{"id":243846,"date":"2025-07-04T17:20:11","date_gmt":"2025-07-04T17:20:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=243846"},"modified":"2025-07-04T17:20:13","modified_gmt":"2025-07-04T17:20:13","slug":"calculate-v-and-ix-in-the-circuit-of-fig-2-79","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/calculate-v-and-ix-in-the-circuit-of-fig-2-79\/","title":{"rendered":"Calculate\u00a0v\u00a0and\u00a0ix\u00a0in the circuit of Fig. 2.79."},"content":{"rendered":"\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-144.png\" alt=\"\" class=\"wp-image-243847\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answers are&nbsp;<strong>v = 10 V<\/strong>&nbsp;and&nbsp;<strong>ix = -2 A<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>To determine the values of the voltage&nbsp;v&nbsp;and the current&nbsp;ix&nbsp;in the given circuit, we can utilize Kirchhoff&#8217;s Voltage Law (KVL). KVL states that the algebraic sum of all the voltages around any closed path, or loop, in a circuit is equal to zero. This circuit contains two distinct loops that can be analyzed independently to find the unknown variables.<\/p>\n\n\n\n<p>First, let&#8217;s analyze the left loop, which consists of the 12 V independent voltage source, the 12 \u03a9 resistor, and the central 2 V element. We will sum the voltages by traversing the loop in a clockwise direction, starting from the bottom-left corner.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Moving upward through the 12 V source, we encounter a voltage rise, so we add\u00a0+12 V.<\/li>\n\n\n\n<li>Moving from left to right across the 12 \u03a9 resistor, we see a voltage drop\u00a0v, as indicated by the polarity signs (+ to -). This is represented as\u00a0-v.<\/li>\n\n\n\n<li>Moving downward through the central element, we go from the positive to the negative terminal, which is a voltage drop of\u00a0\u20132 V.<\/li>\n<\/ol>\n\n\n\n<p>Setting the sum of these voltages to zero according to KVL:<br>12 V &#8211; v &#8211; 2 V = 0<br>10 V &#8211; v = 0<br>Solving for&nbsp;v, we get:<br><strong>v = 10 V<\/strong><\/p>\n\n\n\n<p>Next, we apply KVL to the right loop, which contains the central 2 V element, the 8 V source, and the current-controlled dependent voltage source&nbsp;3ix. We will again traverse clockwise, starting from the bottom of the loop.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Moving upward through the dependent source\u00a03ix, we go from the negative to the positive terminal, resulting in a voltage rise of\u00a0+3ix.<\/li>\n\n\n\n<li>Moving across the 8 V source from right to left, we go from negative to positive, which is a voltage rise of\u00a0+8 V.<\/li>\n\n\n\n<li>Moving downward through the central 2 V element, we have a voltage drop of\u00a0\u20132 V.<\/li>\n<\/ol>\n\n\n\n<p>The KVL equation for the right loop is:<br>3ix + 8 V &#8211; 2 V = 0<br>3ix + 6 V = 0<br>Solving for&nbsp;ix:<br>3ix = -6 V<br><strong>ix = -2 A<\/strong><\/p>\n\n\n\n<p>The negative result for&nbsp;ix&nbsp;signifies that the current&#8217;s actual direction is opposite to the one indicated by the arrow in the diagram. Thus, a current of 2 A flows upwards through the central branch.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-122.jpeg\" alt=\"\" class=\"wp-image-243852\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is: The correct answers are&nbsp;v = 10 V&nbsp;and&nbsp;ix = -2 A. To determine the values of the voltage&nbsp;v&nbsp;and the current&nbsp;ix&nbsp;in the given circuit, we can utilize Kirchhoff&#8217;s Voltage Law (KVL). KVL states that the algebraic sum of all the voltages around any closed path, or loop, in a circuit is [&hellip;]<\/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-243846","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243846","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=243846"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243846\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=243846"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=243846"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=243846"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}