{"id":243942,"date":"2025-07-04T18:08:53","date_gmt":"2025-07-04T18:08:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=243942"},"modified":"2025-07-04T18:08:55","modified_gmt":"2025-07-04T18:08:55","slug":"calculate-v-and-ix-in-the-circuit-of-fig-2-79-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/calculate-v-and-ix-in-the-circuit-of-fig-2-79-2\/","title":{"rendered":"\u00a0Calculate\u00a0v\u00a0and\u00a0ix\u00a0in the circuit of Fig. 2.79."},"content":{"rendered":"\n
 Calculate v and ix in the circuit of Fig. 2.79.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
v = 10 V \u2014 i\u2093 = 10\/3 A<\/p>\n\n\n\n
\n\n\n\n
To determine the values of the voltage v and the current i\u2093 in the provided circuit, we can apply Kirchhoff’s Voltage Law (KVL), which states that the algebraic sum of all voltages around any closed loop in a circuit is equal to zero. This circuit can be analyzed by considering its two distinct loops: the left loop and the right loop.<\/p>\n\n\n\n
1. Analysis of the Left Loop to find v:<\/strong><\/p>\n\n\n\n
The left loop consists of the 12 V independent voltage source, the 12 \u03a9 resistor (with voltage v across it), and the 2 V independent voltage source. We can traverse this loop, starting from the bottom-left corner and moving clockwise.<\/p>\n\n\n\n
    \n
  • Moving up through the 12 V source, we go from the negative to the positive terminal, which is a voltage rise of +12 V.<\/li>\n\n\n\n
  • Moving across the 12 \u03a9 resistor, we follow the direction of the indicated voltage drop\u00a0v\u00a0(from + to -), so we subtract\u00a0v.<\/li>\n\n\n\n
  • Moving down through the 2 V source, we go from the positive to the negative terminal, which is a voltage drop of -2 V.<\/li>\n<\/ul>\n\n\n\n
    Setting up the KVL equation for the left loop:
    +12 V – v – 2 V = 0<\/p>\n\n\n\n
    Now, we can solve for v:
    10 V – v = 0
    v = 10 V<\/strong><\/p>\n\n\n\n
    2. Analysis of the Right Loop to find i\u2093:<\/strong><\/p>\n\n\n\n
    The right loop contains the 2 V independent voltage source, the 8 V independent voltage source, and the 3i\u2093 current-controlled dependent voltage source. We can traverse this loop, starting from the bottom of the 2 V source and moving clockwise.<\/p>\n\n\n\n
      \n
    • Moving up through the 2 V source, we go from negative to positive, which is a voltage rise of +2 V.<\/li>\n\n\n\n
    • Moving across the 8 V source, we go from negative to positive, which is a voltage rise of +8 V.<\/li>\n\n\n\n
    • Moving down through the dependent voltage source, we go from its positive to its negative terminal, which is a voltage drop of -3i\u2093.<\/li>\n<\/ul>\n\n\n\n
      Setting up the KVL equation for the right loop:
      +2 V + 8 V – 3i\u2093 = 0<\/p>\n\n\n\n
      Now, we can solve for i\u2093:
      10 V – 3i\u2093 = 0
      3i\u2093 = 10 V
      i\u2093 = 10\/3 A<\/strong> (or approximately 3.33 A)<\/p>\n\n\n\n
      Therefore, by applying KVL to the two independent loops in the circuit, we find that the voltage v is 10 V and the current i\u2093 is 10\/3 A.<\/p>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
       Calculate v and ix in the circuit of Fig. 2.79. The Correct Answer and Explanation is: v = 10 V \u2014 i\u2093 = 10\/3 A To determine the values of the voltage v and the current i\u2093 in the provided circuit, we can apply Kirchhoff’s Voltage Law (KVL), which states that the algebraic sum of all voltages around any closed loop in a […]<\/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-243942","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243942","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=243942"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243942\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=243942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=243942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=243942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}