 20KVA, 4800 V \u2013 240 V.\u00a0 With the secondary short-circuited, it takes 96 volts to reach the rated primary current (4.2 amps).\u00a0 Determine the %Z and the available fault current.\r\n\r\n%Z = [latex]\\frac{96}{4800}[\/latex]\r\n\r\n%Z = 2<\/strong>\r\n\r\nI(rated secondary) = [latex]\\frac{20000}{240}[\/latex]\r\n\r\nI(rated secondary) = 83 Amps\r\n\r\nI(Short circuit) = [latex]\\frac{83}{.02}[\/latex]\r\n\r\nI(Short circuit) = 4150 Amps<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n

    Attributions<\/h1>\r\nHow to determine the percent impedance of a transformer<\/a> video by The Electric Academy<\/a> is under a Creative Commons Attribution License<\/a>.\r\n\r\nHow to calculate fault current using percent impedance<\/a> video by The Electric Academy<\/a> is under a Creative Commons Attribution License<\/a>.","rendered":"

    The\u00a0percent impedance<\/em>\u00a0(%Z) is the percent of the rated load impedance possessed by a transformer. The percent impedance is important in that it allows us to:<\/p>\n

      \n
    1. Calculate available fault currents (both individual and bank).<\/li>\n
    2. Determine whether two transformers are suitable for paralleling.<\/li>\n<\/ol>\n

      In the last chapter, we learned how to determine the short-circuit voltage of a transformer. We now can use that voltage to determine the percent impedance:<\/p>\n

      Eshort circuit<\/sub>\/ERated<\/sub>\u00a0= %Z<\/strong><\/p>\n

      Video Alert!<\/strong><\/h2>\n