{"id":1636,"date":"2018-03-28T15:03:28","date_gmt":"2018-03-28T07:03:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1636"},"modified":"2023-08-10T00:38:10","modified_gmt":"2023-08-10T00:38:10","slug":"ohms-law-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/ohms-law-problems-and-solutions.htm","title":{"rendered":"Ohm&#8217;s law &#8211; problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Ohm&#8217;s law &#8211; problems and solutions<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">1. <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Based on the table below, the relation between <a href=\"https:\/\/gurumuda.net\/physics\/electric-voltage-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">voltage<\/a> (V), <a href=\"https:\/\/gurumuda.net\/physics\/electric-currents-electric-charges-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">electric current<\/a> (I) and resistance (R) is&#8230;..<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-1637\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Ohms-law-problems-and-solutions-1.png\" alt=\"Ohm's law - problems and solutions 1\" width=\"243\" height=\"70\" \/>Solution :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Ohm&#8217;s law :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">V = I R or R = V \/ I<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">R<sub>1<\/sub> = V \/ I = 1.50 \/ 0.08 = 18.75 Ohm<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">R<sub>2<\/sub> = V \/ I = 2.80 \/ 1.50 = 2.87 Ohm<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">R<sub>3 <\/sub>= V \/ I = 3.99 \/ 2.10 = 1.9 Ohm <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Based on the table, the relationship between V, I and R is the value of R inversely proportional to V and I. That means the larger the V and I, the smaller R.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The value of V \/ I is not constant so it does not illustrate <a href=\"https:\/\/gurumuda.net\/physics\/ohms-law-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">Ohm&#8217;s law<\/a>. According to Ohm&#8217;s law, the value of V \/ I must be constant.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">2. <span lang=\"en-US\">Which graph indicates the relationship between the potential difference with the electric current?<\/span><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1638\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Ohms-law-problems-and-solutions-2.png\" alt=\"Ohm's law - problems and solutions 2\" width=\"264\" height=\"163\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Solution :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">A<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">3. <span lang=\"en-US\">Based on the data in the following <\/span><span lang=\"en-US\">table of the <\/span><span lang=\"en-US\">experiment result, it can be concluded that the electric current &#8230;<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-1639\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/03\/Ohms-law-problems-and-solutions-3.png\" alt=\"Ohm's law - problems and solutions 3\" width=\"238\" height=\"102\" \/>Solution :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Based on the data in the table, the greater the electrical voltage, the greater the electric current. Therefore it is concluded that the electric current is proportional to the voltage.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">4. A device has a 150 Ohm resistance and a 2A electric current passes. The <a href=\"https:\/\/gurumuda.net\/physics\/electric-potential.htm\" target=\"_blank\" rel=\"noopener\">potential difference<\/a> is&#8230;<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Resistance (R) = 150 Ohm<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The electric current (I) = 2 Ampere<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> Voltage (V)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">V = I R<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>V = electric voltage, I = electric current, R = electric resistance<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Voltage :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">V = I R = (2 Ampere)(150 Ohm) = 300 Volt<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">5. <span lang=\"en-US\">The electric current in a resistor wire is 4 A. When both ends are given a potential of 12 Volts. What is the electrical resistance?<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Electric current (I) = 4 Ampere<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Voltage (V) = 12 Volt<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Electric resistance (R)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Resistance :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">R = V \/ I = 12 Volt \/ 4 Ampere = 3 Ohm<\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is Ohm&#8217;s Law?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Ohm&#8217;s Law states that the current (<span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><\/span><\/span><\/span><\/span>) flowing through a conductor between two points is directly proportional to the voltage (<span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span>) across the two points, and inversely proportional to the resistance (<span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span><\/span>) of the conductor. Mathematically, it&#8217;s represented as <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">V<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span><\/span>.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>If a resistor follows Ohm&#8217;s Law, what kind of relationship should we observe between voltage and current?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: We should observe a linear relationship. If you graph voltage against current, the slope of the line would represent the resistance, and the line would pass through the origin.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does increasing the resistance in a circuit affect the current, assuming voltage remains constant?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: If the resistance increases while voltage remains constant, the current in the circuit will decrease, as per Ohm&#8217;s Law.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why might Ohm&#8217;s Law not apply under certain conditions or for certain materials?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Ohm&#8217;s Law applies to &#8220;ohmic&#8221; materials where resistance remains constant with changing voltage. However, for some materials and devices (like diodes, or at very high frequencies or temperatures), resistance may change with voltage or current, making the relationship non-linear.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the unit of resistance, and how is it defined in terms of Ohm&#8217;s Law?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The unit of resistance is the ohm (\u03a9). One ohm is defined as the resistance that will allow one ampere of current to flow when one volt is applied across it.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>If two resistors in series both strictly obey Ohm&#8217;s Law, will the combination also obey Ohm&#8217;s Law?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: Yes, the combination will also obey Ohm&#8217;s Law. The total resistance in series is the sum of the individual resistances, and the current through both will be the same. The voltage across the combination will be the sum of the voltages across each resistor.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the resistance of a wire depend on its length and cross-sectional area?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Thus, a longer wire will have greater resistance, and a thicker wire will have less resistance.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>If the voltage across a resistor is doubled, how does the current change, assuming the resistor remains ohmic?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: If the resistor remains ohmic (and thus follows Ohm&#8217;s Law), doubling the voltage will also double the current.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Can a component with zero resistance be realized in practice?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: In practical everyday scenarios, a component with zero resistance, or a &#8220;superconductor&#8221;, cannot be easily realized. However, certain materials at extremely low temperatures can become superconductors and offer zero resistance to current flow.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the power dissipation of a resistor in terms of Ohm&#8217;s Law?<\/strong><\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer<\/strong>: The power (<span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><\/span><\/span><\/span><\/span>) dissipated by a resistor is given by <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">I<\/span><sup><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/sup><\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span><\/span> or <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">V<\/span><sup><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/sup><\/span><span class=\"mord\">\/<\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span><\/span>, derived from the combination of Ohm&#8217;s Law and the power formula <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span>.<\/span><\/li>\n<\/ul>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Ohm&#8217;s law &#8211; problems and solutions 1. Based on the table below, the relation between voltage (V), electric current (I) and resistance (R) is&#8230;.. Solution : Ohm&#8217;s law : V = I R or R = V \/ I R1 = V \/ I = 1.50 \/ 0.08 = 18.75 Ohm R2 = V \/ &#8230; <a title=\"Ohm&#8217;s law &#8211; problems and solutions\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/ohms-law-problems-and-solutions.htm\" aria-label=\"Read more about Ohm&#8217;s law &#8211; problems and solutions\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_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":"Ohm&#039;s law - problems and solutions","_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":[3],"tags":[],"class_list":["post-1636","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1636","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=1636"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1636\/revisions"}],"predecessor-version":[{"id":8737,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1636\/revisions\/8737"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1636"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1636"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1636"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}