{"id":12248,"date":"2016-01-30T05:21:35","date_gmt":"2016-01-30T04:21:35","guid":{"rendered":"https:\/\/kamerpower.com\/?p=12248"},"modified":"2025-01-29T19:55:24","modified_gmt":"2025-01-29T18:55:24","slug":"sujets-corriges-concours-ensp-yaounde-2012-physique-chimie","status":"publish","type":"post","link":"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/","title":{"rendered":"Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_2 counter-hierarchy ez-toc-counter ez-toc-custom ez-toc-container-direction\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Contents<\/p>\n<label for=\"ez-toc-cssicon-toggle-item-69dbee7192820\" class=\"ez-toc-cssicon-toggle-label\"><span class=\"ez-toc-cssicon\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/label><input type=\"checkbox\"  id=\"ez-toc-cssicon-toggle-item-69dbee7192820\"  aria-label=\"Toggle\" \/><nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/#sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie-ecole-nationale-polytechnique-de-yaounde\" >Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie\u00a0Ecole Nationale Polytechnique de Yaound\u00e9:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/#kamerpower-probleme-1\" >Kamerpower Probl\u00e8me\u00a01<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/#kamerpower-solution-au-probleme-1\" >Kamerpower Solution au probl\u00e8me 1\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/#probleme-2-%e2%80%93-sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie\" >Probl\u00e8me\u00a02 &#8211;\u00a0Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/kamerpower.com\/fr\/sujets-corriges-concours-ensp-yaounde-2012-physique-chimie\/#solution-au-probleme-2-%e2%80%93-sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie\" >Solution au probl\u00e8me 2 &#8211;\u00a0Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie<\/a><\/li><\/ul><\/nav><\/div>\n<h3 class=\"post-title single-title entry-title\" style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie-ecole-nationale-polytechnique-de-yaounde\"><\/span><span style=\"color: #ff0000;\"><strong>Sujets Corrig\u00e9s <a style=\"color: #ff0000;\" href=\"https:\/\/kamerpower.com\/fr\/concours-ensp-2015-ouverture-concours-dentree-en-premiere-annee-du-cycle-des-ingenieurs-de-conception-universite-de-yaounde-1\/\">Concours ENSP Yaounde<\/a> 2012 Physique &#8211; Chimie\u00a0Ecole Nationale Polytechnique de Yaound\u00e9:<\/strong><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<hr \/>\n<h3 class=\"post-title single-title entry-title\" style=\"text-align: center;\"><\/h3>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"kamerpower-probleme-1\"><\/span><strong><span style=\"color: #3366ff;\">Kamerpower Probl\u00e8me\u00a01<\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Sujets Corrig\u00e9s Concours ENSP Yaounde<\/strong>: Un avion dont l\u2019h\u00e9lice a <span style=\"color: #ff0000;\">2 m<\/span> de diam\u00e8tre avance sur une piste \u00e0 la vitesse de <span style=\"color: #ff0000;\">280km\/h<\/span>, alors que son h\u00e9lice tourne \u00e0 une vitesse de <span style=\"color: #ff0000;\">3000trs\/min<\/span>. Quel est le module de la vitesse de l\u2019extr\u00e9mit\u00e9 d\u2019une pale d\u2019h\u00e9lice par rapport \u00e0 un observateur fixe au sol\u00a0? kamerpower.com<\/span><\/p>\n<hr \/>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"kamerpower-solution-au-probleme-1\"><\/span><strong><span style=\"color: #ff0000;\">Kamerpower Solution au probl\u00e8me 1\u00a0<\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Remarquons d\u2019abord que l\u2019h\u00e9lice n\u2019est pas au dessus de l\u2019avion comme pour les h\u00e9licopt\u00e8res,<\/span><\/p>\n<p><span style=\"color: #000000;\">OM = OO<sub>1<\/sub> + O<sub>1<\/sub>M<\/span><\/p>\n<p><span style=\"color: #000000;\">dOM\/dt \u00a0= \u00a0(d<strong>OO<sub>1<\/sub><\/strong>\/dt) \u00a0+ (d<strong>O<sub>1<\/sub><\/strong><strong>M<\/strong>)\/dt<\/span><\/p>\n<p><span style=\"color: #000000;\">V = V<sub>1<\/sub> + V<sub>2<\/sub><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">C&#8217;est aussi clair qu\u2019\u00e0 chaque instant,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u00a0 V<sub>2<\/sub> \u00a0I \u00a0V<sub>2<\/sub> =&gt;<\/span><\/p>\n<p><span style=\"color: #000000;\">V = \u221a(<strong>V<sub>1<\/sub><\/strong>^2 + <strong>V<sub>2<\/sub><\/strong>^2) \u00a0=&gt; V = \u221a(V<sub>1<\/sub>^2 + (\u03c0DN)^2)<\/span><\/p>\n<p><span style=\"color: #000000;\">V = \u00a0\u221a[((280.10^3)\/3600)^2 \u00a0+ 2.\u03c0(3600\/60)^2] \u00a0= 323,6m\/s<\/span><\/p>\n<p style=\"text-align: right;\"><span style=\"text-decoration: underline;\"><span style=\"color: #3366ff; text-decoration: underline;\">v = 323,6 m\/s<\/span><\/span><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"probleme-2-%e2%80%93-sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie\"><\/span><strong><span style=\"color: #3366ff;\">Probl\u00e8me\u00a02 &#8211;<span style=\"color: #000000;\">\u00a0Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie<\/span><\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Un circuit est constitu\u00e9 d\u2019une r\u00e9sistance, d\u2019une bobine et d\u2019un condensateur tel qu\u2019indiqu\u00e9 sur la figure.<\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li><span style=\"color: #000000;\">Calculer les courants <span style=\"color: #ff0000;\">I<sub>1<\/sub>, I<sub>2<\/sub>, I<sub>3<\/sub><\/span> en r\u00e9gime permanent.<\/span><\/li>\n<li><span style=\"color: #000000;\">On remplace E par un g\u00e9n\u00e9rateur de f.e.m sinuso\u00efdale de valeur efficace <span style=\"color: #ff0000;\">100V<\/span> et de pulsation <span style=\"color: #ff0000;\">360 rad\/s<\/span>. reprendre le calcul de<span style=\"color: #ff0000;\"> I<sub>1<\/sub><\/span>.<\/span><\/li>\n<li><span style=\"color: #000000;\">Quel est le d\u00e9phasage entre <span style=\"color: #ff0000;\">I<sub>1<\/sub><\/span> et la tension du g\u00e9n\u00e9rateur\u00a0? Pr\u00e9ciser si le courant est en avance ou en retard.<\/span><\/li>\n<li><span style=\"color: #000000;\">Quelle puissance (active) fournit le g\u00e9n\u00e9rateur\u00a0?<\/span><\/li>\n<li><span style=\"color: #000000;\">Pour une valeur <span style=\"color: #ff0000;\"><em>f<\/em><sub>0<\/sub><\/span> de la fr\u00e9quence, le courant <span style=\"color: #ff0000;\">I<sub>1<\/sub><\/span> et la tension du g\u00e9n\u00e9rateur sont en phase calculer<span style=\"color: #ff0000;\"> <em>f<\/em><sub>0<\/sub><\/span>.<\/span><\/li>\n<\/ol>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-4581\" src=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/04\/une-inductance-de-r\u00e9sistance-interne-Kamerpowers.png\" alt=\"une inductance de r\u00e9sistance interne\" width=\"461\" height=\"231\" \/><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"solution-au-probleme-2-%e2%80%93-sujets-corriges-concours-ensp-yaounde-2012-physique-%e2%80%93-chimie\"><\/span><strong><span style=\"color: #ff0000;\">Solution au probl\u00e8me 2 &#8211;\u00a0<span style=\"color: #000000;\">Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie<\/span><\/span><\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"color: #000000;\">\u00a0 \u00a0<span style=\"color: #3366ff;\"> 1)<\/span> Calcul de <span style=\"color: #ff0000;\"><em>I<\/em><sub>1<\/sub>, <em>I<\/em><sub>2<\/sub>, <em>I<\/em><sub>3<\/sub><\/span> en r\u00e9gime permanent:<\/span><\/p>\n<p><span style=\"color: #000000;\">Tout au long de cet exercice, nous utiliserons la m\u00e9thodes des complexes. Ainsi nous noterons les imp\u00e9dances comme suit\u00a0:\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #ff0000;\">z\u00a0= R<\/span> pour r\u00e9sistance<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #ff0000;\">z<sub>L<\/sub>\u00a0= r + jLw<\/span> pour une inductance de r\u00e9sistance interne<\/span><\/p>\n<p><span style=\"color: #000000;\">\u00a0 Z_{{c}}=\\frac{-j}{cw} pour un condensateur ( et <span style=\"color: #ff0000;\">j<sup>2<\/sup> = -1<\/span> remplace notre <span style=\"color: #ff0000;\">i<\/span> habituel qui dans le cas-ci peut se confondre \u00e0 l\u2019intensit\u00e9).<\/span><\/p>\n<p><span style=\"color: #000000;\">Aussi nous appliquerons pour tout le montage les r\u00e8gles de calcul de l\u2019imp\u00e9dance \u00e9quivalente \u00e0 savoir <span style=\"color: #ff0000;\">zeq = z<sub>1<\/sub> + z<sub>2<\/sub> + z<sub>3<\/sub><\/span> (s\u00e9rie) et \u00a0\\frac{1}{Z_{1}} \u00a0+ \u00a0\\frac{1}{Z_{2}} + \u00a0\\frac{1}{Z_{3}} (en parall\u00e8le).<\/span><\/p>\n<p><span style=\"color: #000000;\">le r\u00e9gime permanent dans la figure pr\u00e9c\u00e9dente. signifie tout simplement que les courants sont constants. I.e longtemps apr\u00e8s la fermeture de l\u2019interrupteur, et\u00a0notre courant est continu. \u00a0le condensateur est compl\u00e8tement charg\u00e9 aussi.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u00a0q = de<\/span><\/p>\n<p><span style=\"color: #000000;\">I<sub>3<\/sub> = dq\/dt = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Et puis\u00a0car tous les I sont constants<\/span><\/p>\n<p><span style=\"color: #000000;\">Ul = rI<sub>2<\/sub> &#8211; L(dI2\/dt) = rI<sub>2<\/sub><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">U<sub>L<\/sub> = rI<sub>2<\/sub>, loi des n\u0153uds I<sub>1<\/sub> + I<sub>2<\/sub>\u00a0= 0 en plus E = (R + r) I<sub>2<\/sub><\/span><\/p>\n<p><span style=\"color: #000000;\">I<sub>1<\/sub> = E\/R+1 = -I<sub>2\u00a0<\/sub>\u00a0 \u00a0 , \u00a0\u00a0 I<sub>3<\/sub>\u00a0= 0<\/span><\/p>\n<p><span style=\"color: #000000;\">I<sub>1<\/sub> \u00a0= 100\/(8+7) = 6,666<\/span><\/p>\n<p><span style=\"color: #000000;\">I<sub>1<\/sub> \u00a0= -I<sub>2<\/sub> = 6,66A<\/span><\/p>\n<p><span style=\"color: #000000;\"><span style=\"color: #3366ff;\">2 )<\/span> Calcul de <span style=\"color: #ff0000;\">I<sub>1<\/sub><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" class=\"aligncenter size-full wp-image-4583\" src=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/04\/l\u2019imp\u00e9dance-\u00e9quivalente-de-tout-circuit-Kamerpowers.png\" alt=\"l\u2019imp\u00e9dance \u00e9quivalente de tout circuit Kamerpower\" width=\"479\" height=\"270\" srcset=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/04\/l\u2019imp\u00e9dance-\u00e9quivalente-de-tout-circuit-Kamerpowers.png 479w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/04\/l\u2019imp\u00e9dance-\u00e9quivalente-de-tout-circuit-Kamerpowers-300x169.png 300w\" sizes=\"(max-width: 479px) 100vw, 479px\" \/><\/p>\n<p style=\"text-align: justify;\">\u00a0<span style=\"color: #000000;\">Nous allons \u00e0 cet effet calculer l\u2019imp\u00e9dance \u00e9quivalente de tout circuit et ainsi on aura \u00a0i<sub>1<\/sub> = e\/Z<sub>eq<\/sub> \u00a0\u00a0. Z<sub>C<\/sub> et Z<sub>L<\/sub> sont en parall\u00e8le et les deux en s\u00e9ries avec B<sub>R<\/sub> donc ;<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Z_{{eq}=Z_{{R}}+Z_{{1}} avec \\frac{1}{Z_{1}} = \\frac{1}{Z_{C}}+\\frac{1}{Z_{L}}<\/span><\/p>\n<p><span style=\"color: #000000;\">Z<sub>1<\/sub> = [ ((r+jLw) &#8211; j\/cw) \/ (r + jLw) &#8211; j\/cw ] \u00a0= \u00a0((-r\/cw)j \u00a0+ \u00a0L\/c) \/ r + j(Lw &#8211; 1\/cw)<\/span><\/p>\n<p class=\"post-title single-title entry-title\" style=\"text-align: center;\"><span style=\"color: #3366ff;\"><strong><a style=\"color: #3366ff;\" href=\"https:\/\/kamerpower.com\/fr\/epreuves-concours-polytechnique-cameroun-pdf\/\">Sujets Corrig\u00e9s Concours ENSP Yaounde 2012 Physique &#8211; Chimie\u00a0Ecole Nationale Polytechnique de Yaound\u00e9:<\/a><\/strong><\/span><\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-5914\" src=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP.jpg\" alt=\"ENSP Concours R\u00e9vision Cameroun\" width=\"289\" height=\"287\" srcset=\"https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-100x100.jpg 100w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-150x150.jpg 150w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-144x144.jpg 144w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-80x80.jpg 80w, https:\/\/kamerpower.com\/wp-content\/uploads\/2015\/05\/LogoENSP-96x96.jpg 96w\" sizes=\"(max-width: 289px) 100vw, 289px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":5914,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_themeisle_gutenberg_block_has_review":false,"footnotes":""},"categories":[1701],"tags":[1712,1727,2078],"class_list":["post-12248","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-epreuves","tag-epreuve-de-concours-dentree-aux-grandes-ecoles-cameroun","tag-epreuve-ensp-concours-cameroun","tag-sujets-corrigees-concours-ensp-yaounde-polytechnique"],"_links":{"self":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12248","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/comments?post=12248"}],"version-history":[{"count":3,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12248\/revisions"}],"predecessor-version":[{"id":67817,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/posts\/12248\/revisions\/67817"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/media\/5914"}],"wp:attachment":[{"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/media?parent=12248"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/categories?post=12248"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kamerpower.com\/fr\/wp-json\/wp\/v2\/tags?post=12248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}