{"id":4271,"date":"2026-05-01T09:38:27","date_gmt":"2026-05-01T07:38:27","guid":{"rendered":"https:\/\/sti2tech.com\/?page_id=4271"},"modified":"2026-05-05T08:43:26","modified_gmt":"2026-05-05T06:43:26","slug":"bts-die-synthese-n3","status":"publish","type":"page","link":"https:\/\/sti2tech.com\/?page_id=4271","title":{"rendered":"BTS – DIE – Synth\u00e8se n\u00b03"},"content":{"rendered":"\t\t
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Synth\u00e8se n\u00b03 : Le courant d’emploi et fictif\u00a0<\/h2>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Objectifs:<\/h3>
  • D\u00e9finir <\/strong>le courant d’emploi et fictif d’une\u00a0installation<\/li>
  • Adapter<\/strong> la m\u00e9thodologie selon les \u00e9l\u00e9ments donn\u00e9es en r\u00e9ponses.<\/li>
  • R\u00e9aliser<\/strong> des calculs avec rigueur<\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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    Partie 1 – Circuit sans Neutre Charg\u00e9<\/strong><\/h4>

    Contexte<\/strong> :<\/p>

    • Une armoire alimente uniquement des charges r\u00e9sistives triphas\u00e9es \u00e9quilibr\u00e9es<\/strong> (pas de neutre charg\u00e9).<\/li><\/ul>

      Donn\u00e9es<\/strong> :<\/p>

      • Charge triphas\u00e9e :
        • Puissance apparente S<\/span>=<\/span><\/span>20<\/span>kVA<\/span><\/span><\/span><\/span><\/span>.<\/span><\/li>
        • Tension U<\/span>=<\/span><\/span>400<\/span>V<\/span><\/span><\/span><\/span><\/span>.<\/span><\/li><\/ul><\/li>
        • Mode de pose<\/strong> : C\u00e2bles enterr\u00e9s.<\/li>
        • Temp\u00e9rature ambiante<\/strong> : 25\u00b0<\/span>C<\/span><\/span><\/span><\/span>.<\/span><\/li>
        • Nombre de conducteurs \u00e0\u00a0<\/strong>proximit\u00e9<\/b>\u00a0: 2 c\u00e2bles (3 phases, pas de neutre).<\/li>
        • Espace entre les conducteur : 25 cm<\/strong><\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
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          Questions<\/strong> :<\/p>

          1. Calculer le courant d’emploi<\/strong> (Ib)\u00a0en utilisant la <\/span>meilleur<\/strong> m\u00e9thode<\/strong>. Justifier votre choix.<\/span><\/li>
          2. D\u00e9terminer le courant fictif<\/strong> (Ifictif)\u00a0en tenant compte :<\/span><\/span><\/span><\/span><\/span>
            • Du mode de pose.<\/li>
            • De la temp\u00e9rature ambiante.<\/li>
            • Du nombre de conducteurs adjacents.<\/li><\/ul><\/li>
            • V\u00e9rifier si un c\u00e2ble de section 10 mm\u00b2<\/strong> est adapt\u00e9 (utiliser les tables de courant admissible).<\/li><\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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              Partie 2 – Circuit avec Neutre Charg\u00e9<\/strong><\/h4>

              Contexte<\/strong> :<\/p>

              • Une armoire \u00e9lectrique alimente un moteur triphas\u00e9<\/strong> et deux \u00e9clairages monophas\u00e9s<\/strong> (branch\u00e9 entre phase et neutre).\u00a0<\/li>
              • Le neutre est donc charg\u00e9.<\/li><\/ul>

                Donn\u00e9es<\/strong> :<\/p>

                • Moteur triphas\u00e9 :
                  • Puissance nominale P<\/span>n<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>15<\/span>kW<\/span><\/span><\/span><\/span>.<\/span><\/li>
                  • Tension U<\/span>=<\/span><\/span>400<\/span>V<\/span><\/span><\/span><\/span><\/span>.<\/span><\/li>
                  • Facteur de puissance cos<\/span>(<\/span>\u03d5<\/span>)<\/span>=<\/span><\/span>0.85<\/span><\/span><\/span><\/span>.<\/span><\/li>
                  • Rendement\u03b7<\/span>=<\/span><\/span>0.9<\/span><\/span><\/span><\/span>.<\/span><\/mo><\/mrow><\/semantics><\/math><\/span><\/span><\/span>

                    \u00a0<\/p><\/li><\/ul><\/li>

                  • \u00c9clairage monophas\u00e9 :
                    • 2 P\u00e9<\/span>c<\/span>l<\/span>ai<\/span>r<\/span>a<\/span>g<\/span>e<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>5<\/span>kW<\/span><\/span><\/span><\/span><\/span><\/li>
                    • Facteur de puissance cos<\/span>(<\/span>\u03d5<\/span>)<\/span>=0,8<\/span><\/span><\/span><\/span>.<\/span>

                      \u00a0<\/p><\/li><\/ul><\/li>

                    • Mode de pose<\/strong> : C\u00e2bles en conduit.<\/li>
                    • Temp\u00e9rature ambiante<\/strong> : 35\u00b0<\/span>C<\/span><\/span><\/span>.<\/span><\/li>
                    • C\u00e2ble en poly\u00e9thyl\u00e8ne r\u00e9ticul\u00e9<\/strong><\/li>
                    • Nombre de conducteurs adjacents<\/strong> : 3 c\u00e2bles (3 phases + neutre).<\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
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                      Questions<\/strong> :<\/p>

                      1. Calculer le courant d’emploi du moteur<\/strong> (Ib_m<\/span>o<\/span>t<\/span>e<\/span>u<\/span>r<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) en utilisant les <\/span>3 m\u00e9thodes<\/strong>
                        • coefficients<\/span><\/li>
                        • puissance apparente<\/span><\/li>
                        • puissance active.<\/span><\/li><\/ul><\/li>
                        • Calculer le courant d’emploi de l’\u00e9clairage<\/strong> (Ib\u00e9c<\/span>l<\/span>ai<\/span>r<\/span>a<\/span>g<\/span>e<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>).<\/span><\/li>
                        • D\u00e9terminer le courant fictif<\/strong> (Ifi<\/span>c<\/span>t<\/span>i<\/span>f<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>) pour le circuit complet en tenant compte :
                          • Du mode de pose.<\/li>
                          • De la temp\u00e9rature ambiante.<\/li>
                          • Du nombre de conducteurs adjacents.<\/li><\/ul><\/li>
                          • Choisir la section de c\u00e2ble adapt\u00e9<\/strong> (section minimale) en utilisant les tables de courant admissible (ex. : norme NF C 15-100).<\/li><\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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                            Documents techniques<\/h2>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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                            Synth\u00e8se n\u00b03 : Le courant d’emploi et fictif\u00a0 Objectifs: D\u00e9finir le courant d’emploi et fictif d’une\u00a0installation Adapter la m\u00e9thodologie selon les \u00e9l\u00e9ments donn\u00e9es en r\u00e9ponses. R\u00e9aliser des calculs avec rigueur Partie 1 – Circuit sans Neutre Charg\u00e9 Contexte : Une armoire alimente uniquement des charges r\u00e9sistives triphas\u00e9es \u00e9quilibr\u00e9es (pas de neutre charg\u00e9). Donn\u00e9es : Charge…<\/p>\n","protected":false},"author":4,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_kad_post_transparent":"","_kad_post_title":"hide","_kad_post_layout":"fullwidth","_kad_post_sidebar_id":"","_kad_post_content_style":"unboxed","_kad_post_vertical_padding":"hide","_kad_post_feature":"hide","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_kad_post_classname":"","footnotes":""},"folder":[22],"class_list":["post-4271","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4271","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4271"}],"version-history":[{"count":10,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4271\/revisions"}],"predecessor-version":[{"id":4286,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4271\/revisions\/4286"}],"wp:attachment":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4271"}],"wp:term":[{"taxonomy":"folder","embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Ffolder&post=4271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}