{"id":89032,"date":"2025-06-01T07:28:05","date_gmt":"2025-06-01T07:28:05","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89032"},"modified":"2025-06-01T07:28:05","modified_gmt":"2025-06-01T07:28:05","slug":"two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/","title":{"rendered":"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com"},"content":{"rendered":"

Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the combi- nation is connected in parallel with a resistance of 36.0 \u03a9. The equivalent resistance of the combination of three resistors is<\/p>\n

[amp_mcq option1=”24.0 \u03a9” option2=”12.0 \u03a9” option3=”9.0 \u03a9” option4=”6.0 \u03a9” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is C) 9.0 \u03a9.
\n<\/section>\n
\nFirst, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances:
\nR_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9.
\nNext, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances:
\n1\/R_eq = 1\/R_series + 1\/R\u2083
\n1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9
\nTo add the fractions, find a common denominator, which is 36.
\n1\/R_eq = (3\/36) + (1\/36) = 4\/36
\n1\/R_eq = 1\/9
\nR_eq = 9.0 \u03a9.
\n<\/section>\n
\nUnderstanding how to combine resistances in series and parallel is fundamental in circuit analysis. Resistances in series add directly, increasing the total resistance. Resistances in parallel combine in a way that the reciprocal of the total resistance is the sum of the reciprocals, resulting in a lower total resistance than the smallest individual resistance.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the combi- nation is connected in parallel with a resistance of 36.0 \u03a9. The equivalent resistance of the combination of three resistors is [amp_mcq option1=”24.0 \u03a9” option2=”12.0 \u03a9” option3=”9.0 \u03a9” option4=”6.0 \u03a9” correct=”option3″] This question was previously asked in UPSC NDA-2 – … <\/p>\n

Detailed SolutionTwo resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1103,1201,1128],"class_list":["post-89032","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1103","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com<\/title>\n<meta name=\"description\" content=\"The correct answer is C) 9.0 \u03a9. First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances: R_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9. Next, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances: 1\/R_eq = 1\/R_series + 1\/R\u2083 1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9 To add the fractions, find a common denominator, which is 36. 1\/R_eq = (3\/36) + (1\/36) = 4\/36 1\/R_eq = 1\/9 R_eq = 9.0 \u03a9.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-\u03c9-and-7-0-\u03c9-are-connected-in-series-and-the-com\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com\" \/>\n<meta property=\"og:description\" content=\"The correct answer is C) 9.0 \u03a9. First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances: R_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9. Next, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances: 1\/R_eq = 1\/R_series + 1\/R\u2083 1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9 To add the fractions, find a common denominator, which is 36. 1\/R_eq = (3\/36) + (1\/36) = 4\/36 1\/R_eq = 1\/9 R_eq = 9.0 \u03a9.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-\u03c9-and-7-0-\u03c9-are-connected-in-series-and-the-com\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:28:05+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com","description":"The correct answer is C) 9.0 \u03a9. First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances: R_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9. Next, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances: 1\/R_eq = 1\/R_series + 1\/R\u2083 1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9 To add the fractions, find a common denominator, which is 36. 1\/R_eq = (3\/36) + (1\/36) = 4\/36 1\/R_eq = 1\/9 R_eq = 9.0 \u03a9.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-\u03c9-and-7-0-\u03c9-are-connected-in-series-and-the-com\/","og_locale":"en_US","og_type":"article","og_title":"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com","og_description":"The correct answer is C) 9.0 \u03a9. First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances: R_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9. Next, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances: 1\/R_eq = 1\/R_series + 1\/R\u2083 1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9 To add the fractions, find a common denominator, which is 36. 1\/R_eq = (3\/36) + (1\/36) = 4\/36 1\/R_eq = 1\/9 R_eq = 9.0 \u03a9.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-\u03c9-and-7-0-\u03c9-are-connected-in-series-and-the-com\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:28:05+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/","name":"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:28:05+00:00","dateModified":"2025-06-01T07:28:05+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is C) 9.0 \u03a9. First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances: R_series = R\u2081 + R\u2082 = 5.0 \u03a9 + 7.0 \u03a9 = 12.0 \u03a9. Next, this series combination (with R_series = 12.0 \u03a9) is connected in parallel with a third resistance (R\u2083 = 36.0 \u03a9). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances: 1\/R_eq = 1\/R_series + 1\/R\u2083 1\/R_eq = 1\/12.0 \u03a9 + 1\/36.0 \u03a9 To add the fractions, find a common denominator, which is 36. 1\/R_eq = (3\/36) + (1\/36) = 4\/36 1\/R_eq = 1\/9 R_eq = 9.0 \u03a9.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-resistances-of-5-0-%cf%89-and-7-0-%cf%89-are-connected-in-series-and-the-com\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"Two resistances of 5.0 \u03a9 and 7.0 \u03a9 are connected in series and the com"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89032","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=89032"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89032\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=89032"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=89032"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=89032"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}