{"id":263323,"date":"2025-07-21T05:51:28","date_gmt":"2025-07-21T05:51:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=263323"},"modified":"2025-07-21T05:51:30","modified_gmt":"2025-07-21T05:51:30","slug":"calculate-%ce%b2-for-two-transistors-for-which-%ce%b10-99-and-0-98","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/21\/calculate-%ce%b2-for-two-transistors-for-which-%ce%b10-99-and-0-98\/","title":{"rendered":"Calculate \u03b2 for two transistors for which \u03b1=0.99 and 0.98"},"content":{"rendered":"\n<p>Calculate \u03b2 for two transistors for which \u03b1=0.99 and 0.98. For collector currents of 10 mA, find the base current of each transistor.<\/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 calculate \u03b2\\beta\u03b2 (current gain) for two transistors given their \u03b1\\alpha\u03b1 values, we use the following relationship between \u03b1\\alpha\u03b1 and \u03b2\\beta\u03b2:\u03b2=\u03b11\u2212\u03b1\\beta = \\frac{\\alpha}{1 &#8211; \\alpha}\u03b2=1\u2212\u03b1\u03b1\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">For the first transistor with \u03b1=0.99\\alpha = 0.99\u03b1=0.99:<\/h3>\n\n\n\n<p>\u03b2=0.991\u22120.99=0.990.01=99\\beta = \\frac{0.99}{1 &#8211; 0.99} = \\frac{0.99}{0.01} = 99\u03b2=1\u22120.990.99\u200b=0.010.99\u200b=99<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">For the second transistor with \u03b1=0.98\\alpha = 0.98\u03b1=0.98:<\/h3>\n\n\n\n<p>\u03b2=0.981\u22120.98=0.980.02=49\\beta = \\frac{0.98}{1 &#8211; 0.98} = \\frac{0.98}{0.02} = 49\u03b2=1\u22120.980.98\u200b=0.020.98\u200b=49<\/p>\n\n\n\n<p>So, the current gains for the two transistors are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For \u03b1=0.99\\alpha = 0.99\u03b1=0.99, \u03b2=99\\beta = 99\u03b2=99<\/li>\n\n\n\n<li>For \u03b1=0.98\\alpha = 0.98\u03b1=0.98, \u03b2=49\\beta = 49\u03b2=49<\/li>\n<\/ul>\n\n\n\n<p>Now, let\u2019s calculate the base current for each transistor, given that the collector current IC=10\u2009mAI_C = 10 \\, \\text{mA}IC\u200b=10mA.<\/p>\n\n\n\n<p>Using the relationship between the collector current ICI_CIC\u200b, the base current IBI_BIB\u200b, and \u03b2\\beta\u03b2:IC=\u03b2\u22c5IBI_C = \\beta \\cdot I_BIC\u200b=\u03b2\u22c5IB\u200b<\/p>\n\n\n\n<p>Solving for IBI_BIB\u200b:IB=IC\u03b2I_B = \\frac{I_C}{\\beta}IB\u200b=\u03b2IC\u200b\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">For the first transistor with \u03b2=99\\beta = 99\u03b2=99:<\/h3>\n\n\n\n<p>IB=10\u2009mA99\u22480.101\u2009mAI_B = \\frac{10 \\, \\text{mA}}{99} \\approx 0.101 \\, \\text{mA}IB\u200b=9910mA\u200b\u22480.101mA<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">For the second transistor with \u03b2=49\\beta = 49\u03b2=49:<\/h3>\n\n\n\n<p>IB=10\u2009mA49\u22480.204\u2009mAI_B = \\frac{10 \\, \\text{mA}}{49} \\approx 0.204 \\, \\text{mA}IB\u200b=4910mA\u200b\u22480.204mA<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For the transistor with \u03b1=0.99\\alpha = 0.99\u03b1=0.99, \u03b2=99\\beta = 99\u03b2=99, and the base current IB\u22480.101\u2009mAI_B \\approx 0.101 \\, \\text{mA}IB\u200b\u22480.101mA.<\/li>\n\n\n\n<li>For the transistor with \u03b1=0.98\\alpha = 0.98\u03b1=0.98, \u03b2=49\\beta = 49\u03b2=49, and the base current IB\u22480.204\u2009mAI_B \\approx 0.204 \\, \\text{mA}IB\u200b\u22480.204mA.<\/li>\n<\/ul>\n\n\n\n<p>This shows how variations in \u03b1\\alpha\u03b1 affect both the current gain \u03b2\\beta\u03b2 and the base current.<\/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-banner6-1298.jpeg\" alt=\"\" class=\"wp-image-263326\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate \u03b2 for two transistors for which \u03b1=0.99 and 0.98. For collector currents of 10 mA, find the base current of each transistor. The Correct Answer and Explanation is: To calculate \u03b2\\beta\u03b2 (current gain) for two transistors given their \u03b1\\alpha\u03b1 values, we use the following relationship between \u03b1\\alpha\u03b1 and \u03b2\\beta\u03b2:\u03b2=\u03b11\u2212\u03b1\\beta = \\frac{\\alpha}{1 &#8211; \\alpha}\u03b2=1\u2212\u03b1\u03b1\u200b For [&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-263323","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263323","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=263323"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263323\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=263323"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=263323"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=263323"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}