{"id":6231,"date":"2022-05-02T13:19:58","date_gmt":"2022-05-02T07:49:58","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=6231"},"modified":"2024-09-30T08:10:12","modified_gmt":"2024-09-30T02:40:12","slug":"cube-resistance-problem-jee-main-jee-advanced","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/cube-resistance-problem-jee-main-jee-advanced\/","title":{"rendered":"Cube Resistance Problem | JEE Main | JEE Advanced"},"content":{"rendered":"\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t

For a cube of equal resistance along its edges, let’s determine the equivalent resistance between<\/p>

  1. two vertices along the body diagonal<\/a><\/li>
  2. two vertices along the face diagonal<\/a><\/li>
  3. two adjacent vertices<\/a><\/li><\/ol>

    Note that we will explore two methods here: (a) Equivalent circuit, utilizing symmetry (b) Potential method<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    1. Cube Resistance Problem | Equivalent resistance between two vertices along the 'body' diagonal<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    Approach 1 : Equivalent circuit, utilizing symmetry<\/h4>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Equivalent resistance between two vertices along the body diagonal is $= \\cfrac{5R}{6}$. The equivalent representations are shown below which will help you understand the approach. Note that vertices $B$, $D$, and $E$ are at the same potential because of symmetry and similarly $C$, $F$ and $H$ are at the same potential.\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    Approach 2 : Current Method <\/h4>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Assume that the current flowing into vertex $A$ is $i$. Now because of the symmetry we can assign the current in different branches as follows<\/p>

    Now, to establish Potential difference, $V_{AG}$, between vertices $A$ and $G$ let’s follow the path $A – B – C – G$ and let’s write down the potential drop<\/p>

    So, $V_{AG} = \\cfrac{i}{3} R +\u00a0\\cfrac{i}{6} R +\u00a0\\cfrac{i}{3} R$ $= \\cfrac{5R}{6} i$<\/p>

    So $R_{eq} = \\cfrac{V_{AG}}{i} = \\cfrac{5R}{6}$\u00a0<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    2. Cube Resistance Problem | Equivalent resistance between two vertices along the 'face' diagonal<\/span><\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    Approach 1 : Equivalent circuit<\/h4>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Equivalent resistance between two vertices along the face diagonal is $= \\cfrac{3R}{4}$.\u00a0The equivalent representations are shown below which will help you understand the approach.\u00a0<\/span><\/p>

    Note that circuit is symmetric about $B-F-H-D$ as inidicated by red dotted line, meaning all these vertices are at the same potential which helps us simplify the circuit as shown below<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    Approach 2 : Potential Method<\/h4>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Having assigned potentials to different vertices (taking into account the symmetry), we will now write the junction rule for vertices $D$, $E$ and $F$, which will give us 3 equations in 3 unknowns, namely, $V_1$, $V_2$ and $V_3$<\/p>

    So, writing the junction rule at $E$ will give us
    $\\cfrac{V-V_2}{R} = \\cfrac{V_2-V_3}{R} + \\cfrac{V_2-V_3}{R}$<\/p>

    At junction $H$, we will have
    $\\cfrac{V_2-V_3}{R} = \\cfrac{V_3-V_1}{R} + \\cfrac{V_3-(V-V_2)}{R}$<\/p>

    At junction $D$, we will have
    $\\cfrac{V-V_1}{R} = \\cfrac{V_1-V_3}{R} + \\cfrac{V_1}{R}$<\/p>

    And writing the equivalent resistance equation for junction $A$ we get
    $\\cfrac{V}{R_{eq}} = \\cfrac{V – V_2}{R} + \\cfrac{V-V_1}{R} + \\cfrac{V-V_1}{R}$ $\\cdots$ (1)<\/p>

    Solving for $V_1$ and $V_2$ in terms of $V$ and substituting them in equation $1$ we get,<\/p>

    $R_{eq} = \\cfrac{3R}{4}$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    3. Cube Resistance Problem | Equivalent circuit between two adjacent vertices<\/span><\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t

    Approach 1 : Equivalent circuit<\/h4>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Equivalent resistance between two adjacent vertices $= \\cfrac{7R}{12}$.\u00a0The equivalent representations are shown below which will help you understand the approach.\u00a0<\/span><\/p>

    Note that branch 2 and 3 have horizontal symmetry (as indicated by dotted red line) which helps us simplify the circuit as shown below<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Cube\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

    For a cube of equal resistance along its edges, determine the equivalent resistance between two vertices (adjacent or along body \/facediagonal)<\/p>\n","protected":false},"author":1,"featured_media":6232,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"0","ocean_second_sidebar":"0","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"0","ocean_custom_header_template":"0","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"0","ocean_menu_typo_font_family":"0","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"on","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"off","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"0","cybocfi_hide_featured_image":"yes","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"off","ocean_gallery_id":[],"footnotes":""},"categories":[16],"tags":[],"class_list":["post-6231","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jee-physics-chapter-wise-approach","entry","has-media"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/6231","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/comments?post=6231"}],"version-history":[{"count":1,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/6231\/revisions"}],"predecessor-version":[{"id":9427,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/6231\/revisions\/9427"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media\/6232"}],"wp:attachment":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media?parent=6231"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/categories?post=6231"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/tags?post=6231"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}