{"id":224193,"date":"2025-06-02T19:01:03","date_gmt":"2025-06-02T19:01:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224193"},"modified":"2025-06-02T19:01:05","modified_gmt":"2025-06-02T19:01:05","slug":"a-45-mh-ideal-inductor-is-connected-in-series-with-a-60-%cf%89-resistor-through-an-ideal-15-v-dc-power-supply-and-an-open-switch-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/a-45-mh-ideal-inductor-is-connected-in-series-with-a-60-%cf%89-resistor-through-an-ideal-15-v-dc-power-supply-and-an-open-switch-2\/","title":{"rendered":"A 45-mH ideal inductor is connected in series with a 60-\u03a9 resistor through an ideal 15-V DC power supply and an open switch"},"content":{"rendered":"\n<p>A 45-mH ideal inductor is connected in series with a 60-\u03a9 resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?<\/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<p>To find the current through an RL (resistor-inductor) circuit after a given time, we use the following formula for the <strong>current in an RL circuit connected to a DC source<\/strong>: I(t)=VR(1\u2212e\u2212t\/\u03c4)I(t) = \\frac{V}{R} \\left(1 &#8211; e^{-t\/\\tau}\\right)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>I(t)I(t) is the current at time tt,<\/li>\n\n\n\n<li>VV is the voltage of the power supply (15 V),<\/li>\n\n\n\n<li>RR is the resistance (60 \u03a9),<\/li>\n\n\n\n<li>LL is the inductance (45 mH = 0.045 H),<\/li>\n\n\n\n<li>\u03c4=LR\\tau = \\frac{L}{R} is the time constant of the RL circuit,<\/li>\n\n\n\n<li>ee is Euler\u2019s number (~2.718),<\/li>\n\n\n\n<li>tt is the time since the switch was closed (7.0 ms = 0.007 s).<\/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: Calculate the time constant \u03c4\\tau<\/h3>\n\n\n\n<p>\u03c4=LR=0.04560=0.00075\u2009s=0.75\u2009ms\\tau = \\frac{L}{R} = \\frac{0.045}{60} = 0.00075 \\, \\text{s} = 0.75 \\, \\text{ms}<\/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: Plug into the formula<\/h3>\n\n\n\n<p>I(t)=1560(1\u2212e\u22120.007\/0.00075)I(t) = \\frac{15}{60} \\left(1 &#8211; e^{-0.007\/0.00075} \\right) I(t)=0.25(1\u2212e\u22129.33)I(t) = 0.25 \\left(1 &#8211; e^{-9.33} \\right)<\/p>\n\n\n\n<p>Since e\u22129.33\u22488.84\u00d710\u22125e^{-9.33} \\approx 8.84 \\times 10^{-5}, we get: I(t)=0.25(1\u22128.84\u00d710\u22125)I(t) = 0.25 \\left(1 &#8211; 8.84 \\times 10^{-5} \\right) I(t)\u22480.25\u00d70.9999116\u22480.249978\u2009AI(t) \\approx 0.25 \\times 0.9999116 \\approx 0.249978 \\, \\text{A}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<p>\\boxed{0.250 \\, \\text{A}} \\] (rounded to three significant figures) &#8212; ### Explanation (300 words): When an inductor is suddenly connected in series with a resistor and a DC power source, the current does not immediately jump to its final value. This is due to the inductor\u2019s property of opposing changes in current. The opposition arises from the self-induced EMF generated when the current through the inductor changes. The inductor essentially &#8220;resists&#8221; the change, causing the current to rise gradually, not instantly. The rate at which current increases is governed by the **time constant** \\( \\tau = \\frac{L}{R} \\), which represents the time it takes for the current to reach about 63.2% of its maximum value. After several time constants, the current gets very close to its steady-state value \\( I = \\frac{V}{R} \\). In this case, with a 45-mH inductor and 60-\u03a9 resistor, the time constant is 0.75 ms. At 7 ms (roughly 9.33 time constants), the current has had plenty of time to rise close to its maximum value. Mathematically, at this point, the exponential decay term \\( e^{-t\/\\tau} \\) becomes nearly zero. Therefore, the current approaches: \\[ I = \\frac{V}{R} = \\frac{15}{60} = 0.25 \\, \\text{A}<\/p>\n\n\n\n<p>This shows that by 7 ms, the current has effectively reached steady state, and the inductor is now acting like a short circuit, allowing full current to flow as though only the resistor were in the circuit.<\/p>\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-banner5-42.jpeg\" alt=\"\" class=\"wp-image-224194\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A 45-mH ideal inductor is connected in series with a 60-\u03a9 resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later? The Correct Answer and Explanation is: To find the current through an RL (resistor-inductor) [&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-224193","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224193","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=224193"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224193\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224193"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224193"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224193"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}