{"id":9006,"date":"2022-03-30T15:29:06","date_gmt":"2022-03-30T18:29:06","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=9006"},"modified":"2023-08-06T21:00:55","modified_gmt":"2023-08-07T00:00:55","slug":"explicacao-e-exercicios-sobre-resistores-e-leis-de-ohm","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/explicacao-e-exercicios-sobre-resistores-e-leis-de-ohm\/","title":{"rendered":"Explica\u00e7\u00e3o e Exerc\u00edcios sobre Resistores e Leis de Ohm"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Fala, pessoal, tudo belezinha? Nesta aula, vamos revisar o conte\u00fado sobre resistores e as leis de Ohm e, em seguida, vamos resolver alguns exerc\u00edcios juntos. Acompanhe!<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Resistor el\u00e9trico e resist\u00eancia el\u00e9trica<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">O resistor \u00e9 o elemento do circuito que transforma energia el\u00e9trica em energia t\u00e9rmica. Por sua vez, a resist\u00eancia el\u00e9trica \u00e9 uma grandeza f\u00edsica associada \u00e0 dificuldade para a passagem de corrente el\u00e9trica. Ela \u00e9 medida em ohm (\u03a9).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Tens\u00e3o el\u00e9trica<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Tens\u00e3o el\u00e9trica ou ddp \u00e9 a grandeza escalar que mostra a rela\u00e7\u00e3o entre a energia el\u00e9trica (J) e a quantidade de carga el\u00e9trica (C). Ela \u00e9 representada pela f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/s9Yu1WjzzJcPi6lj6y2K-T0NEyVJEemgpTDQdE3xvYLOFwX9Mt8f6fLX4GWg4i9tsRx-Z8m0wQ0KPBGzanEc6Q7R0jc4t8sxp8gDHH3zHUZfLihTLYXX3yaEwUcdjamA0b_3Ip7U\" alt=\"\"\/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Primeira Lei de Ohm<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Veja a figura:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/gmUoOgYH9HcGLJTUsragPHmngF1scW3BdeaKpoX4Q4RVFZK9t2S_hKO-2Ecj-K-zxAVdWnGqlT-tL1p1qD5O513gFg6BGhQ6gzS0NtnhGA7MAjHSFD50O_n0gFdrUHnHLKkMpq2a\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">A primeira lei de Ohm \u00e9 representada pela f\u00f3rmula:&nbsp;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>U<\/strong><strong><sub>AB<\/sub><\/strong><strong> = R . i<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Vale lembrar que isso se aplica tanto para resistores \u00f4hmicos (resist\u00eancia constante) ou n\u00e3o. Nesse caso, teremos um gr\u00e1fico representando por uma reta que passa pela origem:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/5StsoyYH1RTQNhrp12lCQQS4QNPk2pNU3aTAzpKTmpta8iKQMl67-ic1ZtLkTC_E13AwCl9L9yQvTOE5YUphmBDiME4DF1Jm3c7ljVlVLen4ftgNm7c3HFP_lAFpquYLqFoeWnUv\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Caso o resistor seja do tipo n\u00e3o \u00f4hmico, o gr\u00e1fico ser\u00e1 uma curva, pois a resist\u00eancia vai variar de acordo com a ddp:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/6o-yWdoFuYPSSJAGyscey8lmqX8xNu-oCLT9E4uJMTOqkeErlsb_FCfhG__aNstflLfzJTnvP4BovYJYTUyljgQEjlS4nfpfgl_mmt7zfla80iWugr_QrzzosFD7CZsetzXbLRSg\" alt=\"\"\/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Segunda Lei de Ohm<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">A segunda lei determina a resist\u00eancia de um peda\u00e7o de feio, que vai depender do comprimento (L) e da grossura (A) do resistor. \u00c9 representada pela seguinte f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/tEZHTb7a_X_rXBwkevttWtxW8kwKcRbDKgjVKa4REr5Lss9vojmE2QqdmTS40bd9gYipN6cfYH07cA-BZoP_8n0fnWg2nZzPf_8apsppiB-Gbgo-PWQfGEOrLtX8G5Vgn8Aku_oE\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">\u03c1: resistividade do material.<br>L: comprimento do condutor<br>A: \u00e1rea da se\u00e7\u00e3o transversal do condutor<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Importante: quando o fio for cil\u00edndrico, a \u00e1rea (A) ser\u00e1: <strong>A = \u03c0 . r\u00b2<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios sobre resistores e leis de Ohm<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 1<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Fuvest) O aquecimento de um forno el\u00e9trico \u00e9 baseado na convers\u00e3o de energia el\u00e9trica em energia t\u00e9rmica em um resistor. A resist\u00eancia R do resistor desse forno, submetido a uma diferen\u00e7a de potencial V constante, varia com a sua temperatura T. Na figura a seguir \u00e9 mostrado o gr\u00e1fico da fun\u00e7\u00e3o R(T) = R<\/strong><strong><sub>0<\/sub><\/strong><strong> &#8211; \u03b1(T &#8211; T<\/strong><strong><sub>0<\/sub><\/strong><strong>), sendo R<\/strong><strong><sub>0<\/sub><\/strong><strong> o valor da resist\u00eancia na temperatura T<\/strong><strong><sub>0<\/sub><\/strong><strong> e \u03b1 uma constante.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/JVj_X-rO2A8RgpGewOj-XIX3j7oyDrge4wssptdLoPIVwQKNgI0ZvOLPvDIIg3IrlKkLHw5mgZJl_Aujwvq6ndWCMjsvOlagELSdX696NuzgKH1513DPsBNz3nPSXMdZn_6L2SOh\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Ao se ligar o forno, com o resistor a 20 \u00baC,a corrente \u00e9 10 A. Ao atingir a temperatura T<\/strong><strong><sub>M<\/sub><\/strong><strong>, a corrente \u00e9 5 A.&nbsp;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Determine a<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>a) constante \u03b1;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para isso, teremos que usar um pouco de Matem\u00e1tica. Essa constante \u03b1 nada mais \u00e9 do que o coeficiente angular da reta. Geralmente, isso \u00e9 visto quando temos um gr\u00e1fico e a express\u00e3o y = b + ax, em que a \u00e9 a tangente da reta do gr\u00e1fico, formando um \u00e2ngulo, que vamos chamar de \u03b8.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Portanto, \u03b1 ser\u00e1 numericamente igual \u00e0 tangente de \u03b8. Ent\u00e3o:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u03b1 = tg\u03b8<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/fFeUHH9ePTuC7aAk1av1t4AS-U5Nbnxj5GI4no1Jd7as9x2xvWerTR1cuTpk7JW_nCf-GgqedRm3n-D6Zh6aju8L0-Zblde6JFwPnbC1VNBNYc4WmDFzuWMRvhKEC1tqPG99HuNQ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">\u03b1 = 0,06 \u03a9\/\u00baC<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>b) diferen\u00e7a de potencial V;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Vamos usar a primeira lei de ohm:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">V = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que o enunciado nos deu o valor da corrente de 10 A quando temos 20 \u00baC. Pelo gr\u00e1fico, neste ponto, temos uma resist\u00eancia de 12 ohms. Ent\u00e3o:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">V = 12 . 10<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">V = 120 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>c) temperatura T<\/strong><strong><sub>M<\/sub><\/strong><strong>;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Pelo enunciado, teremos a temperatura T<sub>M<\/sub> quando a corrente valer 5 A. Repare tamb\u00e9m que a ddp que descobrimos (V) \u00e9 constante: Ent\u00e3o:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">V = R . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">120 = R . 5<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R = 24 \u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Agora, basta olhar o gr\u00e1fico e identificar a temperatura. Portanto, T<sub>M<\/sub> = 220 \u00baC<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>d) pot\u00eancia P dissipada no resistor na temperatura \u2122<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nesta quest\u00e3o, podemos utilizar:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = R . i\u00b2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 24 . 5\u00b2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 600 W<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 2<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(UFPR) A ind\u00fastria eletr\u00f4nica busca produzir e aperfei\u00e7oar dispositivos com propriedades el\u00e9tricas adequadas para as mais diversas aplica\u00e7\u00f5es. O gr\u00e1fico acima ilustra o comportamento el\u00e9trico de tr\u00eas dispositivos eletr\u00f4nicos quando submetidos a uma tens\u00e3o de opera\u00e7\u00e3o V entre seus terminais, de modo que por eles circula uma corrente i.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/pUyApl5-Nhu8doNCpi6bH0Ucf3ff-g0Psn1H9EPNd87neiumNuK2_XUbPD99r5hBZpjRa1mjLYh5b2cFoCDxudEtkvXTajGSPwwhsADWIftigy2Vumo9wH5JaWAUWspVEgs5YxAq\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Com base na figura acima, assinale a alternativa correta.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) O dispositivo D<sub>1<\/sub> \u00e9 n\u00e3o \u00f4hmico na faixa de -30 a +30 V e sua resist\u00eancia vale 0,2 k\u03a9.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) O dispositivo D<sub>2<\/sub> \u00e9 \u00f4hmico na faixa de -20 a +20 V e sua resist\u00eancia vale 6 k\u03a9.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) O dispositivo D<sub>3<\/sub> \u00e9 \u00f4hmico na faixa de -10 a +10 V e sua resist\u00eancia vale 0,5 k\u03a9.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) O dispositivo D<sub>1<\/sub> \u00e9 \u00f4hmico na faixa de -30 a +30 V e sua resist\u00eancia vale 6 k\u03a9.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) O dispositivo D<sub>3<\/sub> \u00e9 n\u00e3o \u00f4hmico na faixa de -10 a +10 V e sua resist\u00eancia vale 0,5 k\u03a9<\/p>\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 pelas alternativas que o enunciado nos pede para descobrir quais dispositivos s\u00e3o \u00f4hmicos ou n\u00e3o. Como vimos, para isso, o gr\u00e1fico deve ser um reta e passar pela origem. Nesse caso, podemos descartar a parte negativa do gr\u00e1fico.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que D<sub>1<\/sub> \u00e9 \u00f4hmico, mas n\u00e3o s\u00f3 ele. H\u00e1 uma faixa de ddp em que D<sub>3<\/sub> tamb\u00e9m \u00e9 \u00f4hmico, pois faz um reta.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Isso posto, vamos calcular a resist\u00eancia de cada um deles. Cuidado, pois o gr\u00e1fico est\u00e1 em mA.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para D<sub>1<\/sub>:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = R<sub>1<\/sub> . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">30 = R<sub>1<\/sub> . 5.10<sup>-3<\/sup><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R<sub>1<\/sub> = 6.10<sup>3<\/sup> \u03a9 ou 6 k\u03a9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Repare que j\u00e1 acertamos a quest\u00e3o.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: D<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 3<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(EsPCEx) Um fio de cobre possui uma resist\u00eancia R. Um outro fio de cobre, com o triplo do comprimento e a metade da \u00e1rea da se\u00e7\u00e3o transversal do fio anterior, ter\u00e1 uma resist\u00eancia igual a:&nbsp;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 2R\/3.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 3R\/2.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 2R.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 3R.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 6R.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Note que os fios t\u00eam o mesmo material, ou seja, a mesma resistividade (\u03c1). Nesta quest\u00e3o, temos que comparar duas situa\u00e7\u00f5es.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Na primeira:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/tEZHTb7a_X_rXBwkevttWtxW8kwKcRbDKgjVKa4REr5Lss9vojmE2QqdmTS40bd9gYipN6cfYH07cA-BZoP_8n0fnWg2nZzPf_8apsppiB-Gbgo-PWQfGEOrLtX8G5Vgn8Aku_oE\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Depois:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/bZLYQBNMJl44Om-8Q9xIFlaPgL7tT7LuYnP17-_VKfy1ecKtOwdf7VBtJMCMRjAOEkwKCe7OLSmw-vneR-xoTu_H0-CBh2-w41sXPKl7VPKxo4-FTUDk4Fi2V-4ePvhKZGADPD4F\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Para resolver essa bagun\u00e7a, temos que calcular separando o que \u00e9 letra do que \u00e9 n\u00famero. Ent\u00e3o:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/z3LVkeY3tJgPnt_XQQanZqWj9Xh5W_brIAvYuio3Mib5Tsp91JDg9tTgplqklQdc8tFXmuiXn8tOdWgr4OQ_N0wetPkp5SJftLbvullakLDFaeIkYVdok-2-KykDJe7aGYIFivSO\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Observe que basta substituir:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">R\u2019 = 6R<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: E<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">Para aprender mais: <\/p>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe title=\"Resistor Leis de Ohm - F\u00edsica - Eletrodin\u00e2mica - Aula 3 - Professor Pinguim\" width=\"800\" height=\"450\" src=\"https:\/\/www.youtube.com\/embed\/MqNwY-4JQGU?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/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 resistores e leis de Ohm. 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<figure class=\"wp-block-image\"><a href=\"https:\/\/curso.professorpinguim.com.br\/pt\/comprar\/produto\/curso-gratuito-matematica-para-aprender-fisica\" target=\"_blank\" rel=\"noreferrer noopener\"><img decoding=\"async\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/02\/curso-fisica-para-matematica-professor-pinguim.png\" alt=\"Curso de Matem\u00e1tica para aprender F\u00edsica - Professor Pinguim\" class=\"wp-image-7950\"\/><\/a><\/figure>\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\/category\/eletricidade\/\" target=\"_blank\">Pot\u00eancia e energia el\u00e9trica<\/a><br>\u2705\u00a0<a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/exercicios-resolvidos-sobre-campo-eletrico-uniforme-ceu\/\" target=\"_blank\">Exerc\u00edcios sobre Campo El\u00e9trico Uniforme<\/a><br>\u2705\u00a0<a href=\"https:\/\/professorpinguim.com.br\/blog\/questoes-resolvidas-sobre-carga-eletrica-e-processos-de-eletrizacao\/\" target=\"_blank\" rel=\"noreferrer noopener\">Exerc\u00edcios sobre carga el\u00e9trica e eletriza\u00e7\u00e3o<\/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 revisar o conte\u00fado sobre resistores e as leis de Ohm e, em seguida, vamos resolver alguns exerc\u00edcios juntos. Acompanhe!<\/p>\n","protected":false},"author":1,"featured_media":9008,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[746,729,770,742],"tags":[388,762,211,707,821,822,820],"class_list":["post-9006","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-eletricidade","category-eletrostatica","category-gravitacao","category-hidrostatica","tag-eletricidade","tag-eletrostatica","tag-exercicios","tag-gravitacao","tag-leis-de-ohm","tag-resistor-eletrico","tag-resistores"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Explica\u00e7\u00e3o e Exerc\u00edcios sobre Resistores e Leis de Ohm - 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\/explicacao-e-exercicios-sobre-resistores-e-leis-de-ohm\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Explica\u00e7\u00e3o e Exerc\u00edcios sobre Resistores e Leis de Ohm - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Nesta aula, vamos revisar o conte\u00fado sobre resistores e as leis de Ohm e, em seguida, vamos resolver alguns exerc\u00edcios juntos. 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