{"id":206618,"date":"2025-04-14T19:03:14","date_gmt":"2025-04-14T19:03:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=206618"},"modified":"2025-04-14T19:03:18","modified_gmt":"2025-04-14T19:03:18","slug":"copper-and-aluminum-are-being-considered-for-a-high-voltage-transmission-line-that-must-carry-a-current-of-60-0-a","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/14\/copper-and-aluminum-are-being-considered-for-a-high-voltage-transmission-line-that-must-carry-a-current-of-60-0-a\/","title":{"rendered":"Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.0 A"},"content":{"rendered":"\n

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.0 A. The resistance per unit length is to be 0.150 \u03a9 \/km. The densities of copper and aluminum are 3960 and 2600 kg\/m3, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length), for a copper cable and (c) \/ and (d) \u03bb for an aluminum cable.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(a) Current Density ( J ) and (b, d) Mass per Unit Length ( \\lambda )<\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Current ( I = 60.0 \\, \\text{A} )<\/li>\n\n\n\n
  • Resistance per unit length ( R\/L = 0.150 \\, \\Omega\/\\text{km} = 0.150 \\times 10^{-3} \\, \\Omega\/\\text{m} )<\/li>\n\n\n\n
  • Resistivity:<\/li>\n\n\n\n
  • Copper: ( \\rho_{\\text{Cu}} = 1.68 \\times 10^{-8} \\, \\Omega\\cdot\\text{m} )<\/li>\n\n\n\n
  • Aluminum: ( \\rho_{\\text{Al}} = 2.82 \\times 10^{-8} \\, \\Omega\\cdot\\text{m} )<\/li>\n\n\n\n
  • Densities:<\/li>\n\n\n\n
  • Copper: ( \\rho_{\\text{m,Cu}} = 8960 \\, \\text{kg\/m}^3 )<\/li>\n\n\n\n
  • Aluminum: ( \\rho_{\\text{m,Al}} = 2600 \\, \\text{kg\/m}^3 )<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-Step Calculations<\/h3>\n\n\n\n

    Common Step: Area of cross-section<\/h4>\n\n\n\n

    Use resistance per unit length ( R\/L = \\rho \/ A )
    Thus, ( A = \\rho \/ (R\/L) )<\/p>\n\n\n\n

      \n
    • Copper Area:<\/strong>
      [
      A_{\\text{Cu}} = \\frac{1.68 \\times 10^{-8}}{0.150 \\times 10^{-3}} = 1.12 \\times 10^{-4} \\, \\text{m}^2
      ]<\/li>\n\n\n\n
    • Aluminum Area:<\/strong>
      [
      A_{\\text{Al}} = \\frac{2.82 \\times 10^{-8}}{0.150 \\times 10^{-3}} = 1.88 \\times 10^{-4} \\, \\text{m}^2
      ]<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      (a) Current Density ( J = I\/A )<\/h3>\n\n\n\n
        \n
      • Copper:<\/strong>
        [
        J_{\\text{Cu}} = \\frac{60}{1.12 \\times 10^{-4}} = 5.36 \\times 10^5 \\, \\text{A\/m}^2
        ]<\/li>\n\n\n\n
      • Aluminum:<\/strong>
        [
        J_{\\text{Al}} = \\frac{60}{1.88 \\times 10^{-4}} = 3.19 \\times 10^5 \\, \\text{A\/m}^2
        ]<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        (b) and (d) Mass per Unit Length ( \\lambda = A \\times \\text{density} )<\/h3>\n\n\n\n
          \n
        • Copper:<\/strong>
          [
          \\lambda_{\\text{Cu}} = 1.12 \\times 10^{-4} \\times 8960 = 1.00 \\, \\text{kg\/m}
          ]<\/li>\n\n\n\n
        • Aluminum:<\/strong>
          [
          \\lambda_{\\text{Al}} = 1.88 \\times 10^{-4} \\times 2600 = 0.489 \\, \\text{kg\/m}
          ]<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Summary<\/h3>\n\n\n\n
          Material<\/th>( J ) (A\/m\u00b2)<\/th>( \\lambda ) (kg\/m)<\/th><\/tr><\/thead>
          Copper<\/td>( 5.36 \\times 10^5 )<\/td>1.00<\/td><\/tr>
          Aluminum<\/td>( 3.19 \\times 10^5 )<\/td>0.489<\/td><\/tr><\/tbody><\/table><\/figure>\n","protected":false},"excerpt":{"rendered":"

          Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.0 A. The resistance per unit length is to be 0.150 \u03a9 \/km. The densities of copper and aluminum are 3960 and 2600 kg\/m3, respectively. Compute (a) the magnitude J of the current density and (b) the mass […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-206618","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206618","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=206618"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206618\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206618"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206618"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206618"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}