{"id":9089,"date":"2022-04-06T09:00:00","date_gmt":"2022-04-06T12:00:00","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=9089"},"modified":"2023-08-06T21:00:51","modified_gmt":"2023-08-07T00:00:51","slug":"associacao-de-resistores-serie-e-paralelo-com-exercicios-resolvidos","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/associacao-de-resistores-serie-e-paralelo-com-exercicios-resolvidos\/","title":{"rendered":"Associa\u00e7\u00e3o de Resistores S\u00e9rie e Paralelo com exerc\u00edcios resolvidos"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Fala, pessoal, tudo certo? Nesta aula, vamos resolver juntos algumas quest\u00f5es sobre associa\u00e7\u00e3o de resistores, tanto em s\u00e9rie quanto em paralelo. Antes, como sempre, vamos revisar os principais conceitos e f\u00f3rmulas. Acompanhe!<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Associa\u00e7\u00e3o de resistores<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Em s\u00e9rie:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/kbjAszocG7DoTObjSwK5SJjMNUlJVZAV9bEIH9WxwM-7NWYm1Bl8SHS4K-Z-Dw9JnXIp9__Ja27RY8N4-EAKJcxg55jqCMNsU69OdemlX7gL2o02D_fX5x8uMZadqnwNdf2vvHrZ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Quando temos resistores em s\u00e9rie, n\u00e3o importa o n\u00famero, todos s\u00e3o percorridos pela mesma corrente (i). Para identificar resistores em s\u00e9rie, n\u00e3o haver\u00e1 n\u00f3 (encontro de dois ou mais fios) na figura, como vemos acima.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Por sua vez, a diferen\u00e7a de potencial (U) do sistema ser\u00e1 a soma de todos os resistores:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AD<\/sub> = U<sub>AB<\/sub> + U<sub>BC<\/sub> + U<sub>CD<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A resist\u00eancia equivalente ser\u00e1 igual \u00e0 soma de todas as outras:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>eq<\/sub> = R, + R<sub>2<\/sub> + R<sub>3<\/sub>\u2026<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Em paralelo:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/mqttsGvy7esIIbHQ-jzcu_N7XTysiQIQDlaeHohGKeIBojoy8zyDz_ALvuUHTMdtY0LIzN1JiMp0nxcA3AFJaZ72qnpjMeZldw2CGxb6VMAzYC_ha6lPGz4N4shEQ-APhBcB19md\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Na associa\u00e7\u00e3o em paralelo, temos todos os resistores ligados no mesmo par de pontos. Na figura acima, seria o ponto de A para B.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ao contr\u00e1rio do que acontece com a associa\u00e7\u00e3o em s\u00e9rie, a ddp (U) ser\u00e1 igual por todo o sistema:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = U<sub>1<\/sub> = U<sub>2<\/sub> = U<sub>3<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Por sua vez, a corrente total (i) ser\u00e1 igual \u00e0 soma de todas as correntes que passam pelos resistores.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i<sub>total<\/sub> = i<sub>1<\/sub> + i<sub>2<\/sub> + i<sub>3<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para calcular a resist\u00eancia equivalente, utilizaremos:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/6VIT_gB52-nJaKiHanH8LStzbbB4zhEKtQPebcSP4Ss4h2NAG3l1RSlX_2AOrsggN_famioWlrg4UxTAXxvF5mzJBWfNwTYOjF2geefX2G4UUv1q62iBAbYB5jTG8cbf0eozHNy2\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Caso tenhamos apenas dois resistores, podemos usar o produto pela soma:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/_00vcwBj9xwZPc0iTtNn2TY51F7J1lrM9FGSDGpKG9n3ejwvKzCg1FbckZGtVYIuyHLrC-ow6atg7C2XpT2e5O9Q8QThBCQM2Vbonw0onPio629RrwLwuhssNwprcwgr9Tm7sPYw\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">J\u00e1 no caso de termos v\u00e1rios resistores iguais:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/3B9Jj5xK3wxq9EvQqww_ty1cIsW_b-Fn895Kc8p8eFbBflX1G0i1r28ZZHPgrebWWS-VRu3ysmWBrU9DdHPFQQTAoY3UPpODFAp0m-BIlTBTGA5RKZdyrq-UW1l2QhvpX8uoAgPn\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">R: valor de um dos resistores.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">N: quantidade de resistores.<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios sobre associa\u00e7\u00e3o de resistores<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 1<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A associa\u00e7\u00e3o de resistores abaixo recebe uma tens\u00e3o de 36 volts entre os terminais A e D.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/erRdGWswhqKZdiWTDvTNikGakuylejH1BdwWhAml1fEvgqivA3697pUQKTlHIW-pyImSRxeM63PwsByo5kEGKmih11YTge2e8aoBybXEawh2HYMOqVXT4GAAIOw2Szy19osP8kcO\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>a) Determina a intensidade da corrente que percorre os resistores.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Perceba que \u00e9 a resist\u00eancia equivalente do sistema que recebe os 36 volts. Isso posto, temos que calcul\u00e1-la. Como est\u00e3o em s\u00e9rie, basta som\u00e1-las:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>eq<\/sub> = 2 + 4 + 16<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>eq<\/sub> = 12 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ent\u00e3o, vamos utilizar a primeira lei de Ohm:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AD<\/sub> = R<sub>eq<\/sub> . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">36 = 12 . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i = 3 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>b) Calcule as tens\u00f5es el\u00e9tricas entre os pontos A e B, B e C e C e D.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Temos que usar a primeira lei de ohm para cada um deles, sempre lembrando que, nesse caso, a corrente \u00e9 constante:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = 2. 3 = 6 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>BC<\/sub> = 4 . 3 = 12 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>CD<\/sub> = 6 . 3 = 18 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que deu certo, pois a soma das tens\u00f5es deu a ddp total (36) do enunciado.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 2<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>No esquema a seguir, dois resistores est\u00e3o ligados em paralelo. Determine as intensidades de correntes i<\/strong><strong><sub>2<\/sub><\/strong><strong> e i<\/strong><strong><sub>T<\/sub><\/strong><strong>.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/2Gp_hJQDFJF15UQtpVbSq9tceGnLw9K7yQcK4H8BSvFjhZAsS0oRp47TTRwpkx_orJ15Py2abUR8Pyy5rqzXjRKnEZca455fG-WI82EvUOtSudrnJVehz5Sz_mAhIFy0YDCb5AuT\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Geralmente, h\u00e1 diversas formas de resolver esse tipo de quest\u00e3o. Vamos calcular a ddp de A para B utilizando o resistor de cima:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = 5 . 2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = 10 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Agora, podemos utilizar essa mesma ddp no resistor de baixo:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">10 = 20 . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i<sub>2<\/sub> = 0,5 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para calcular a corrente total, vamos lembrar que temos que somar todas as correntes:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i<sub>T<\/sub> = i<sub>1<\/sub> + i<sub>2<\/sub><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i<sub>T<\/sub> = 2 + 0,5<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i<sub>T<\/sub> = 2,5 A<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 3<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A figura mostra a liga\u00e7\u00e3o de dois resistores em s\u00e9rie, calcule a intensidade da corrente i2 e a tens\u00e3o el\u00e9trica entre os pontos A e B.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/Hs2iuzj3cgDcuMTJVAUDQBEYI4IrTMmvASghRCkVkTRd3qydC5hdBli3OOiDl7GhFtoZNehQ9czU9LhnIihU94uT7Tt3cL8Fv72k8UQzIZbhmG_YIkJF4wcMk6GRNECzD5OEn4MR\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">A corrente i<sub>2<\/sub> ser\u00e1 igual \u00e0 corrente i<sub>1<\/sub>, pois esses resistores est\u00e3o em s\u00e9rie. Ent\u00e3o, i<sub>2<\/sub> = 3 A.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">J\u00e1 para calcular a tens\u00e3o el\u00e9trica entre A e B, podemos calcular a resist\u00eancia equivalente primeiro. Neste caso, basta somarmos:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>eq<\/sub> = 12 + 24 = 36 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Em seguida:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = R<sub>eq<\/sub> . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = 36 . 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U<sub>AB<\/sub> = 108 V<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 4<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Unesp) Dentro de uma caixa com terminais A e B, existe uma associa\u00e7\u00e3o de resistores. A corrente que atravessa a caixa em fun\u00e7\u00e3o da tens\u00e3o aplicada nos terminais A e B \u00e9 dada pela tabela.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/NvuzDjSzYm9a78UqtO2pQCSVDvC6tqdXlrq-fbBhYNGsGp2YUX0JPM0ZvHUr0BH_ATMIk9K0gLFMKGhfgjEG2y0PUlKnAwabCjuKEWWO8t-tNnOjHkeSSRHP5LfTvb2NKPI7i0ha\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A caixa poderia conter<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/2YnPOasGpxhdMqtCMcghQdDTZldWywLu8YyEmqc_C9FFt4Rw9FC9RITFTBuIIMjKQs6ZPx3dZBkfktTuP5zISRaaTGiWYkv9dvmCvC5sZBXcv0CIE2lZTBPJdnZDzKdwt0dA986p\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que temos os dados para utilizar a f\u00f3rmula de sempre:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">3 = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R = 3 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Volte para a tabela e veja que o resultado ser\u00e1 sempre igual, ou seja, a resist\u00eancia \u00e9 constante. No entanto, repare que o enunciado nos diz que dentro da caixa h\u00e1 v\u00e1rios resistores. Ou seja, este resultado \u00e9 na verdade a resist\u00eancia equivalente do circuito.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ent\u00e3o, temos que identificar em qual alternativa a resist\u00eancia equivalente d\u00e1 3 \u03a9. Vamos fazendo um a um at\u00e9 encontrar a alternativa:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a)<img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/3B9Jj5xK3wxq9EvQqww_ty1cIsW_b-Fn895Kc8p8eFbBflX1G0i1r28ZZHPgrebWWS-VRu3ysmWBrU9DdHPFQQTAoY3UPpODFAp0m-BIlTBTGA5RKZdyrq-UW1l2QhvpX8uoAgPn\" width=\"92\" height=\"37\"><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/2yOoX003HakL7iVp9Ss1hZFs8hH--eufFZrtR8e6RLhljFNbdkiQxwgyTAkkibqe2Euq4o0JXd3n8UO3gtvjZnCkdb6_YIadNjpSGhEdxFZKfZrgq6utK6_M_DHFWaagzk2UBnBR\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">b) <img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/E-vQ5jGI9cOA6BVkcXK2tADsKU9MgzhepfXZALNJvx-Oo9fmRkOMNSxEFVqnwixXxk9bnyloLyrh2jFzkf6Efl5c2RzH_OGx9LRPk2iM2TUjMUIQEzL5UBW_VOlHwAiiozECSdcZ\" width=\"260\" height=\"45\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) Veja que, nesta alternativa, temos dois resistores em paralelo seguidos de um em s\u00e9rie. Repare, portanto, que a resist\u00eancia dos resistores em paralelo \u00e9 igual a 1. Ela deve ser somada \u00e0 resist\u00eancia do seguinte. Ou seja:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Rc = 1 + 2 = 3 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 5<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A figura mostra tr\u00eas resistores \u00f4hmicos de resist\u00eancias 8 \u03a9, 2 \u03a9 e 1,6 \u03a9, conectados em paralelo por fios de resist\u00eancia desprez\u00edvel a uma fonte de tens\u00e3o que fornece 8 V entre os pontos A e B<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/VSnKUw1K-QFQhG3sHgr5-EAx2rsFJLt1VPW07CRLfsZEecIrbLX4_yKJNgb_3IZvMoDKnbeUeFC7YMtvlSP3mlpImReYcSr9JqdSrNJSsI-t3FLiZoSGqzOkiEiRuTDgsUhCJI_m\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>a) Determine a intensidade da corrente el\u00e9trica, i, que parte da bateria.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Aqui, voc\u00ea pode calcular a resist\u00eancia equivalente e aplicar a primeira lei de Ohm ou calcular a corrente de cada um e som\u00e1-las no final. Vamos calcular a equivalente:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/GaV7rhPQKivZ-q6utjsvNa92dLFRDYp-4cVVd4I9DcBMRJpP1RxpvvFiZB2f4OOoq5IJ6uUWTQiqOiuE4v3fkDS5Sg3dqOJDZMeSPgKpt6h6z217hKFu4AnbyXnLikQCvhdZ_e9g\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/qtF0q8CgsBhMT3UhMzp_H1Pok7xQ4mCUjS5bBHLEFYesYbRGfW-k5dA3QrAz2MWjnkR4rYRr8uP0gC7DJSTtw7Hr3A4Ks3cIVE00g_Pblp7zPbYedhw7x89IBikClsvN5aO44T89\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/U3QMtmQjFgK3WY6MHhe3cyVRyql0tmCPMQD6HP7fsHboKdct3dwPSWgaw0xuIwdwRvSXd9oWVQy6-n59XxgaBTkT6H6A0Wm_EfYIRwEoMrcMbbYt5FHqRWsgD6xKTvKOn-bY5hg6\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>eq<\/sub> = 0,8 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Agora, vamos \u00e0 nossa f\u00f3rmula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = R<sub>eq<\/sub> . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">8 = 0,8 . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i = 10 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>b) Calcule as intensidades das correntes el\u00e9tricas i<\/strong><strong><sub>1<\/sub><\/strong><strong>, i<\/strong><strong><sub>2<\/sub><\/strong><strong> e i<\/strong><strong><sub>3<\/sub><\/strong><strong>.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para calcular a corrente para cada um, devemos utilizar a mesma f\u00f3rmula para cada um:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/oz6wmqGPIzHiuEa9HMyw1pX1JCvgS5bnKAT3bjeT6w1wDh7A_nJ8giSwH2rkdvCtjwCTS4tfb_79NSP4lcpMo4E__0VeQr58wnkxXWG-YxT8XSrpUcZc3gcYmX-wxtXAhLsyGOQa\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/fk8rzLq2BH8xPlESkKgV6hxrlC5GaVmELz-Hoc__5tphbgajKyLpWa44vKSMmrDQ3Ch0Oq40bMgfMB-rt6KftuSBuP2lc5PaLT1jX-jBPwj95BTOYcJMiabzSnw3Xb3dsPkgx960\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/wlkEDcjsnibpAcKS08ZpKo88hgGk4uW3O9qzfe7nbpvanKTORqcnkXzXFOv_Ik3PyADVZ8B376H2BGRwH7XZaO9s-Shu_x__9dTqK_uSnNpG_jJkRITL0KiB3Qo-3ZI4aPwvTsiH\" alt=\"\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/5oq-cpK9nWE6fP-Moij1xcLsOWE7CEDoNuOxOxroGhiE7LAi3oBIZv14-rZiN41r9WU18FYnwlJpoZT7YKURo-anXu2y-xh_gDQZcK1i6_hgz-yEQx05VJnE8KIcaGEVHkpz3ijQ\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">Espero que voc\u00ea tenha entendido um pouco melhor como resolver exerc\u00edcios sobre associa\u00e7\u00e3o de resistores em s\u00e9rie e em paralelo. E se quiser ajuda para melhorar seu n\u00edvel de F\u00edsica em outras mat\u00e9rias,<a href=\"https:\/\/professorpinguim.com.br\/\"> entre em contato comigo e escolha o curso de F\u00edsica mais adequado para voc\u00ea<\/a>!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>SAIBA MAIS<\/strong><br>\u2705\u00a0<a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/questoes-resolvidas-sobre-carga-eletrica-e-processos-de-eletrizacao\/\" target=\"_blank\">Carga el\u00e9trica e eletriza\u00e7\u00e3o<\/a><br>\u2705\u00a0<a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/exercicios-resolvidos-de-campo-eletrico\/\" target=\"_blank\">Exerc\u00edcios sobre campo el\u00e9trico<\/a><br>\u2705\u00a0<a href=\"https:\/\/professorpinguim.com.br\/blog\/potencial-eletrico-energia-e-trabalho-exercicios-com-gabarito\/\" target=\"_blank\" rel=\"noreferrer noopener\">Potencial el\u00e9trico, energia potencial e trabalho da for\u00e7a el\u00e9trica<\/a><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Me acompanhe nas redes sociais: curta a minha&nbsp;<a href=\"https:\/\/www.facebook.com\/professorpinguim\/\"><strong>p\u00e1gina no Facebook<\/strong><\/a><strong>,<\/strong>&nbsp;me&nbsp;siga no&nbsp;<a href=\"https:\/\/www.instagram.com\/professorpinguim\/\"><strong>Instagram<\/strong><\/a><strong>,<\/strong>&nbsp;se inscreva no&nbsp;<a href=\"https:\/\/www.youtube.com\/c\/PinguimV%C3%ADdeoF%C3%ADsica\/?sub_confirmation=1\"><strong>Youtube<\/strong><\/a>&nbsp;e participe do meu canal oficial no&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/telegram\"><strong>Telegram<\/strong><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Nesta aula, vamos resolver juntos algumas quest\u00f5es sobre associa\u00e7\u00e3o de resistores, tanto em s\u00e9rie quanto em paralelo. Antes, como sempre, vamos revisar os principais conceitos e f\u00f3rmulas. Acompanhe!<\/p>\n","protected":false},"author":1,"featured_media":9090,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[746,729,770,742],"tags":[825,755,827,826],"class_list":["post-9089","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-eletricidade","category-eletrostatica","category-gravitacao","category-hidrostatica","tag-associacao-de-resistores","tag-conceitos-e-formulas","tag-paralelo","tag-serie"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Associa\u00e7\u00e3o de Resistores S\u00e9rie e Paralelo com exerc\u00edcios resolvidos - Professor Pinguim | F\u00edsica<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/professorpinguim.com.br\/blog\/associacao-de-resistores-serie-e-paralelo-com-exercicios-resolvidos\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Associa\u00e7\u00e3o de Resistores S\u00e9rie e Paralelo com exerc\u00edcios resolvidos - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Nesta aula, vamos resolver juntos algumas quest\u00f5es sobre associa\u00e7\u00e3o de resistores, tanto em s\u00e9rie quanto em paralelo. 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