{"id":3806,"date":"2018-02-15T11:03:37","date_gmt":"2018-02-15T11:03:37","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3806"},"modified":"2024-08-20T16:06:04","modified_gmt":"2024-08-20T16:06:04","slug":"resolucao-comentada-eletrodinamica-circuitos-com-geradores-e-aparelhos-de-medidas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/eletrodinamica-circuitos-com-geradores-e-aparelhos-de-medidas\/resolucao-comentada-eletrodinamica-circuitos-com-geradores-e-aparelhos-de-medidas\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada – ELETRODIN\u00c2MICA"},"content":{"rendered":"

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

ELETRODIN\u00c2MICA<\/b><\/span><\/span><\/span><\/p>\n

Circuitos com Geradores e Aparelhos de Medidas<\/b><\/span><\/span><\/span><\/p>\n

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

01- (UNESP-SP)<\/b><\/span><\/span><\/span><\/p>\n

Quando R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=6\u2126\u00a0 —\u00a0 observe as sequ\u00eancias abaixo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 4=1,2\/i\u00a0 —\u00a0 i=0,3A\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=R.i=2.0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=0,6V<\/b><\/span><\/span><\/span><\/p>\n

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

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

02-(UFSCAR-SP)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>–<\/b><\/span><\/span><\/span>\u00a0O \u201cfio terra\u201d indica que o potencial no ponto B \u00e9 nulo (V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0\u00ad)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=1 + 2 + 4\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=7\u2126\u00a0 —\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 7=14\/i\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>.i\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=4.2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 0=8\u00a0 — <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=8V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>.i\u00a0 —\u00a0 0 \u2013 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=2.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>= – 4V\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

Obs: Se a polaridade do gerador fosse invertida, o sentido da corrente tamb\u00e9m seria invertido e os valores seriam V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= – 8V e V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=2V<\/b><\/span><\/span><\/span><\/p>\n

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

No fio neutro o potencial \u00e9 nulo\u00a0 —\u00a0\u00a0\u00a0 diferen\u00e7a de potencial (tens\u00e3o) entre os pontos A e P (quatro l\u00e2mpadas de cima) \u00e9 VA\u00a0\u2013 VP=110 \u2013 0=110V, e nas duas l\u00e2mpadas de baixo (entre P e B) \u00e9 VP\u00a0\u2013 VB=0 \u2013 (-110)=110V.\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

04-(Mackenzie-SP)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Como os dois geradores est\u00e3o ligados corretamente, eles forncem uma ddp U=(1,5 + 1,5)=3V\u00a0 —\u00a0 c\u00e1lculo da resist\u00eancia de cada l\u00e2mpada\u00a0 —\u00a0 P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 0,5=(1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 R=2\u2126\u00a0 —\u00a0 Chave aberta (figura <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

1)\u00a0 —\u00a0\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 6 + 2r=(1,5 + 1,5)\/0,3\u00a0 —\u00a0 2r=4\u00a0 —\u00a0 r=2\u2126\u00a0 —\u00a0 chave fechada (figura 2)\u00a0 —\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 6 + 2=3\/i\u00a0 —\u00a0 i=3\/8=0,375\u00a0 —\u00a0 i=375mA\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

05-(UFRGS-RS)<\/b><\/span><\/span><\/span><\/p>\n

As tr\u00eas pilhas em s\u00e9rie fornecem uma fem total de E=4,5V\u00a0 —\u00a0 como todos os elementos est\u00e3o em s\u00e9rie a tens\u00e3o total fornecida (E=4,5V0 \u00e9 igual\u00a0 \u00e0 soma das tens\u00f5es parciais\u00a0 —\u00a0 E= U<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0+U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 4,5=4 + U<\/b><\/span><\/span><\/span>R\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 0,5=R.i\u00a0 —\u00a0 0,5=1.i\u00a0 —\u00a0 i=0,5 A\u00a0 —\u00a0 pot\u00eancia do motor\u00a0 —\u00a0 P=U.i=4.0,5\u00a0 —\u00a0 P=2,0W\u00a0 —<\/b><\/span><\/span><\/span>–\u00a0 <\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

06-(UFSC-SC)<\/b><\/span><\/span><\/span><\/p>\n

I. Falsa\u00a0 —\u00a0 l\u00e2mpadas id\u00eanticas em s\u00e9rie\u00a0 —\u00a0 percorridas pela mesma corrente\u00a0 —\u00a0 brilham por igual<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 veja 01<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 brilham por igual\u00a0 —\u00a0 mesma ddp e mesma corrente<\/b><\/span><\/span><\/span><\/p>\n

IV. Correta\u00a0 —\u00a0 \u00e9 percorrida por maior corrente, pois tem menor resist\u00eancia<\/b><\/span><\/span><\/span><\/p>\n

V. Falsa\u00a0 —\u00a0 a l\u00e2mpada B apaga, entra em curto circuito<\/b><\/span><\/span><\/span><\/p>\n

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

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

Observe que o volt\u00edmetro que indica V=2,0V est\u00e1 ligado apenas nos terminais de um resistor R e que, tirando-o e curto- circuitando o amper\u00edmetro voc\u00ea obter\u00e1 os esquema a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

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

Esquematizando o circuito:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

09-(UFC-CE))<\/b><\/span><\/span><\/span><\/p>\n

Curto-circuitando os amper\u00edmetros, pois s\u00e3o ideais\u00a0 <\/b><\/span><\/span><\/span>—\u00a0\u00a0chave aberta (figura 1)\u00a0 —\u00a0 passa corrente apenas por<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

apenas por R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e os dois amper\u00edmetros indicam a mesma corrente\u00a0 —\u00a0 i\u2019=E\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 fechando a chave (figura 2) a resist\u00eancia do resistor equivalente fica menor que R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e, portanto, a corrente total aumenta (A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0indica mais que antes)\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0indica o mesmo valor (i\u2019=E\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

10-(UEG-GO)<\/b><\/span><\/span><\/span><\/p>\n

a) Est\u00e1 correta pois o amper\u00edmetro est\u00e1 ligado corretamente, ou seja, em s\u00e9rie, e indicar\u00e1 a corrente el\u00e9trica no circuito independente de estar do lado esquerdo ou direito da bateria.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 veja a).<\/b><\/span><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 observe que o amper\u00edmetro est\u00e1 ligado de maneira errada, ou seja, em paralelo com a l\u00e2mpada.<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 \u00e9 a mesma situa\u00e7\u00e3o que a de c).<\/b><\/span><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 veja explica\u00e7\u00f5es anteriores.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

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

11-(UFRGS-RS)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

O Volt\u00edmetro deve estar ligado em paralelo com R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e o amper\u00edmetro em s\u00e9rie com ele\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\u00a0Resolu\u00e7\u00e3o Comentada ELETRODIN\u00c2MICA Circuitos com Geradores e Aparelhos de Medidas \u00a0 01- (UNESP-SP) Quando R3=6\u2126\u00a0 —\u00a0 observe as sequ\u00eancias abaixo Req=U\/i\u00a0 —\u00a0 4=1,2\/i\u00a0 —\u00a0 i=0,3A\u00a0 —\u00a0 UAB=R.i=2.0,3\u00a0 —\u00a0\u00a0UAB=0,6V R- A \u00a0 02-(UFSCAR-SP) \u00a0–\u00a0O \u201cfio terra\u201d indica que o potencial no ponto B \u00e9 nulo (VB=0\u00ad)\u00a0 —\u00a0 Req=1 + 2 + 4\u00a0 —\u00a0 Req=7\u2126\u00a0 —\u00a0 — Req=U\/i\u00a0 —\u00a0 7=14\/i\u00a0 —\u00a0 i=2<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3804,"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-3806","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3806","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=3806"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3806\/revisions"}],"predecessor-version":[{"id":10635,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3806\/revisions\/10635"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3804"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}