{"id":8847,"date":"2022-03-23T09:00:00","date_gmt":"2022-03-23T12:00:00","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=8847"},"modified":"2023-08-06T21:00:59","modified_gmt":"2023-08-07T00:00:59","slug":"potencia-e-energia-eletrica-exercicios-resolvidos","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/potencia-e-energia-eletrica-exercicios-resolvidos\/","title":{"rendered":"Pot\u00eancia e Energia El\u00e9trica: Exerc\u00edcios resolvidos"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Nesta aula, vamos resolver juntos algumas quest\u00f5es sobre pot\u00eancia e energia el\u00e9trica. Antes, no entanto, vamos revisar alguns conceitos e f\u00f3rmulas importantes. Acompanhe!<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Pot\u00eancia e energia el\u00e9trica<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Vamos lembrar que quando falamos em pot\u00eancia, estamos nos referindo \u00e0 rapidez na transforma\u00e7\u00e3o da energia. Na Eletrodin\u00e2mica, teremos a seguinte f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/zCljjviHgCxZUi_-9Led5lLg2ABNYNMFj1xcslx4V3xB5MxCqwbnloUabDOENrejIhSffhEHfLsFt063BxmTZatugCH65YVF3HpUa6HZFYjejMKrskSStCeFH_IxewPZfLpj_zbT\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Nesta mesma f\u00f3rmula, podemos isolar a energia el\u00e9trica. Assim teremos:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E<sub>el<\/sub> = Pot . \u0394t<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Quando n\u00e3o soubermos a pot\u00eancia de um aparelho bipolo, precisaremos da ddp (diferen\u00e7a de potencial ou tens\u00e3o):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Pot = U . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">J\u00e1 quando nos depararmos com um resistor, podemos utilizar:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = R . i\u00b2<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/xZcaX7_5i6iPsjUT4nE5uXpb__xFQ9DwWk5PV6nb6kJMAbtXAhkrHvWbYAqyjdAKzo1MDoqLUAEg9B97IA7hwrRciays4HX9BrP-sOqnbHJ8ciV10nELAd2BbpQQ38W60zWcfUlA\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios sobre pot\u00eancia e energia el\u00e9trica<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 1<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(UFRRJ) Um chuveiro, quando ligado a uma diferen\u00e7a de potencial constante de 110 V, desenvolve uma pot\u00eancia de 2.200 W. Esse mesmo chuveiro fica ligado nessa d.d.p. todos os dias durante apenas 20 minutos. Nesse caso:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>a) qual \u00e9 a energia, em joules, consumida pelo chuveiro durante 10 minutos?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Quando a quest\u00e3o nos pedir uma resultado em joule, lembre-se de que joule \u00e9 igual a w.s. Portanto, tudo o que estiver em minutos deve ser convertido.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Sendo assim:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E = P . \u0394t<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E = 2200 . 600<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E = 1320000 = 1,32.10<sup>6 <\/sup>J<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>b) se 1 kWh custa R$0,20 (vinte centavos), qual \u00e9 a despesa em reais com esse chuveiro durante um m\u00eas (30 dias)?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Temos que calcular a energia em kWh. Deixarei as unidades para facilitar a compreens\u00e3o. Veja:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E = P . \u0394t<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/BfeYxys9K7lnpXg-IeHOCvYRIXXLIeC77eAFhS-5Gg48UKHyK5r-kemLPMOqdB9797I78WQ8hpMdSdxKoF6Dc2pQEtCCA-wnLTmFOxX0o0espXju2Pgsw149Vv9dA1yh_bGgxzxW\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">E = 2,2 . 5 = 11 KWh<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para calcular o custo, basta multiplicar pelo pre\u00e7o: 11 . 0,2 = R$ 2,20<\/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>Um chuveiro de resist\u00eancia constante foi fabricado para funcionar em Santa Catarina, a uma diferen\u00e7a de potencial de 220V e pot\u00eancia de 6000W. Se o mesmo chuveiro for, por engano, ligado a um diferen\u00e7a de potencial de 110V, qual ser\u00e1 sua nova pot\u00eancia?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Quando ligado um chuveiro na tomada, a ddp geralmente \u00e9 constante. Por\u00e9m, quando trocamos de tomada, a energia que ele dissipa n\u00e3o \u00e9 a mesma. E como um chuveiro el\u00e9trico \u00e9 um resistor, vamos usar:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/xZcaX7_5i6iPsjUT4nE5uXpb__xFQ9DwWk5PV6nb6kJMAbtXAhkrHvWbYAqyjdAKzo1MDoqLUAEg9B97IA7hwrRciays4HX9BrP-sOqnbHJ8ciV10nELAd2BbpQQ38W60zWcfUlA\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Ent\u00e3o, primeiro a quest\u00e3o nos disse que a pot\u00eancia era 6000:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/qzcgxG8A3H6-ianOJ2s9M_F3G2GholVN50uoiTEdW4PqREG4N0VyGU5d9XjLj5C0xDb6ub96LqZakWemc4gQgGeskoltPR1yaav0MigJmiL2oXcKFpcTL-pDFyBHgnBF4c5zwHb0\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">S\u00f3 que o chuveiro foi ligado em uma ddp de 110 e queremos descobrir a nova pot\u00eancia:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/sjYqKOGQR9VLCPpV14zJzSHSWkcbUJ5AQL40AnYq1wGyRbIz039ehfceQSoeRfl09gCRKL6yRwjGDHa5644UZZIXlQMCQxje9Ux2xEOLCz9cBaOq5jXp48aPA7E65XiakKzCFHgV\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que 220 \u00e9 o dobro de 110. Ent\u00e3o, podemos isolar a resist\u00eancia e igualar as duas f\u00f3rmulas:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/_ykkitoz7YxcPTveUnif3beoLPJklwkdOGyacWEgDadav6ze7-FPsXL_-felVphaU-0ho5XPmS3Ud8K9BNcZYWH8u6uYEPH2X8VL8brknAKylLg9OaZ59yau867wJdb_cWrYUGji\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">4P = 6000<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 1500 W<\/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>(UFJF) Um computador \u00e9 ligado a um no-break, que, basicamente, \u00e9 um sistema armazenador de energia. Quando falta energia el\u00e9trica, o no-break entra em funcionamento, fazendo com que o computador permane\u00e7a funcionando por mais certo tempo. Determine o tempo m\u00e1ximo que o computador fica ligado, ap\u00f3s faltar energia el\u00e9trica, sabendo que a pot\u00eancia do computador \u00e9 de 500W (ou seja, 0,5 kW) e que a energia m\u00e1xima do no-break \u00e9 de 2 kWh.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 4h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 5h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 10h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 0,4h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 0,5h<\/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\">Para essa quest\u00e3o, vamos usar:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/zCljjviHgCxZUi_-9Led5lLg2ABNYNMFj1xcslx4V3xB5MxCqwbnloUabDOENrejIhSffhEHfLsFt063BxmTZatugCH65YVF3HpUa6HZFYjejMKrskSStCeFH_IxewPZfLpj_zbT\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">No entanto, cuidado com as unidades. Veja que temos 500 W e 2 kWh e queremos descobrir o tempo. Ent\u00e3o, temos que transformar o &#8216;k&#8217;:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/wLzj9xcGGIaoYWdA-kgoqg5AyiYfIN3JrUXsJNE36DUVLVMZhRIU0Cd5MbN89d8JdU8GR3N0zDX1yxnvmrGZPAgOYehFd1Nj-qcyYJdfwtWRtQlu5eh2eTllHm1G7iahKxeGfUEj\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">\u0394t = 4 horas<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: A<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 4<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(UFPE) Um chuveiro el\u00e9trico funciona a uma pot\u00eancia de 3600W. Qual o consumo mensal de energia em kwh, se ele \u00e9 usado durante 15 minutos diariamente? Considere o m\u00eas com 30 dias.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 27<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 25<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 23<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 21<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 19<\/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\">Direto:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">E = P . \u0394t<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Veja que a quest\u00e3o nos pede em kWh, ou seja, dividir a pot\u00eancia por 1000. A convers\u00e3o deve ser feita de minutos para horas e multiplicado por 30 dias:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/vPRhS8HzwDYHPP2rI_XD-nKu5C8zOWsF8Bx1WJA3drZuK-GDhbi7dOjguMw1fsVDOJqhbrMcWmVUsRf4o4ZUR68BkoUuTvevNJSpqkstJ6KlxCFDjXnbVeyNgyJX4Kk_2EmVzDX_\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">E = 27 kWh<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: A<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 5<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Fatec) No an\u00fancio promocional de um ferro de passar roupas a vapor, \u00e9 explicado que, em funcionamento, o aparelho borrifa constantemente 20 g de vapor de \u00e1gua a cada minuto, o que torna mais f\u00e1cil o ato de passar roupas. Al\u00e9m dessa explica\u00e7\u00e3o, o an\u00fancio informa que a pot\u00eancia do aparelho \u00e9 1440 W e que sua tens\u00e3o de funcionamento \u00e9 de 110 V.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Jorge comprou um desses ferros e, para utiliz\u00e1-lo, precisa comprar tamb\u00e9m uma extens\u00e3o de fio que conecte o aparelho a uma \u00fanica tomada de 110 V dispon\u00edvel no c\u00f4modo em que passa roupas. As cinco extens\u00f5es que encontra \u00e0 venda suportam as intensidades de correntes m\u00e1ximas de 5 A, 10 A, 15 A, 20 A e 25 A, e seus pre\u00e7os aumentam proporcionalmente \u00e0s respectivas intensidades.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Sendo assim, a op\u00e7\u00e3o que permite o funcionamento adequado de seu ferro de passar em pot\u00eancia m\u00e1xima, sem danificar a extens\u00e3o de fio e que seja a de menor custo para Jorge, ser\u00e1 a que suporta o m\u00e1ximo de<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 5 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 10 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 15 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 20 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 25 A<\/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\">Vamos reunir os dados que temos:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 1440 W<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">U = 110 V<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A \u00fanica coisa que vamos usar de todas as informa\u00e7\u00f5es dadas \u00e9 que:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = U . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1440 = 110 . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i \u2245 13,1 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 6<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(IFBA) Um disjuntor \u00e9 um dispositivo eletromec\u00e2nico destinado a proteger circuitos contra a sobrecarga e o superaquecimento. Pretende-se dimensionar um disjuntor para proteger um ambiente cuja rede el\u00e9trica fornece uma tens\u00e3o de 120 V e possui uma l\u00e2mpada de 60 W, um ar condicionado de 1000 W e um computador de 140 W. Este ambiente ficar\u00e1 mais bem protegido, considerando-se a toler\u00e2ncia de 30%, com um disjuntor de:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 30 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 22 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 20 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 13 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 10 A<\/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 que temos que calcular a pot\u00eancia total que o circuito tem que aguentar e considerar 30% a mais.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Lembre-se de que para calcular a pot\u00eancia total, basta somar todas as pot\u00eancias que temos:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 60 + 1000 + 140<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = 1200 W<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Como queremos saber o disjuntor que precisamos, temos que calcular a pot\u00eancia que o sistema precisa:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P = U . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1200 = 120 . i<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">i = 10 A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">N\u00e3o se esque\u00e7a da toler\u00e2ncia de 30%:<\/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\">Para aprender mais: <\/h3>\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=\"Pot\u00eancia El\u00e9trica Energia El\u00e9trica - F\u00edsica - Eletrodin\u00e2mica - Aula 2 - Professor Pinguim\" width=\"800\" height=\"450\" src=\"https:\/\/www.youtube.com\/embed\/1nqpt0Pngjw?start=447&amp;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\">E se quiser ajuda para melhorar seu n\u00edvel de F\u00edsica em outras mat\u00e9rias,\u00a0<a href=\"https:\/\/professorpinguim.com.br\/\"><strong>entre em contato comigo e escolha o curso de F\u00edsica mais adequado para voc\u00ea<\/strong><\/a><strong>!<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/conteudo.professorpinguim.com.br\/curso-de-fisica-gratuito\" target=\"_blank\" rel=\"noreferrer noopener\"><img decoding=\"async\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/05\/curso-fisica-professor-pinguim.png\" alt=\"Curso de F\u00edsica Gr\u00e1tis do Professor Pinguim\" class=\"wp-image-11364\"\/><\/a><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>SAIBA MAIS<\/strong><br>\u2705&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/blog\/energia-potencial-gravitacional\/\" target=\"_blank\" rel=\"noreferrer noopener\">Energia potencial gravitacional<\/a><br>\u2705&nbsp;<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 CEU<\/a><br>\u2705&nbsp;<a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/potencial-eletrico-energia-e-trabalho-exercicios-com-gabarito\/\" target=\"_blank\">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 pot\u00eancia e energia el\u00e9trica. Antes, no entanto, vamos revisar alguns conceitos e f\u00f3rmulas importantes. Acompanhe!<\/p>\n","protected":false},"author":1,"featured_media":8849,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[746,729,744,770,742],"tags":[812,211,813,777],"class_list":["post-8847","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-eletricidade","category-eletrostatica","category-exercicios-resolvidos","category-gravitacao","category-hidrostatica","tag-energia-eletrica","tag-exercicios","tag-potencia-e-energia-eletrica","tag-potencial-eletrico"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Pot\u00eancia e Energia El\u00e9trica: 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\/potencia-e-energia-eletrica-exercicios-resolvidos\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Pot\u00eancia e Energia El\u00e9trica: Exerc\u00edcios resolvidos - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Nesta aula, vamos resolver juntos algumas quest\u00f5es sobre pot\u00eancia e energia el\u00e9trica. 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