{"id":250040,"date":"2025-07-10T07:07:37","date_gmt":"2025-07-10T07:07:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=250040"},"modified":"2025-07-10T07:07:39","modified_gmt":"2025-07-10T07:07:39","slug":"the-switch-in-the-circuit-has-been-in-position-a-for-a-long-time","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/the-switch-in-the-circuit-has-been-in-position-a-for-a-long-time\/","title":{"rendered":"The switch in the circuit has been in position &#8220;a&#8221; for a long time."},"content":{"rendered":"\n<p>The switch in the circuit has been in position &#8220;a&#8221; for a long time. At time t = 0, it moves instantaneously to position &#8220;b&#8221;. For t \u2265 0+, find: v0 (t) = 11. 80 1\/1 points + 12. -40 1\/1 points e( 13. -1000 1\/1 points t) V i0 (t) = 14. 0<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>It seems like you&#8217;re dealing with an RL circuit where the switch changes position at time t=0t = 0t=0, transitioning from position &#8220;a&#8221; to position &#8220;b&#8221;. We need to calculate the voltage and current as functions of time after the switch moves. This type of circuit is typically analyzed using differential equations for the circuit components.<\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The initial voltage v0(t)v_0(t)v0\u200b(t) and initial current i0(t)i_0(t)i0\u200b(t).<\/li>\n\n\n\n<li>The voltage for t\u22650t \\geq 0t\u22650 is expressed as v0(t)=11.80\u221240e\u22121000tv_0(t) = 11.80 &#8211; 40 e^{-1000t}v0\u200b(t)=11.80\u221240e\u22121000t volts.<\/li>\n\n\n\n<li>The current for t\u22650t \\geq 0t\u22650 is i0(t)=14.0i_0(t) = 14.0i0\u200b(t)=14.0 A.<\/li>\n<\/ul>\n\n\n\n<p><strong>Steps for analysis:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Initial condition at t=0t = 0t=0:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The voltage across the inductor or resistor at t=0t = 0t=0 must be calculated from the function v0(t)v_0(t)v0\u200b(t). At t=0t = 0t=0, v0(0)=11.80\u221240e0=11.80\u221240=\u221228.20v_0(0) = 11.80 &#8211; 40 e^{0} = 11.80 &#8211; 40 = -28.20v0\u200b(0)=11.80\u221240e0=11.80\u221240=\u221228.20 V.<\/li>\n\n\n\n<li>The current i0(t)i_0(t)i0\u200b(t) is a constant after t=0t = 0t=0, so i0(t)=14.0i_0(t) = 14.0i0\u200b(t)=14.0 A for all t\u22650t \\geq 0t\u22650.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Exponential term interpretation:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The term \u221240e\u22121000t-40 e^{-1000t}\u221240e\u22121000t suggests an exponentially decaying voltage over time, which is common in RL circuits as the inductor resists sudden changes in current.<\/li>\n\n\n\n<li>The rate of decay is determined by the time constant \u03c4=LR\\tau = \\frac{L}{R}\u03c4=RL\u200b, which controls how quickly the current or voltage approaches its steady-state value.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Current through the inductor:<\/strong>\n<ul class=\"wp-block-list\">\n<li>For an RL circuit, the current is governed by the equation i(t)=i0(1\u2212e\u2212t\/\u03c4)i(t) = i_0 \\left(1 &#8211; e^{-t\/\\tau} \\right)i(t)=i0\u200b(1\u2212e\u2212t\/\u03c4), where i0i_0i0\u200b is the final steady-state current.<\/li>\n\n\n\n<li>The current i0(t)i_0(t)i0\u200b(t) is given as constant, suggesting that the circuit is already in a steady state, and the voltage will approach a constant value after a certain time.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Conclusion:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The voltage decays over time, and the current remains constant after t=0t = 0t=0. This behavior is characteristic of RL circuits where the inductor initially opposes changes in current but eventually settles to a steady state.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Thus, v0(t)=11.80\u221240e\u22121000tv_0(t) = 11.80 &#8211; 40 e^{-1000t}v0\u200b(t)=11.80\u221240e\u22121000t and i0(t)=14.0i_0(t) = 14.0i0\u200b(t)=14.0 A represent the voltage and current for t\u22650t \\geq 0t\u22650, respectively.<\/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-banner6-81.jpeg\" alt=\"\" class=\"wp-image-250041\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The switch in the circuit has been in position &#8220;a&#8221; for a long time. At time t = 0, it moves instantaneously to position &#8220;b&#8221;. For t \u2265 0+, find: v0 (t) = 11. 80 1\/1 points + 12. -40 1\/1 points e( 13. -1000 1\/1 points t) V i0 (t) = 14. 0 The [&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-250040","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250040","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=250040"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250040\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=250040"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=250040"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=250040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}