{"id":6967,"date":"2014-12-10T19:48:06","date_gmt":"2014-12-10T19:48:06","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=6967"},"modified":"2018-01-27T23:09:57","modified_gmt":"2018-01-27T23:09:57","slug":"circuit-analysis-the-maximum-power-transfer-theorem","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/circuit-analysis-the-maximum-power-transfer-theorem\/","title":{"rendered":"Circuit analysis:  The maximum power transfer theorem"},"content":{"rendered":"<h1>From time to time, a tutor might get asked questions about electric circuits. \u00a0In the context of tutoring or just for general interest, the maximum power transfer theorem \u00a0is nice.<\/h1>\n<p>I last studied electronics about twelve years ago. \u00a0Ideas from it return to mind now and then. \u00a0Researching for my Nov 10 article on <a href=\"?p=6353\">auto batteries,<\/a> I read a remark that whatever the battery&#8217;s resistance was, the starter should match it to receive maximum possible power.  I recognized the idea as a case of the maximum power transfer theorem:<\/p>\n<p><em>Whatever the resistance of the surrounding circuit, the load resistor should match it in order to receive maximum power.<\/em><\/p>\n<p>Here&#8217;s a proof using calculus:<\/p>\n<p>Let&#8217;s imagine a series circuit with peripheral resistance $R$ and load resistance aR, where a&#8805;0.  Since the two resistances are in series, R<sub>total<\/sub>=R+aR.<\/p>\n<p>Now, since V=IR, we have I=V\/R.  In particular,<\/p>\n<p>I=V\/(R+aR)<\/p>\n<p>Furthermore, the power dissipated by a resistor is given by P=I^2R.  Therefore, the power in the load resistor of our circuit is<\/p>\n<p>P=(V\/(R+aR))^2(aR)=aRV^2\/(R+aR)^2<\/p>\n<p>To find the value of a that gives the maximum value of P, we take the derivative dP\/da, set it equal to zero, and solve for a.<\/p>\n<p>To take the derivative dP\/da, we use the quotient rule:<\/p>\n<p>dP\/da=((R + aR)^2(RV^2) &#8211; aRV^2(2(R + aR)R))\/(R + aR)^4<\/p>\n<p>Set dP\/da to zero and solve for a:<\/p>\n<p>0=((R + aR)^2(RV^2) &#8211; aRV^2(2(R + aR)R))\/(R + aR)^4<\/p>\n<p>Multiply both sides by (R + aR)^4<\/p>\n<p>0=(R + aR)^2(RV^2) &#8211; 2aV^2R^2(R + aR)<\/p>\n<p>Factor R + aR:<\/p>\n<p>0=(R +aR)[(R + aR)RV^2 &#8211; 2aV^2R^2]<\/p>\n<p>Divide out R + aR from both sides:<\/p>\n<p>0=(R + aR)RV^2 &#8211; 2aV^2R^2<\/p>\n<p>Factor out RV^2:<\/p>\n<p>0=RV^2[(R + aR) &#8211; 2aR]<\/p>\n<p>Divide out RV^2:<\/p>\n<p>0=R + aR &#8211; 2aR<\/p>\n<p>Factor out R:<\/p>\n<p>0=R(1 + a -2a)<\/p>\n<p>Divide out R:<\/p>\n<p>0=1 + a -2a<\/p>\n<p>Simplify:<\/p>\n<p>0=1 -a<\/p>\n<p>Finally, <\/p>\n<p>a=1<\/p>\n<p>Let&#8217;s recall that, in our circuit, the peripheral resistance is R, while the load resistance is aR.  We now find that for maximum power, a=1.  It follows that the load resistance should be 1R=R, the same as the peripheral resistance, for maximum power.<\/p>\n<p>The maximum power transfer theorem, while many never encounter it, is a fundamental part of everyday life for many others.  Anticipating what we may need to know in the future is often a challenge&#8230;.<\/p>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC. <\/p>\n","protected":false},"excerpt":{"rendered":"<p>From time to time, a tutor might get asked questions about electric circuits. \u00a0In the context of tutoring or just for general interest, the maximum power transfer theorem \u00a0is nice. I last studied electronics about twelve years ago. \u00a0Ideas from &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/circuit-analysis-the-maximum-power-transfer-theorem\/\"> <span class=\"screen-reader-text\">Circuit analysis:  The maximum power transfer theorem<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[342,297,7],"tags":[343],"class_list":["post-6967","post","type-post","status-publish","format-standard","hentry","category-electric-circuits","category-electronics","category-physics","tag-maximum-power-transfer-theorem"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6967","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=6967"}],"version-history":[{"count":36,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6967\/revisions"}],"predecessor-version":[{"id":28852,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6967\/revisions\/28852"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=6967"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=6967"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=6967"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}