{"id":4899,"date":"2021-06-27T15:53:09","date_gmt":"2021-06-27T22:53:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=4899"},"modified":"2021-06-27T15:53:09","modified_gmt":"2021-06-27T22:53:09","slug":"kirchhoffs-second-rule","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/kirchhoffs-second-rule.htm","title":{"rendered":"Kirchhoffs second rule","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Kirchhoff&#8217;s second rule states that the change in electric potential on the circumference of a closed circuit is zero. Kirchhoff&#8217;s second rule is based on the law of conservation of energy, which states that energy is eternal.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-4901\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/11\/Kirchhoff\u2019s-second-rule-1.png\" alt=\"Kirchhoff\u2019s second rule 1\" width=\"154\" height=\"98\" \/>To better understand this, imagine the electric charge moving in a closed circuit, as in the figure. When an electric charge passes through an <a href=\"https:\/\/gurumuda.net\/physics\/electric-resistance.htm\">electrical resistance<\/a> (R), the <a href=\"https:\/\/gurumuda.net\/physics\/electric-potential-energy-problems-and-solutions.htm\">electrical potential energy<\/a> is reduced because it is used on these resistances. If the electric charge passes through another electrical resistance, the electric potential energy decreases again because it is used again on the resistance. Furthermore, when the electric charge passes through the voltage source from a low potential to a high potential, the electric potential energy increases. When it returns to its original point, the electric potential energy is the same as before, where the change in electrical potential energy is zero. When applying <a href=\"https:\/\/gurumuda.net\/physics\/kirchhoff-law-problems-and-solutions.htm\">Kirchhoff<\/a>&#8216;s second rule to an electrical circuit, we use the change in electrical voltage, not the change in electrical potential energy.<\/span><\/span><!--more--><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-4902\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/11\/Kirchhoff\u2019s-second-rule-2.png\" alt=\"Kirchhoff\u2019s second rule 2\" width=\"138\" height=\"131\" \/>The circuit as in the figure above can be analyzed using Ohm&#8217;s law (V = I R) and the equation for series resistance or parallel resistance. More complex circuits such as the figure can be analyzed using Kirchhoff&#8217;s first rule and Kirchhoff&#8217;s second rule.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Here are some significant rules when using Kirchhoff&#8217;s second rule to analyze a circuit.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>First<\/i><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">, choose the current direction, whether clockwise or counterclockwise. If the answer is negative, the actual direction of the electric current is in the direction of the current decided. If the answer is positive, the actual direction of the electric current is the same as the direction of the selected current.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>Second<\/i><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">, when passing through a voltage source from a high potential to a low potential (to -), there is a decrease in electrical voltage so that the voltage is negative (\u0394V = -\u03b5). Conversely, if it moves from a low potential to a high potential (- to), there is an increase in electrical voltage so that the voltage is positive (\u0394V = \u03b5).<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>Third<\/i><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">, when it goes through electrical resistance, if the loop is in the direction of the electric current, the voltage is negative (V = &#8211; I R). Conversely, if the loop direction is opposite to the direction of the electric current, the electric voltage is positive (\u0394V = I R).<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Sample problem 1:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-4903\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/11\/Kirchhoff\u2019s-second-rule-3.png\" alt=\"Kirchhoff\u2019s second rule 3\" width=\"153\" height=\"98\" \/>R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 200 \u03a9 and R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 300 \u03a9. \u03b5 = 12 Volt. Calculate the electric current flowing in the circuit using <a href=\"https:\/\/en.wikipedia.org\/wiki\/Gustav_Kirchhoff\">Kirchhoff<\/a>&#8216;s second rule!<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Solution:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The direction of the clockwise direction.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; I R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> \u2013 I R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + \u03b5 = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 200 I \u2013 300 I + 12 = 0.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 500 I + 12 = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 500 I = &#8211; 12<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">I = 12 \/ 500 <\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">I = 0.024 A<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">If calculated using Ohm&#8217;s law and the formula for the combination of resistors in series:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">R = R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 200 + 300 = 500 <\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u03a9.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Equation of Ohm&#8217;s law: <\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">V = I R<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">I = V \/ R = 12 \/ 500 = 0.024 A.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Sample problem 2:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 100 \u03a9, R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 200 \u03a9, R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">3<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 300 \u03a9, \u03b5<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 9 Volt, \u03b5<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 12 Volt. Calculate the magnitude and direction of the electric current in the circuit!<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Solution:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><img loading=\"lazy\" decoding=\"async\" class=\"alignleft size-full wp-image-4904\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/11\/Kirchhoff\u2019s-second-rule-4.png\" alt=\"Kirchhoff\u2019s second rule 4\" width=\"157\" height=\"126\" \/>Select the direction of electric current and the direction of the loop is the same as the clockwise direction.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; I R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> \u2013 I R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> \u2013 <\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u03b5<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> \u2013 I R<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">3<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + <\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u03b5<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 100 I \u2013 200 I \u2013 9 \u2013 300 I + 12 = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 100 I \u2013 200 I \u2013 300 I + 3 = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 600 I + 3 = 0<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">&#8211; 600 I = &#8211; 3<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">I = 3 \/ 600<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">I = 0.005 A<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Electric current is positive so that the direction is the same as the direction chosen.<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Kirchhoff&#8217;s second rule states that the change in electric potential on the circumference of a closed circuit is zero. Kirchhoff&#8217;s second rule is based on the law of conservation of energy, which states that energy is eternal. To better understand this, imagine the electric charge moving in a closed circuit, as in the figure. When &#8230; <a title=\"Kirchhoffs second rule\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/kirchhoffs-second-rule.htm\" aria-label=\"Read more about Kirchhoffs second rule\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"Kirchhoffs second rule states that the change in electric potential on the circumference of a closed circuit is zero","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Kirchhoffs second rule","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[2],"tags":[],"class_list":["post-4899","post","type-post","status-publish","format-standard","hentry","category-basic-physics-tutorials"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4899","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=4899"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4899\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=4899"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=4899"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=4899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}