The correct answer is i(t) = 52e^(-1.5t) – 26e^(-3t) A<\/strong>.<\/p>\n\n\n\n Explanation<\/strong><\/p>\n\n\n\n To determine the inductor current i(t) for t>0, we follow a three step process: analyze the circuit for t<0 to find initial conditions, characterize the circuit for t>0, and use the initial conditions to find the final solution.<\/p>\n\n\n\n First, for t<0, the switch has been closed for a long time, meaning the circuit is in a DC steady state. In this state, the inductor acts as a short circuit (a wire), and the capacitor acts as an open circuit. The inductor shorts out the parallel 2R resistor, so all the current passing through the first resistor R flows through the inductor. The initial inductor current is i(0) = V\/R = 26V \/ 1 ohm = 26 A. The capacitor is in parallel with the inductor (which is a short circuit), so its initial voltage is v_c(0) = 0 V. Due to their physical properties, inductor current and capacitor voltage cannot change instantaneously, so i(0+) = 26 A and v_c(0+) = 0 V.<\/p>\n\n\n\n Second, for t>0, the switch opens, disconnecting the voltage source and the first resistor. The remaining circuit is a source free parallel RLC circuit consisting of the 2R resistor, the inductor L, and the capacitor C. The response of this circuit is described by a second order differential equation. We must determine the type of response by calculating the neper frequency (alpha) and the resonant frequency (omega_0). For a parallel RLC circuit, alpha = 1 \/ (2 * (2R) * C) = 1 \/ (4RC) = 1 \/ (4 * 1 * 1\/9) = 2.25. The resonant frequency is omega_0 = 1 \/ sqrt(LC) = 1 \/ sqrt(2 * 1\/9) \u2248 2.12. Since alpha > omega_0, the circuit is overdamped.<\/p>\n\n\n\n Third, the general solution for an overdamped response is i(t) = A1e^(s1<\/em>t) + A2e^(s2<\/em>t). The roots are s1,2 = -alpha \u00b1 sqrt(alpha^2 – omega_0^2), which calculates to s1 = -1.5 and s2 = -3. We use the initial conditions to find the constants A1 and A2. In the circuit given below, V=26 V and R=1 ohm. The switch in the circuit has been closed for a long time but is opened at t =0. Determine i(t) for t>0. The Correct Answer and Explanation is: The correct answer is i(t) = 52e^(-1.5t) – 26e^(-3t) A. Explanation To determine the inductor current i(t) for […]<\/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-258121","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258121","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=258121"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258121\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258121"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258121"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
From i(0) = 26 A, we get A1 + A2 = 26.
The second condition comes from di(0)\/dt = v_c(0) \/ L = 0 V \/ 2 H = 0. The derivative is di\/dt = -1.5A1<\/em>e^(-1.5t) – 3A2<\/em>e^(-3t), so at t=0, we have -1.5A1 – 3<\/em>A2 = 0. Solving these two simultaneous equations gives A1 = 52 and A2 = -26. Substituting these constants yields the final expression for the current.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-876.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"