{"id":223555,"date":"2025-06-02T04:03:14","date_gmt":"2025-06-02T04:03:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223555"},"modified":"2025-06-02T04:03:16","modified_gmt":"2025-06-02T04:03:16","slug":"a-series-lr-circuit-contains-an-emf-source-of-14v-having-no-internal-resistance-a-resistor-a-34h-inductor-having-no-appreciable-resistance-and-a-switch","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/a-series-lr-circuit-contains-an-emf-source-of-14v-having-no-internal-resistance-a-resistor-a-34h-inductor-having-no-appreciable-resistance-and-a-switch\/","title":{"rendered":"A series LR circuit contains an emf source of 14V having no internal resistance, a resistor, a 34H inductor having no appreciable resistance, and a switch"},"content":{"rendered":"\n<p>A series LR circuit contains an emf source of 14V having no internal resistance, a resistor, a 34H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, calculate the resistance of the resistor.<\/p>\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<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>emf E=14\u2009VE = 14 \\, V (source voltage)<\/li>\n\n\n\n<li>Inductance L=34\u2009HL = 34 \\, H<\/li>\n\n\n\n<li>Time t=4.0\u2009st = 4.0 \\, s<\/li>\n\n\n\n<li>Voltage across inductor at t=4.0\u2009st = 4.0 \\, s is 80% of its maximum value<\/li>\n\n\n\n<li>No internal resistance in the emf source and negligible resistance in the inductor<\/li>\n\n\n\n<li>Need to find resistance RR<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understand the circuit behavior<\/h3>\n\n\n\n<p>When the switch is closed in an LRLR series circuit, the current builds up according to: i(t)=ER(1\u2212e\u2212RLt)i(t) = \\frac{E}{R} \\left(1 &#8211; e^{-\\frac{R}{L}t}\\right)<\/p>\n\n\n\n<p>The voltage across the inductor is: VL(t)=LdidtV_L(t) = L \\frac{di}{dt}<\/p>\n\n\n\n<p>Taking derivative of i(t)i(t): didt=ER\u22c5RLe\u2212RLt=ELe\u2212RLt\\frac{di}{dt} = \\frac{E}{R} \\cdot \\frac{R}{L} e^{-\\frac{R}{L}t} = \\frac{E}{L} e^{-\\frac{R}{L}t}<\/p>\n\n\n\n<p>Therefore: VL(t)=L\u00d7ELe\u2212RLt=Ee\u2212RLtV_L(t) = L \\times \\frac{E}{L} e^{-\\frac{R}{L}t} = E e^{-\\frac{R}{L}t}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Voltage across the inductor at t=4t = 4 s<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The maximum voltage across the inductor just after closing the switch (at t=0t=0) is equal to the emf E=14\u2009VE = 14 \\, V.<\/li>\n\n\n\n<li>At t=4\u2009st=4 \\, s, the voltage across the inductor is 80% of its maximum:<\/li>\n<\/ul>\n\n\n\n<p>VL(4)=0.8\u00d714=11.2\u2009VV_L(4) = 0.8 \\times 14 = 11.2 \\, V<\/p>\n\n\n\n<p>From the equation above: VL(t)=Ee\u2212RLtV_L(t) = E e^{-\\frac{R}{L} t}<\/p>\n\n\n\n<p>So, 11.2=14e\u2212R34\u00d7411.2 = 14 e^{-\\frac{R}{34} \\times 4}<\/p>\n\n\n\n<p>Divide both sides by 14: 0.8=e\u22124R340.8 = e^{-\\frac{4R}{34}}<\/p>\n\n\n\n<p>Take natural logarithm: ln\u2061(0.8)=\u22124R34\\ln(0.8) = -\\frac{4R}{34} \u22120.2231=\u22124R34-0.2231 = -\\frac{4R}{34}<\/p>\n\n\n\n<p>Multiply both sides by \u22121-1: 0.2231=4R340.2231 = \\frac{4R}{34}<\/p>\n\n\n\n<p>Solve for RR: R=0.2231\u00d7344=7.58544=1.896\u2009\u03a9R = \\frac{0.2231 \\times 34}{4} = \\frac{7.5854}{4} = 1.896 \\, \\Omega<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final answer:<\/strong><\/h3>\n\n\n\n<p>R\u22481.9\u2009\u03a9\\boxed{R \\approx 1.9 \\, \\Omega}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>When an LR circuit is energized by closing the switch, the current through the circuit increases gradually from zero toward its maximum value determined by Ohm&#8217;s law, Imax=ERI_{max} = \\frac{E}{R}. The inductor initially opposes the sudden change in current, and the voltage across it is maximum (equal to the emf) at the instant the switch is closed.<\/p>\n\n\n\n<p>Over time, as the current increases, the voltage across the inductor decreases exponentially, following the relation: VL=Ee\u2212RLtV_L = E e^{-\\frac{R}{L}t}<\/p>\n\n\n\n<p>This exponential decay is governed by the time constant \u03c4=LR\\tau = \\frac{L}{R}. The time constant indicates how quickly the current reaches its steady-state value and how fast the inductor&#8217;s voltage decays.<\/p>\n\n\n\n<p>Given that at 4 seconds the inductor&#8217;s voltage is 80% of its initial value, we use this exponential decay to find the resistance. By taking the natural logarithm of the voltage ratio, we can solve for RR.<\/p>\n\n\n\n<p>Physically, a higher resistance RR causes the current to rise more slowly and the voltage across the inductor to decay more quickly, because the time constant \u03c4\\tau becomes smaller. Conversely, a smaller RR means a slower current increase and slower decay in VLV_L.<\/p>\n\n\n\n<p>In this problem, the resistance is approximately 1.9\u2009\u03a91.9\\, \\Omega, which balances the inductance and time given the voltage decay observed.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner9-5.jpeg\" alt=\"\" class=\"wp-image-223556\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A series LR circuit contains an emf source of 14V having no internal resistance, a resistor, a 34H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, calculate the resistance of the resistor. The Correct Answer and [&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-223555","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223555","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=223555"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223555\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223555"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223555"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}