{"id":84773,"date":"2025-06-01T02:49:50","date_gmt":"2025-06-01T02:49:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=84773"},"modified":"2025-06-01T02:49:50","modified_gmt":"2025-06-01T02:49:50","slug":"a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/","title":{"rendered":"A wire of copper having length l and area of cross-section A is taken"},"content":{"rendered":"

A wire of copper having length l and area of cross-section A is taken and a current I is flown through it. The power dissipated in the wire is P. If we take an aluminum wire having same dimensions and pass the same current through it, the power dissipated will be<\/p>\n

[amp_mcq option1=”P” option2=”< P” option3=”> P” option4=”2P” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-1 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section.
\nFor the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P.
\nFor the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A).
\nTo compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper.
\nSince P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire.
\n<\/section>\n
\n– Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI).
\n– Resistance depends on the material’s resistivity and the wire’s dimensions (R = \u03c1l\/A).
\n– Aluminum has higher resistivity than copper.
\n<\/section>\n
\nCopper is widely used in electrical wiring due to its low resistivity and ductility. Aluminum is also used, particularly in high-voltage transmission lines, as it is lighter and cheaper than copper, despite having a higher resistivity. The difference in resistivity directly impacts the power loss due to heating (Joule heating) for the same current and dimensions.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A wire of copper having length l and area of cross-section A is taken and a current I is flown through it. The power dissipated in the wire is P. If we take an aluminum wire having same dimensions and pass the same current through it, the power dissipated will be [amp_mcq option1=”P” option2=”< P” … <\/p>\n

Detailed SolutionA wire of copper having length l and area of cross-section A is taken<\/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":[1087],"tags":[1114,1201,1128],"class_list":["post-84773","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-1","tag-1114","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nA wire of copper having length l and area of cross-section A is taken<\/title>\n<meta name=\"description\" content=\"The power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section. For the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P. For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A). To compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper. Since P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire. - Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = \u03c1l\/A). - Aluminum has higher resistivity than copper.\" \/>\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\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A wire of copper having length l and area of cross-section A is taken\" \/>\n<meta property=\"og:description\" content=\"The power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section. For the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P. For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A). To compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper. Since P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire. - Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = \u03c1l\/A). - Aluminum has higher resistivity than copper.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T02:49:50+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A wire of copper having length l and area of cross-section A is taken","description":"The power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section. For the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P. For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A). To compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper. Since P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire. - Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = \u03c1l\/A). - Aluminum has higher resistivity than copper.","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\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/","og_locale":"en_US","og_type":"article","og_title":"A wire of copper having length l and area of cross-section A is taken","og_description":"The power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section. For the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P. For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A). To compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper. Since P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire. - Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = \u03c1l\/A). - Aluminum has higher resistivity than copper.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T02:49:50+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/","name":"A wire of copper having length l and area of cross-section A is taken","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T02:49:50+00:00","dateModified":"2025-06-01T02:49:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The power dissipated in a wire is given by P = I\u00b2R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity of the material, l is the length, and A is the area of cross-section. For the copper wire, P = I\u00b2 * R_copper = I\u00b2 * (\u03c1_copper * l \/ A). We are given this power is P. For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I\u00b2 * R_aluminum = I\u00b2 * (\u03c1_aluminum * l \/ A). To compare P_aluminum and P, we need to compare the resistivities of aluminum (\u03c1_aluminum) and copper (\u03c1_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, \u03c1_aluminum > \u03c1_copper. Since P_aluminum is proportional to \u03c1_aluminum (for fixed I, l, A) and P is proportional to \u03c1_copper, and \u03c1_aluminum > \u03c1_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire. - Power dissipated in a resistor is proportional to its resistance (P = I\u00b2R, or P = V\u00b2\/R, or P = VI). - Resistance depends on the material's resistivity and the wire's dimensions (R = \u03c1l\/A). - Aluminum has higher resistivity than copper.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-copper-having-length-l-and-area-of-cross-section-a-is-taken\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-1\/"},{"@type":"ListItem","position":3,"name":"A wire of copper having length l and area of cross-section A is taken"}]},{"@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\/84773","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=84773"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/84773\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=84773"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=84773"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=84773"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}