{"id":105,"date":"2017-12-09T14:17:05","date_gmt":"2017-12-09T13:17:05","guid":{"rendered":"http:\/\/pbelaire.free.fr\/wordpress\/?page_id=105"},"modified":"2017-12-09T14:17:05","modified_gmt":"2017-12-09T13:17:05","slug":"exercices-corriges","status":"publish","type":"page","link":"https:\/\/ordi-tech.fr\/index.php\/exercices-corriges\/","title":{"rendered":"Exercices corrig\u00e9s"},"content":{"rendered":"<p align=\"justify\"><span style=\"font-family: Verdana;\"><em>Voici, avant d&rsquo;aller plus loin, une s\u00e9rie de petits exercices pour vous permettre de v\u00e9rifier si vous avez bien compris les notions vues en \u00e9lectricit\u00e9. Il s&rsquo;agit de probl\u00e8mes tr\u00e8s simples et il n&rsquo;y a aucun \u00ab\u00a0pi\u00e8ge\u00a0\u00bb! Les r\u00e9ponses sont donn\u00e9es en bas de page.<\/em><\/span><\/p>\n<p><b><span style=\"font-family: Verdana; font-size: large;\">Exercices :<\/span><\/b><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\">Dessiner un sch\u00e9ma \u00e9lectrique<\/span><\/li>\n<\/ul>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Une\u00a0<strong>pile de 4,5 V<\/strong>\u00a0alimente une\u00a0<strong>r\u00e9sistance de 220 ohms<\/strong>\u00a0et une\u00a0<strong>DEL rouge<\/strong>\u00a0mont\u00e9es en s\u00e9rie, via un\u00a0<strong>interrupteur<\/strong>. Dessinez le sch\u00e9ma de ce circuit.<\/span><\/p>\n<p><a href=\"#rep1\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\">Calculer une r\u00e9sistance \u00e9quivalente<\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">Soit le sch\u00e9ma ci-dessous. Calculez la r\u00e9sistance \u00e9quivalente entre \u00ab\u00a0A\u00a0\u00bb et \u00ab\u00a0C\u00a0\u00bb.<\/span><\/p>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo02.gif\" alt=\"calcul R \u00e9quivalente\" width=\"362\" height=\"148\" \/><\/span><\/p>\n<p><a href=\"#rep2\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\">Appliquer la loi d&rsquo;Ohm<\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">Vous disposez d&rsquo;une\u00a0<strong>pile de 9 V<\/strong>, de\u00a0<strong>deux r\u00e9sistances de 1 kilo-ohms<\/strong>\u00a0et d&rsquo;une\u00a0<strong>DEL rouge<\/strong>. Comment r\u00e9unir ces \u00e9l\u00e9ments de mani\u00e8re \u00e0 ce que le courant qui traverse la DEL soit d&rsquo;environ\u00a0<strong>15 mA<\/strong>? Dessinez le sch\u00e9ma du circuit.<\/span><br \/>\n<span style=\"font-family: Verdana;\"><em><strong>Rappel:<\/strong>\u00a0la tension de seuil d&rsquo;une DEL rouge est \u00e9gale \u00e0 1,6 V.<\/em><\/span><br \/>\n<a href=\"#rep3\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\">Appliquer la loi des noeuds<\/span><\/li>\n<\/ul>\n<table border=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td><span style=\"font-family: Verdana;\">Soit le sch\u00e9ma ci-contre.<\/span><span style=\"font-family: Verdana;\">Quelle est la valeur du\u00a0<strong>courant<\/strong>\u00a0I?<\/span><br \/>\n<a href=\"#rep4\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/td>\n<td><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo04.gif\" alt=\"loi des noeuds\" width=\"410\" height=\"235\" \/><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\">Calculer un pont diviseur de tension<\/span><\/li>\n<\/ul>\n<table border=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo05.gif\" alt=\"pont diviseur\" width=\"286\" height=\"256\" \/><\/span><\/td>\n<td valign=\"top\">\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Soit le sch\u00e9ma ci-contre. Quelle est la valeur de la\u00a0<strong>d.d.p.<\/strong>\u00a0entre la masse (0 V) et le point not\u00e9 X?<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Prenez les groupes de valeurs A) puis B).<\/span><\/p>\n<p align=\"justify\"><a href=\"#rep5\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\">Appliquer la loi de Joule<\/span><\/li>\n<\/ul>\n<table border=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td valign=\"top\">\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Quelle est la\u00a0<strong>puissance<\/strong>\u00a0P dissip\u00e9e par la r\u00e9sistance\u00a0<strong>R1<\/strong>?<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\"><em><strong>Rappel:<\/strong>\u00a0la tension de seuil d&rsquo;une DEL rouge est \u00e9gale \u00e0 1,6 V.<\/em><\/span><\/p>\n<p align=\"justify\"><a href=\"#rep6\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<\/td>\n<td><span style=\"font-family: Verdana;\"><em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo06.gif\" alt=\"loi de Joule\" width=\"241\" height=\"213\" \/><\/em><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>\n<p align=\"left\"><span style=\"font-family: Verdana;\">Tension continue et tension variable<\/span><\/p>\n<\/li>\n<\/ul>\n<table border=\"0\" cellpadding=\"2\">\n<tbody>\n<tr>\n<td><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo07.gif\" alt=\"tension continue et tension variable\" width=\"383\" height=\"260\" \/><\/span><\/td>\n<td valign=\"top\">\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Sur le graphe ci-contre, on a trac\u00e9 deux courbes, qui repr\u00e9sentent l&rsquo;\u00e9volution dans le temps d&rsquo;une tension U1 (en bleu) et d&rsquo;une tension U2 (en rouge). L&rsquo;axe des x est celui du temps, en secondes; l&rsquo;axe des y est celui des tensions, en volts.<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">U1 et U2 sont-elles des tensions variables ou continues?<\/span><\/p>\n<p align=\"justify\"><a href=\"#rep7\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\">Caract\u00e9ristique d&rsquo;un dip\u00f4le<\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">On branche un g\u00e9n\u00e9rateur aux bornes d&rsquo;un dip\u00f4le quelconque et on fait varier la tension U fournie par le g\u00e9n\u00e9rateur. A l&rsquo;aide d&rsquo;un multim\u00e8tre, on rel\u00e8ve les valeurs suivantes (U la tension du g\u00e9n\u00e9rateur, I le courant dans le dip\u00f4le):<\/span><\/p>\n<div align=\"center\">\n&nbsp;<\/p>\n<table border=\"1\" cellpadding=\"2\">\n<tbody>\n<tr>\n<td align=\"center\" bgcolor=\"#00FFFF\"><span style=\"font-family: Verdana;\"><strong>U<\/strong><\/span><\/td>\n<td align=\"center\" bgcolor=\"#FFFF00\"><span style=\"font-family: Verdana;\"><strong>I<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td bgcolor=\"#00FFFF\"><span style=\"font-family: Verdana;\"><strong>3 V<\/strong><\/span><\/td>\n<td bgcolor=\"#FFFF00\"><span style=\"font-family: Verdana;\"><strong>30 mA<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td bgcolor=\"#00FFFF\"><span style=\"font-family: Verdana;\"><strong>4,5 V<\/strong><\/span><\/td>\n<td bgcolor=\"#FFFF00\"><span style=\"font-family: Verdana;\"><strong>45 mA<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td bgcolor=\"#00FFFF\"><span style=\"font-family: Verdana;\"><strong>6 V<\/strong><\/span><\/td>\n<td bgcolor=\"#FFFF00\"><span style=\"font-family: Verdana;\"><strong>60 mA<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td bgcolor=\"#00FFFF\"><span style=\"font-family: Verdana;\"><strong>9 V<\/strong><\/span><\/td>\n<td bgcolor=\"#FFFF00\"><span style=\"font-family: Verdana;\"><strong>90 mA<\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;\n<\/p><\/div>\n<p><span style=\"font-family: Verdana;\">Tracez la\u00a0<strong>caract\u00e9ristique<\/strong>\u00a0de ce dip\u00f4le, c&rsquo;est-\u00e0-dire la courbe repr\u00e9sentant la variation du courant (I) en fonction de la tension (U) \u00e0 ses bornes, soit I = f (U).<\/span><br \/>\n<span style=\"font-family: Verdana;\">Que pouvez-vous conclure?<\/span><br \/>\n<span style=\"font-family: Verdana;\">Si on inverse le sens de branchement du dip\u00f4le, on rel\u00e8ve exactement les m\u00eames valeurs. Que pouvez-vous conclure?<\/span><\/p>\n<p align=\"left\"><a href=\"#rep8\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\">Tension alternative<\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">Voici l&rsquo;allure d&rsquo;une\u00a0<strong>tension alternative<\/strong>, telle qu&rsquo;on peut la visualiser \u00e0 l&rsquo;aide d&rsquo;un instrument appel\u00e9 oscilloscope:<\/span><\/p>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_exo09.gif\" alt=\"tension alternative\" width=\"559\" height=\"237\" \/><\/span><\/p>\n<p><span style=\"font-family: Verdana;\">Quelle est la valeur de la tension\u00a0<strong>maximale<\/strong>\u00a0(on tension cr\u00eate)? Quelle est la valeur de la tension dite\u00a0<strong>\u00ab\u00a0efficace\u00a0\u00bb\u00a0<\/strong>(ou \u00ab\u00a0rms\u00a0\u00bb)?<\/span><br \/>\n<a href=\"#rep9\"><span style=\"font-family: Verdana;\">Voir la solution.<\/span><\/a><\/p>\n<hr \/>\n<p align=\"left\"><span style=\"font-family: Verdana;\"><strong>Solutions aux probl\u00e8mes :<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep1\"><\/a>Dessiner un sch\u00e9ma \u00e9lectrique<\/strong><\/span><\/li>\n<\/ul>\n<table border=\"0\">\n<tbody>\n<tr>\n<td><span style=\"font-family: Verdana;\">Votre sch\u00e9ma devrait ressembler \u00e0 ceci:<\/span><\/td>\n<td><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige01.gif\" alt=\"corrig\u00e9 exo 1\" width=\"440\" height=\"226\" \/><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep2\"><\/a>Calculer une r\u00e9sistance \u00e9quivalente<\/strong><\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">Observez que le dip\u00f4le entre \u00ab\u00a0A\u00a0\u00bb et \u00ab\u00a0B\u00a0\u00bb est en s\u00e9rie avec le dip\u00f4le entre \u00ab\u00a0B\u00a0\u00bb et \u00ab\u00a0C\u00a0\u00bb. A l&rsquo;int\u00e9rieur du dip\u00f4le BC, R2 et R3 sont en s\u00e9rie et ce dip\u00f4le R2-R3 est en \/\/ avec R4.<\/span><\/p>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige02.gif\" alt=\"solution exo 2\" width=\"711\" height=\"235\" \/><\/strong><\/span><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep3\"><\/a>Appliquer la loi d&rsquo;Ohm<\/strong><\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">La solution passe par la mise en \/\/ des deux r\u00e9sistances, comme ci-dessous:<\/span><\/p>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige03.gif\" alt=\"solution exo 3\" width=\"689\" height=\"194\" \/><\/span><\/p>\n<p><span style=\"font-family: Verdana;\">Avec une seule r\u00e9sistance, on aurait une intensit\u00e9 (insuffisante) de 7,4 mA. En montant les r\u00e9sistances en s\u00e9rie, ce serait pire: le courant ne serait plus que de 3,7 mA!<\/span><\/p>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep4\"><\/a>Appliquer la loi des noeuds<\/strong><\/span><\/li>\n<\/ul>\n<table border=\"0\">\n<tbody>\n<tr>\n<td valign=\"top\">\n<p align=\"justify\"><span style=\"font-family: Verdana;\">La solution est toute simple, puisque la somme des courants I1, I2 et I3 est \u00e9gale au courant entrant I, qui est \u00e9gal au courant sortant I.<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\"><em>Attention: la valeur de I3 est exprim\u00e9e en amp\u00e8re!<\/em><\/span><\/p>\n<\/td>\n<td><span style=\"font-family: Verdana;\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige04.gif\" alt=\"corrig\u00e9 exo 4\" width=\"582\" height=\"232\" \/><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep5\"><\/a>Calculer un pont diviseur de tension<\/strong><\/span><\/li>\n<\/ul>\n<table border=\"0\">\n<tbody>\n<tr>\n<td>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Lorsque R1 = R2, on a toujours une division de U par 2.<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">La diff\u00e9rence de potentiel (d.d.p.) entre le point x et le point a est bien entendu \u00e9gale \u00e0 4,5 V (soit 9 V &#8211; 4,5 V) dans le premier cas et \u00e0 4 V (12 V &#8211; 8 V) dans le second.<\/span><\/p>\n<\/td>\n<td><span style=\"font-family: Verdana;\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige05.gif\" alt=\"solution exo 5\" width=\"545\" height=\"258\" \/><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><span style=\"font-family: Verdana;\"><strong><a name=\"rep6\"><\/a>Appliquer la loi de Joule<\/strong><\/span><\/li>\n<\/ul>\n<p><span style=\"font-family: Verdana;\">La solution au probl\u00e8me implique de calculer au pr\u00e9alable la valeur de la d.d.p. aux bornes de R1 et la valeur du courant I1. On observera que I = I1 + I2 (loi des noeuds), qui sont de m\u00eame valeur puisque R1 et R2 sont ici de m\u00eame valeur.<\/span><\/p>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige06.gif\" alt=\"solution probl\u00e8me 6\" width=\"556\" height=\"288\" \/><\/strong><\/span><\/p>\n<ul>\n<li>\n<p align=\"left\"><span style=\"font-family: Verdana;\"><strong><a name=\"rep7\"><\/a>Tension continue et tension variable<\/strong><\/span><\/p>\n<\/li>\n<\/ul>\n<p align=\"left\"><span style=\"font-family: Verdana;\">U1 (en bleu) est une\u00a0<strong>tension continue<\/strong>, qui reste constante (donc invariable). U2 (en rouge) est une\u00a0<strong>tension variable<\/strong>, puisqu&rsquo;elle n&rsquo;est pas constante: sa valeur varie sans cesse dans le temps. On peut ajouter que U2 est une tension\u00a0<strong>alternative<\/strong>\u00a0(tant\u00f4t positive, tant\u00f4t n\u00e9gative), de forme\u00a0<strong>sinuso\u00efdale<\/strong>.<\/span><\/p>\n<ul>\n<li>\n<p align=\"left\"><span style=\"font-family: Verdana;\"><strong><a name=\"rep8\"><\/a>Caract\u00e9ristique d&rsquo;un dip\u00f4le<\/strong><\/span><\/p>\n<\/li>\n<\/ul>\n<table border=\"0\">\n<tbody>\n<tr>\n<td><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige08.gif\" alt=\"caract\u00e9ristique\" width=\"276\" height=\"252\" \/><\/span><\/td>\n<td>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">On obtient la courbe ci-contre, qui est une droite. On peut donc conclure que la caract\u00e9ristique de ce dip\u00f4le est\u00a0<strong>lin\u00e9aire<\/strong>. (En extrapolant, on voit qu&rsquo;elle passe par l&rsquo;origine des axes.)<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">Apr\u00e8s inversion du sens de branchement du dip\u00f4le, on constate que la courbe est identique: la caract\u00e9ristique est donc\u00a0<strong>sym\u00e9trique<\/strong>.<\/span><\/p>\n<p align=\"justify\"><span style=\"font-family: Verdana;\">On peut conclure que ce dip\u00f4le n&rsquo;est pas polaris\u00e9 et sa caract\u00e9ristique est celle d&rsquo;une r\u00e9sistance.<\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>\n<p align=\"left\"><span style=\"font-family: Verdana;\"><strong><a name=\"rep9\"><\/a>Tension alternative<\/strong><\/span><\/p>\n<\/li>\n<\/ul>\n<p align=\"center\"><span style=\"font-family: Verdana;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pbelaire.free.fr\/images\/image_corrige09.gif\" alt=\"tension alternative: Umax, Urms\" width=\"586\" height=\"279\" \/><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Voici, avant d&rsquo;aller plus loin, une s\u00e9rie de petits exercices pour vous permettre de v\u00e9rifier si vous avez bien compris les notions vues en \u00e9lectricit\u00e9. Il s&rsquo;agit de probl\u00e8mes tr\u00e8s simples et il n&rsquo;y a aucun \u00ab\u00a0pi\u00e8ge\u00a0\u00bb! Les r\u00e9ponses sont donn\u00e9es en bas de page. Exercices : Dessiner un sch\u00e9ma \u00e9lectrique Une\u00a0pile de 4,5 V\u00a0alimente &hellip; <a href=\"https:\/\/ordi-tech.fr\/index.php\/exercices-corriges\/\" class=\"more-link\">Continuer la lecture de <span class=\"screen-reader-text\">Exercices corrig\u00e9s<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-105","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/pages\/105","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/comments?post=105"}],"version-history":[{"count":0,"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/pages\/105\/revisions"}],"wp:attachment":[{"href":"https:\/\/ordi-tech.fr\/index.php\/wp-json\/wp\/v2\/media?parent=105"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}