{"id":195584,"date":"2025-02-28T13:14:37","date_gmt":"2025-02-28T13:14:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195584"},"modified":"2025-02-28T13:15:06","modified_gmt":"2025-02-28T13:15:06","slug":"three-resistors-connected-in-parallel-have-individual-values-of-4-0-6-0-and-10-0-ohms-respectively","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/three-resistors-connected-in-parallel-have-individual-values-of-4-0-6-0-and-10-0-ohms-respectively\/","title":{"rendered":"Three Resistors Connected In Parallel Have Individual Values Of 4.0, 6.0, And 10.0 Ohms Respectively"},"content":{"rendered":"\n

Three Resistors Connected In Parallel Have Individual Values Of 4.0, 6.0, And 10.0 Ohms Respectively. If This Combination Is Connected In Series With A 12.0V Battery And A 2.0- Ohm Resistor, 2.02 12 V 4.0.2 36.00 310.00 What Is The Equivalent Resistance? (Select) What Is The Current In The 10-Ohm Resistor? (Select] What Is Power Produced By The Battery? 1 [<\/p>\n\n\n\n

A) 1.82<\/p>\n\n\n\n

3.94<\/p>\n\n\n\n

22.0<\/p>\n\n\n\n

2.55<\/p>\n\n\n\n

B)0.39<\/p>\n\n\n\n

0.59<\/p>\n\n\n\n

0.65<\/p>\n\n\n\n

0.29<\/p>\n\n\n\n

C)3.0<\/p>\n\n\n\n

12<\/p>\n\n\n\n

37<\/p>\n\n\n\n

24<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Solution Explanation<\/strong><\/h3>\n\n\n\n

We have three resistors connected in parallel with values:<\/p>\n\n\n\n

    \n
  • ( R_1 = 4.0 )\u03a9<\/li>\n\n\n\n
  • ( R_2 = 6.0 )\u03a9<\/li>\n\n\n\n
  • ( R_3 = 10.0 )\u03a9<\/li>\n<\/ul>\n\n\n\n

    This parallel combination is then connected in series<\/strong> with a ( 2.0 )\u03a9 resistor and a 12.0V<\/strong> battery.<\/p>\n\n\n\n

    Step 1: Find the Equivalent Resistance of the Parallel Resistors<\/strong><\/h4>\n\n\n\n

    The formula for equivalent resistance<\/strong> in a parallel circuit is:<\/p>\n\n\n\n

    [
    \\frac{1}{R_{\\text{eq}}} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\frac{1}{R_3}
    ]<\/p>\n\n\n\n

    [
    \\frac{1}{R_{\\text{eq}}} = \\frac{1}{4.0} + \\frac{1}{6.0} + \\frac{1}{10.0}
    ]<\/p>\n\n\n\n

    [
    = 0.25 + 0.1667 + 0.10
    ]<\/p>\n\n\n\n

    [
    = 0.5167
    ]<\/p>\n\n\n\n

    [
    R_{\\text{eq}} = \\frac{1}{0.5167} \\approx 1.93\u03a9
    ]<\/p>\n\n\n\n

    Step 2: Find the Total Resistance in the Circuit<\/strong><\/h4>\n\n\n\n

    Since the parallel combination is in series<\/strong> with a 2.0\u03a9 resistor<\/strong>, the total resistance is:<\/p>\n\n\n\n

    [
    R_{\\text{total}} = R_{\\text{eq}} + R_{\\text{series}}
    ]<\/p>\n\n\n\n

    [
    R_{\\text{total}} = 1.93\u03a9 + 2.0\u03a9 = 3.93\u03a9
    ]<\/p>\n\n\n\n

    Thus, the equivalent resistance<\/strong> is 3.94\u03a9 (rounded)<\/strong>.<\/p>\n\n\n\n

    Step 3: Find the Total Current Using Ohm\u2019s Law<\/strong><\/h4>\n\n\n\n

    Ohm\u2019s Law states:<\/p>\n\n\n\n

    [
    V = IR
    ]<\/p>\n\n\n\n

    [
    I_{\\text{total}} = \\frac{V}{R_{\\text{total}}} = \\frac{12.0V}{3.94\u03a9} \\approx 3.05A
    ]<\/p>\n\n\n\n

    Step 4: Find the Current in the 10\u03a9 Resistor<\/strong><\/h4>\n\n\n\n

    Using the current division formula for parallel circuits:<\/p>\n\n\n\n

    [
    I_{10\u03a9} = I_{\\text{total}} \\times \\frac{R_{\\text{eq}}}{R_3}
    ]<\/p>\n\n\n\n

    [
    I_{10\u03a9} = 3.05A \\times \\frac{1.93\u03a9}{10\u03a9} \\approx 0.59A
    ]<\/p>\n\n\n\n

    Thus, the current in the 10\u03a9 resistor<\/strong> is 0.59A<\/strong>.<\/p>\n\n\n\n

    Step 5: Find the Power Produced by the Battery<\/strong><\/h4>\n\n\n\n

    Power is given by:<\/p>\n\n\n\n

    [
    P = VI
    ]<\/p>\n\n\n\n

    [
    P = 12V \\times 3.05A = 36.6W
    ]<\/p>\n\n\n\n

    Thus, the power produced by the battery<\/strong> is 37W (rounded)<\/strong>.<\/p>\n\n\n\n

    Final Answers<\/strong><\/h3>\n\n\n\n
      \n
    1. Equivalent Resistance<\/strong>: 3.94\u03a9<\/strong><\/li>\n\n\n\n
    2. Current in the 10\u03a9 Resistor<\/strong>: 0.59A<\/strong><\/li>\n\n\n\n
    3. Power Produced by the Battery<\/strong>: 37W<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Now, I will generate an image representing the circuit diagram.<\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      I’ve provided the schematic circuit diagram as per the calculations. Let me know if you need any modifications or further explanations!<\/p>\n","protected":false},"excerpt":{"rendered":"

      Three Resistors Connected In Parallel Have Individual Values Of 4.0, 6.0, And 10.0 Ohms Respectively. If This Combination Is Connected In Series With A 12.0V Battery And A 2.0- Ohm Resistor, 2.02 12 V 4.0.2 36.00 310.00 What Is The Equivalent Resistance? (Select) What Is The Current In The 10-Ohm Resistor? (Select] What Is Power […]<\/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-195584","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195584","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=195584"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195584\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}