{"id":3783,"date":"2018-02-15T10:06:57","date_gmt":"2018-02-15T10:06:57","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3783"},"modified":"2024-08-20T16:04:10","modified_gmt":"2024-08-20T16:04:10","slug":"resolucao-comentada-eletrodinamica-corrente-eletrica-energia-eletrica-potencia-eletrica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/eletrodinamica-corrente-eletrica-energia-eletrica-potencia-eletrica\/resolucao-comentada-eletrodinamica-corrente-eletrica-energia-eletrica-potencia-eletrica\/","title":{"rendered":"Eletrodin\u00e2mica Corrente el\u00e9trica \u2013 Energia el\u00e9trica \u2013 Pot\u00eancia el\u00e9trica – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span><\/p>\n

Eletrodin\u00e2mica<\/b><\/span><\/span><\/span><\/p>\n

Corrente el\u00e9trica \u2013 Energia el\u00e9trica \u2013 Pot\u00eancia el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

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

01-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

Para voc\u00ea acender a l\u00e2mpada ela deve ser submetida a uma diferen\u00e7a de potencial (tens\u00e3o, voltagem) e, consequentemente percorrida por corrente el\u00e9trica\u00a0 —\u00a0 os dois p\u00f3los da l\u00e2mpada s\u00e3o a parte <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

inferior da rosca (ponto I) e a parte lateral da l\u00e2mpada (ponto L)\u00a0 —\u00a0 os dois polos da pilha est\u00e3o indicados na figura\u00a0 —\u00a0 para que a l\u00e2mpada acenda o ponto L deve estar ligados a um dos p\u00f3los da pilha e o ponto I da l\u00e2mpada ao outro polo da pilha, o que ocorre nas situa\u00e7\u00f5es 1, 3 e 7\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

02-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

Quanto maior o n\u00famero de equipamentos el\u00e9tricos, maior o consumo de energia (III)\u00a0 —\u00a0 observe na express\u00e3o W=Po.\u0394t que o consumo de energia W depende tamb\u00e9m da pot\u00eancia do equipamento Po\u00a0(I) e do tempo \u0394t de funcionamento (II)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

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

03-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Observe que em cada uma das recomenda\u00e7\u00f5es voc\u00ea est\u00e1 tentando diminuir o efeito Joule, ou seja, a transforma\u00e7\u00e3o de energia el\u00e9trica em t\u00e9rmica\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

04-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

A energia el\u00e9trica \u00e9 expressa por W=Po.\u0394t, medida em megawatts.hora\u00a0(MW.h) e n\u00e3o megawatts por hora (MW\/h)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

05-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

Observe na tabela que na vers\u00e3o U=220V, a pot\u00eancia m\u00e1xima da torneira \u00e9 de P=5500W\u00a0 —\u00a0 c\u00e1lculo da resist\u00eancia R da torneira, quando est\u00e1 funcionando na pot\u00eancia m\u00e1xima\u00a0 —\u00a0\u00a0 P=U<\/b><\/span><\/span><\/span>2\/<\/b><\/span><\/span><\/sup><\/span>R\u00a0 —\u00a0 5500=(220)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 R=8,8\u2126\u00a0 —\u00a0 quando voc\u00ea ligar essa torneira de R=8,8\u2126 na tens\u00e3o nominal de U\u2019=127V, a nova pot\u00eancia el\u00e9trica dissipada ser\u00e1\u00a0 —\u00a0 P\u2019=U\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=(127)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/8,8\u00a0 —\u00a0 P\u2019=1830W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

06-(ENEM-MEC)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0O rel\u00f3gio indicou um consumo de 2.563 kwh no m\u00eas anterior e de 2.783 kwh na \u00faltima leitura\u00a0 —\u00a0 a diferen\u00e7a entre essas duas medidas determina um consumo de 220 kwh no per\u00edodo de um m\u00eas\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 kwh – R$ 0,20\u00a0 —\u00a0 220 kwh \u2013 C\u00a0 —\u00a0 C=0,20×220\u00a0 —\u00a0 C= R$ 44,00\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

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

07<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0(UFSM-RS)<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria, leia-a atentamente a seguir:<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Corrente el\u00e9trica (i)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>Num fio met\u00e1lico condutor, os el\u00e9trons livres n\u00e3o est\u00e3o em repouso e seus movimentos s\u00e3o totalmente desordenados. Para orient\u00e1-los estabelece-se entre dois pontos desse condutor uma diferen\u00e7a de potencial (ddp), que origina um campo el\u00e9trico (\"\"), respons\u00e1vel pela orienta\u00e7\u00e3o do movimento desses el\u00e9trons livres. Sendo a carga de um el\u00e9tron negativa, eles se movem em sentido contr\u00e1rio ao do campo el\u00e9trico. Observe na figura, que, devido \u00e0 diferen\u00e7a de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

potencial (V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>), os el\u00e9trons livres (portadores de carga) s\u00e3o repelidos pelo p\u00f3lo negativo , de potencial V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0da bateria (gerador) e atra\u00eddos pelo p\u00f3lo positivo V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, deslocando-se no sentido anti-hor\u00e1rio.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Sentido da corrente el\u00e9trica<\/b><\/span><\/span><\/span>\u00a0\u2013<\/b><\/span><\/span><\/span>\u00a0por conven\u00e7\u00e3o<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>o sentido da corrente el\u00e9trica \u00e9 contr\u00e1rio ao sentido do movimento dos portadores de carga negativa.\u00a0<\/b><\/span><\/span><\/span>Assim, num condutor s\u00f3lido, o sentido da corrente el\u00e9trica \u00e9 contr\u00e1rio<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ao do movimento dos el\u00e9trons livres, nos l\u00edquidos, contr\u00e1rio ao movimento dos \u00edons negativos e nos gases, contr\u00e1rio ao do movimento dos el\u00e9trons.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Intensidade da corrente el\u00e9trica (i)<\/b><\/span><\/span><\/span>\u00a0\u2013 Seja um condutor met\u00e1lico de se\u00e7\u00e3o transversal S, sendo percorrido por<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

uma corrente el\u00e9trica. Se, por<\/b><\/span><\/span><\/span>\u00a0S<\/b><\/span><\/span><\/span>, passam\u00a0<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>\u00a0el\u00e9trons (portadores de carga) num intervalo de tempo (<\/b><\/span><\/span><\/span>\u0394t<\/b><\/span><\/span><\/span>), o m\u00f3dulo da carga el\u00e9trica total (<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>) que flui por S nesse intervalo de tempo ser\u00e1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0Q=n.e<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 onde e \u00e9 o\u00a0 m\u00f3dulo da carga el\u00e9trica elementar (de um el\u00e9tron) e n \u00e9 um n\u00famero inteiro.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Define-se a intensidade de corrente el\u00e9trica como sendo a grandeza escalar\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0i=Q\/\u0394t<\/b><\/span><\/span><\/span>. Esta express\u00e3o fornece a intensidade m\u00e9dia de corrente el\u00e9trica mas tamb\u00e9m pode medir corrente el\u00e9trica constante.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>A unidade de corrente el\u00e9trica no SI \u00e9 o amp\u00e8re ( em homenagem ao f\u00edsico e matem\u00e1tico franc\u00eas Andr\u00e9 Marie Amp\u00e8re \u201c1775-1836\u201d), de s\u00edmbolo\u00a0<\/b><\/span><\/span><\/span>A.<\/b><\/span><\/span><\/span><\/p>\n

I=Q\/\u0394t<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 1 amp\u00e8re=1 coulomb\/1 segundo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>1A=1C\/1s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

No exerc\u00edcio\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q=n.e=1018.1,6.10-19\u00a0 —\u00a0 Q=1,6.10-1C\u00a0 —\u00a0 i=Q\/\u0394t=0,16\/1\u00a0 —\u00a0 i=0,16 A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-(UNIFESP-SP)<\/b><\/span><\/span><\/span><\/p>\n

A primeira afirma\u00e7\u00e3o \u00e9 correta e a segunda \u00e9 falsa, pois o ar pode ser condutor conforme a terceira afirmativa\u00a0 —\u00a0 i=Q\/\u0394t\u00a0 —\u00a0 10.000=20\/\u0394t\u00a0 —\u00a0 \u0394t=20\/10.000=0,002s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

09-(FUVEST-SP)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0i=Q\/\u0394t\u00a0 —\u00a0 300.000=Q\/0,5\u00a0 —\u00a0 Q=150.000C\u00a0 —\u00a0 fra\u00e7\u00e3o f=Q\/QTerra=150.000\/600.000\u00a0 —\u00a0 f=1\/4\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

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

10-(UPE-PE)<\/b><\/span><\/span><\/span><\/p>\n

i=\u0394Q\/\u0394t\u00a0 —\u00a0 0,3= \u0394Q\/120\u00a0 —\u00a0 \u0394Q=36C\u00a0 —\u00a0 Q=n.e\u00a0 —\u00a0 36=n.1,6.10-19\u00a0 —\u00a0 n=2,25.1020\u00a0el\u00e9trons\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

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

11-(UFPE-PE)<\/b><\/span><\/span><\/span><\/p>\n

Em todo gr\u00e1fico corrente el\u00e9trica x tempo a carga el\u00e9trica Q \u00e9 numericamente igual \u00e0 \u00e1rea <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

compreendida entre a reta<\/b><\/span><\/span><\/span>\u00a0representativa e o eixo do tempo (figura)\u00a0 —\u00a0 \u00e1rea de um trap\u00e9zio\u00a0 —\u00a0 Q=(B + b).h\/2=(4 + 1).4\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0Q=10C<\/b><\/span><\/span><\/span><\/p>\n

R- B.<\/b><\/span><\/span><\/span><\/p>\n

12-(UECE-CE)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0C\u00e1lculo da corrente el\u00e9trica fornecida pela bateria\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 60=i.12 \u00a0—\u00a0 i=5 A\u00a0 —\u00a0 a carga Q fornecida pela bateria \u00e9 a mesma que o sistema de som recebe, fornecendo 5 A de corrente durante\u00a0 —\u00a0 i=Q\/\u0394t\u00a0 —\u00a0 5= 60\/\u0394t\u00a0 —\u00a0 \u0394t=60\/5\u00a0 —\u00a0 \u0394t=12h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

13-(UFJF-MG)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0P=iU\u00a0 —\u00a0 7.200=i.120\u00a0 —\u00a0\u00a0i=60 A\u00a0 —\u00a0 Wtotal=4kW.2h<\/b><\/span><\/span><\/span> + 6kW.2h + 2kW.2h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=24kWh\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 W=24.30 =720kWh\u00a0 —\u00a0 pre\u00e7o\u00a0 —\u00a0 p=720.0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0p=R$ 360,00<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

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

14-(PUC-RJ)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Dados: P = 4.500 W\u00a0 —\u00a0 \u00a0U = 110 V\u00a0 —\u00a0 P = i U\u00a0 —\u00a0 \u00a0i = P\/U=4.500\/110=40,9 A\u00a0 —\u00a0 portanto o disjuntor escolhido dever\u00e1 ser o de 45 A, que \u00e9 o valor mais pr\u00f3ximo do acima do calculado\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

15- (UNIFOR-CE)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

A tens\u00e3o alternada produzida pelos geradores na usina hidrel\u00e9trica \u00e9 relativamente baixa. Assim, para que se possa abastecer diferentes centros utilizando linhas de transmiss\u00e3o, essa tens\u00e3o \u00e9 aumentada at\u00e9 centenas ou milhares de kV por meio de transformadores. Ao atingir os centros de consumo, a tens\u00e3o \u00e9 reduzida, por exemplo, a algumas dezenas de kV, pelos transformadores das subesta\u00e7\u00f5es e distribu\u00edda para o p\u00fablico.\u00a0 A tens\u00e3o de transmiss\u00e3o \u00e9 elevada e a corrente diminu\u00edda porque assim se podem utilizar fios condutores mais finos (mais leves, mais baratos) j\u00e1 que, para que a pot\u00eancia seja a mesma (P=Ui) se voc\u00ea aumentar a tens\u00e3o (U) voc\u00ea deve diminuir a corrente i.<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u00a0 Eletrodin\u00e2mica Corrente el\u00e9trica \u2013 Energia el\u00e9trica \u2013 Pot\u00eancia el\u00e9trica \u00a0 01-(ENEM-MEC) Para voc\u00ea acender a l\u00e2mpada ela deve ser submetida a uma diferen\u00e7a de potencial (tens\u00e3o, voltagem) e, consequentemente percorrida por corrente el\u00e9trica\u00a0 —\u00a0 os dois p\u00f3los da l\u00e2mpada s\u00e3o a parte inferior da rosca (ponto I) e a parte lateral da l\u00e2mpada (ponto L)\u00a0 —\u00a0 os<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3781,"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-3783","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3783","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=3783"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3783\/revisions"}],"predecessor-version":[{"id":10634,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3783\/revisions\/10634"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3781"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}