{"id":1862,"date":"2016-01-25T01:10:54","date_gmt":"2016-01-25T01:10:54","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1862"},"modified":"2024-08-19T14:10:58","modified_gmt":"2024-08-19T14:10:58","slug":"circuitos-compostos","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/circuitos-compostos\/","title":{"rendered":"Circuitos Compostos"},"content":{"rendered":"

\"\"\u00a0<\/span><\/p>\n

Caracter\u00edsticas dos elementos de um circuito el\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Considere o <\/b><\/span><\/span><\/span>circuito el\u00e9trico <\/b><\/span><\/span><\/span>esquematizado na figura abaixo onde<\/b><\/span><\/span><\/span>, A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, A<\/b><\/span><\/span><\/span>3, .\u00a0<\/b><\/span><\/span><\/sub><\/span>. . . . . . ., A<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span>\u00a0s\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

bipolos.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Defini\u00e7\u00f5es dos principais elementos de um circuito el\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0 <\/span>n\u00f3<\/b><\/span><\/span><\/span>\u00a0da rede <\/b><\/span><\/span><\/span>\"\" <\/b><\/span><\/span><\/span>ponto de interliga\u00e7\u00e3o <\/b><\/span><\/span><\/span>de pelo menos <\/b><\/span><\/span><\/span>dois condutores<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>S\u00e3o n\u00f3s os pontos A, B e C.<\/b><\/span><\/span><\/span><\/p>\n

\"\" \u00a0<\/span>ramo<\/b><\/span><\/span><\/span>\u00a0da rede \"\"<\/b><\/span><\/span><\/span> <\/b>trecho da rede <\/b><\/span><\/span><\/span>compreendido <\/b><\/span><\/span><\/span>entre dois n\u00f3s consecutivos<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>S\u00e3o ramos os trechos: BFDA, AEC e BGHC, AB e BC.<\/b><\/span><\/span><\/span><\/p>\n

\"\" \u00a0<\/span>malha<\/b><\/span><\/span><\/span>\u00a0da rede<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>toda cadeia de ramos <\/b><\/span><\/span><\/span>que forma um <\/b><\/span><\/span><\/span>circuito fechado. <\/b><\/span><\/span><\/span>Assim, <\/b><\/span><\/span><\/span>todo circuito poligonal fechado constitui uma malha. S\u00e3o malhas: ABFDA, AECBA e BGHCB.<\/b><\/span><\/span><\/span><\/p>\n

\"\" \u00a0<\/span>Primeira lei de Kirchhoff ou lei dos n\u00f3s<\/b><\/span><\/span><\/span>: \u201cA <\/b><\/span><\/span><\/span>soma alg\u00e9brica <\/b><\/span><\/span><\/span>das <\/b><\/span><\/span><\/span>correntes que chegam a um n\u00f3 <\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 <\/b><\/span><\/span><\/span>soma alg\u00e9brica das correntes que saem do mesmo n\u00f3\u201d.<\/b><\/span><\/span><\/span><\/p>\n

Exemplo num\u00e9rico\u00a0 \"\" <\/b><\/span><\/span><\/span>\u00a0 determine a <\/b><\/span><\/span><\/span>intensidade da corrente i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0do trecho de circuito abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\" Segunda lei de Kirchhoff ou lei das malhas<\/b><\/span><\/span><\/span>: \u201c<\/b><\/span><\/span><\/span>Percorrendo uma malha num mesmo sentido<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>\u00e9 nula a soma alg\u00e9brica das tens\u00f5es encontradas em cada elemento do circuito\u201d<\/b><\/span><\/span><\/span><\/p>\n

Baseado na <\/b><\/span><\/span><\/span>lei acima, escolhendo um sentido de percurso para cada malha<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>partindo de um ponto, retornando ao mesmo ponto e igualando esta soma alg\u00e9brica de tens\u00f5es a zero, voc\u00ea pode estabelecer as seguintes conven\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Conven\u00e7\u00e3o para o sinal de E:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Conven\u00e7\u00e3o para o sinal de R.i<\/b><\/span><\/span><\/span><\/p>\n

Quando o <\/b><\/span><\/span><\/span>sentido da corrente i coincidir com o sentido do percurso escolhido para a malha,<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>sinal de R.i ser\u00e1 positivo e quando o sentido da corrente i n\u00e3o coincidir com o sentido de percurso escolhido para a malha, o sinal R.i ser\u00e1 negativo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

Sequ\u00eancia de opera\u00e7\u00f5es para resolu\u00e7\u00e3o de exerc\u00edcios sobre circuitos compostos<\/b><\/span><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Assinalar com letras mai\u00fasculas os n\u00f3s da rede.<\/b><\/span><\/span><\/span><\/p>\n

2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Atribuir a cada ramo uma corrente com sentido arbitr\u00e1rio<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

3<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Atribuir polaridade arbitr\u00e1ria a eventuais bipolos que n\u00e3o as tenham.<\/b><\/span><\/span><\/span><\/p>\n

4<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\" <\/b><\/span><\/span><\/span>Aplicar a primeira lei de Kirchhoff a cada n\u00f3.<\/b><\/span><\/span><\/span><\/p>\n

5<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Atribuir a cada malha independente um sentido arbitr\u00e1rio\u00a0(hor\u00e1rio ou anti-hor\u00e1rio) de percurso<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

6<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Aplicar a segunda lei de Kirchhoff, ou seja, percorrer cada malha obedecendo ao percurso fixado e \u00e0s conven\u00e7\u00f5es vistas anteriormente.<\/b><\/span><\/span><\/span><\/p>\n

7<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Resolver o sistema de equa\u00e7\u00f5es obtidas em 4<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0e 6<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

8<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Interpretar os resultados, lembrando que:<\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>deve-se inverter o sentido das correntes que resultarem negativas.<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Inverter as polaridades das tens\u00f5es que resultarem negativas.<\/b><\/span><\/span><\/span><\/p>\n

9<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span><\/span>Para fornecer as respostas deve-se, depois de corrigir no esquema os sentidos das correntes e das polaridades das tens\u00f5es, expressa-las por n\u00fameros positivos.<\/b><\/span><\/span><\/span><\/p>\n

\"\" Deve-se <\/b><\/span><\/span><\/span>observar<\/b><\/span><\/span><\/span>, no <\/b><\/span><\/span><\/span>c\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>diferen\u00e7a de potencial<\/b><\/span><\/span><\/span>\u00a0entre dois pontos quaisquer de um circuito, que, se quiser determinar a ddp (U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0– V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) entre dois pontos A e B, deve-se seguir o circuito de A at\u00e9 B obtendo determinado valor em volt e que, se seguir em sentido contr\u00e1rio obter\u00e1 resultado sim\u00e9trico.<\/b><\/span><\/span><\/span><\/p>\n

\"\" Deve-se lembrar que, <\/b><\/span><\/span><\/span>num gerador, o sentido da corrente \u00e9 do polo negativo para o polo positivo, ao contr\u00e1rio do receptor.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Exemplos de exerc\u00edcios sobre a aplica\u00e7\u00e3o da sequ\u00eancia acima, ou seja, como proceder para resolver exerc\u00edcios sobre circuitos compostos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Dado o <\/b><\/span><\/span><\/span>ramo do circuito abaixo, <\/b><\/span><\/span><\/span>determine a <\/b><\/span><\/span><\/span>diferen\u00e7a de potencial entre os pontos A e D, sabendo que a corrente flui de A para D e tem intensidade de 3 A.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Percorrendo o ramo no sentido de A para D <\/b><\/span><\/span><\/span>e observando as <\/b><\/span><\/span><\/span>conven\u00e7\u00f5es de sinais para E e para R.i (veja informa\u00e7\u00f5es acima)\u00a0 \"\"<\/b><\/span><\/span><\/span>\u00a0 U<\/b><\/span><\/span><\/span>AD <\/b><\/span><\/span><\/sub><\/span>= V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>D <\/b><\/span><\/span><\/sub><\/span>= <\/b><\/span><\/span><\/span>+(2.3) – 6 +(3.3) + 2 +(2.3)\u00a0 + 2 +(4.3)\u00a0 \"\"<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>AD <\/b><\/span><\/span><\/sub><\/span>= V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>= +31V\u00a0 \"\"<\/b><\/span><\/span><\/span>\u00a0 se voc\u00ea <\/b><\/span><\/span><\/span>tivesse percorrido o ramo de D para A, <\/b><\/span><\/span><\/span>obedecendo as mesmas conven\u00e7\u00f5es<\/b><\/span><\/span><\/span>, voc\u00ea teria obtido U<\/b><\/span><\/span><\/span>DA <\/b><\/span><\/span><\/sub><\/span>= V<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/sub><\/span>= – 31V<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0No <\/b><\/span><\/span><\/span>circuito esquematizado<\/b><\/span><\/span><\/span>, determine <\/b><\/span><\/span><\/span>a intensidade e o sentido da corrente el\u00e9trica <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>diferen\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de potencial entre A e B. Estabele\u00e7a para a corrente e para o percurso o sentido hor\u00e1rio.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Esse assunto \u00e9 <\/b><\/span><\/span><\/span>trabalhoso e \u00e9 importante que voc\u00ea resolva no m\u00ednimo alguns exerc\u00edcios<\/b><\/span><\/span><\/span>, analisando atentamente as resolu\u00e7\u00f5es. Selecionei alguns que s\u00e3o os de n\u00fameros <\/b><\/span><\/span><\/span>03, 06, 11, 16, 20 e 22.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Confira os exerc\u00edcios sugeridos<\/span><\/a><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

\u00a0 Caracter\u00edsticas dos elementos de um circuito el\u00e9trico \u00a0Considere o circuito el\u00e9trico esquematizado na figura abaixo onde, A1, A2, A3, .\u00a0. . . . . . ., A12\u00a0s\u00e3o bipolos. Defini\u00e7\u00f5es dos principais elementos de um circuito el\u00e9trico \u00a0 n\u00f3\u00a0da rede ponto de interliga\u00e7\u00e3o de pelo menos dois condutores. S\u00e3o n\u00f3s os pontos A, B e C. \u00a0ramo\u00a0da rede trecho da<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1676,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1862","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1862","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1862"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1862\/revisions"}],"predecessor-version":[{"id":1869,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1862\/revisions\/1869"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1676"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1862"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}