{"id":210416,"date":"2024-11-21T16:10:18","date_gmt":"2024-11-21T16:10:18","guid":{"rendered":"https:\/\/www.electricity-magnetism.org\/?p=210416"},"modified":"2024-11-21T16:10:18","modified_gmt":"2024-11-21T16:10:18","slug":"comment-analyser-un-circuit-serie-parallele","status":"publish","type":"post","link":"https:\/\/www.electricity-magnetism.org\/fr\/comment-analyser-un-circuit-serie-parallele\/","title":{"rendered":"Comment analyser un circuit s\u00e9rie-parall\u00e8le ?"},"content":{"rendered":"<p class=\"sidekick\">Analysez un circuit s\u00e9rie-parall\u00e8le en thermal engineering : m\u00e9thodes simples pour comprendre le fonctionnement, identifier les composants et r\u00e9soudre les probl\u00e8mes.<\/p>\n<h2>Comment analyser un circuit s\u00e9rie-parall\u00e8le ?<\/h2>\n<p>Les circuits s\u00e9rie-parall\u00e8le combinent des \u00e9l\u00e9ments de circuits en s\u00e9rie et en parall\u00e8le. Analyser ces circuits peut sembler complexe, mais en suivant quelques \u00e9tapes m\u00e9thodiques, cela devient beaucoup plus simple.<\/p>\n<h2>\u00c9tapes pour analyser un circuit s\u00e9rie-parall\u00e8le<\/h2>\n<p><u1><\/p>\n<li><strong>\u00c9tape 1 :<\/strong> Identifier les sous-circuits en s\u00e9rie et en parall\u00e8le.<\/li>\n<li><strong>\u00c9tape 2 :<\/strong> Simplifier chaque sous-circuit en parall\u00e8le.<\/li>\n<li><strong>\u00c9tape 3 :<\/strong> R\u00e9duire le circuit \u00e0 une seule r\u00e9sistance \u00e9quivalente.<\/li>\n<li><strong>\u00c9tape 4 :<\/strong> Utiliser les lois d\u2019Ohm et de Kirchhoff pour r\u00e9soudre le circuit.<\/li>\n<p><\/u1><\/p>\n<h2>\u00c9quivalents de r\u00e9sistances en s\u00e9rie et en parall\u00e8le<\/h2>\n<p>Tout d&rsquo;abord, comprenons comment combiner les r\u00e9sistances en s\u00e9rie et en parall\u00e8le :<\/p>\n<p><u1><\/p>\n<li><strong>R\u00e9sistances en s\u00e9rie :<\/strong> La r\u00e9sistance \u00e9quivalente (R<sub>eq<\/sub>) est la somme des r\u00e9sistances individuelles.<br \/>\n        <br \/>R<sub>eq<\/sub> = R<sub>1<\/sub> + R<sub>2<\/sub> + R<sub>3<\/sub> + &#8230;<\/li>\n<li><strong>R\u00e9sistances en parall\u00e8le :<\/strong> La r\u00e9sistance \u00e9quivalente (R<sub>eq<\/sub>) est d\u00e9termin\u00e9e par la formule :<br \/>\n        \\[<br \/>\n        \\frac{1}{R_{eq}} = \\frac{1}{R_{1}} + \\frac{1}{R_{2}} + \\frac{1}{R_{3}} + &#8230;<br \/>\n        \\]<br \/>\n        ou en utilisant une fraction :<br \/>\n        <br \/>\\[<br \/>\n        R_{eq} = \\frac{1}{\\left(\\frac{1}{R_{1}} + \\frac{1}{R_{2}} + \\frac{1}{R_{3}} + &#8230; \\right)}<br \/>\n        \\]\n    <\/li>\n<p><\/u1><\/p>\n<h2>Application des lois de Kirchhoff<\/h2>\n<p>Les lois de Kirchhoff sont essentielles pour r\u00e9soudre les circuits complexes. Voici les deux lois :<\/p>\n<p><u1><\/p>\n<li><strong>La loi des n\u0153uds de Kirchhoff (LNK) :<\/strong> La somme des courants entrant dans un n\u0153ud est \u00e9gale \u00e0 la somme des courants sortant de ce n\u0153ud.<\/li>\n<li><strong>La loi des mailles de Kirchhoff (LMK) :<\/strong> La somme des tensions autour de toute boucle ferm\u00e9e est nulle.<\/li>\n<p><\/u1><\/p>\n<p>En appliquant ces lois, nous pouvons examiner chaque n\u0153ud et chaque boucle pour trouver les courants et les tensions inconnus.<\/p>\n<h2>Exemple pratique<\/h2>\n<p>Imaginez un circuit avec trois r\u00e9sistances (R<sub>1<\/sub>, R<sub>2<\/sub> et R<sub>3<\/sub>) o\u00f9 :<\/p>\n<p><o1><\/p>\n<li>R<sub>1<\/sub> est en s\u00e9rie avec une combinaison parall\u00e8le de R<sub>2<\/sub> et R<sub>3<\/sub>.<\/li>\n<p><\/o1><\/p>\n<p>Voici comment analyser ce circuit :<\/p>\n<p><o1><\/p>\n<li>Tout d&rsquo;abord, trouver la r\u00e9sistance \u00e9quivalente de R<sub>2<\/sub> et R<sub>3<\/sub> en parall\u00e8le :<br \/>\n        \\[<br \/>\n        \\frac{1}{R_{eq_{23}}} = \\frac{1}{R_{2}} + \\frac{1}{R_{3}}<br \/>\n        \\]<br \/>\n        Ensuite, nous avons :<br \/>\n        \\[ R_{eq_{23}} = \\frac{R_{2} * R_{3}}{R_{2} + R_{3}} \\]\n    <\/li>\n<li>Maintenant, ajoutez R<sub>1<\/sub> en s\u00e9rie :<br \/>\n        \\[ R_{eq_{total}} = R_{1} + R_{eq_{23}} \\]\n    <\/li>\n<p><\/o1><\/p>\n<p>En fin de compte, compte tenu de la r\u00e9sultante R<sub>eq_total<\/sub>, vous pouvez appliquer la loi d\u2019Ohm pour trouver la tension et le courant \u00e0 travers chaque r\u00e9sistance.<\/p>\n<p>Analyser un circuit s\u00e9rie-parall\u00e8le peut sembler intimidant au d\u00e9but, mais avec une approche syst\u00e9matique et une compr\u00e9hension des principes fondamentaux, cela devient une t\u00e2che que l&rsquo;on peut ma\u00eetriser facilement.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Analysez un circuit s\u00e9rie-parall\u00e8le en thermal engineering : m\u00e9thodes simples pour comprendre le fonctionnement, identifier les composants et r\u00e9soudre les probl\u00e8mes.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_generate-full-width-content":"","footnotes":""},"categories":[98],"tags":[],"class_list":["post-210416","post","type-post","status-publish","format-standard","hentry","category-des-questions","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-50"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - 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