{"id":3800,"date":"2018-02-15T10:57:13","date_gmt":"2018-02-15T10:57:13","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3800"},"modified":"2024-08-20T16:01:03","modified_gmt":"2024-08-20T16:01:03","slug":"resolucao-comentada-caracteristicas-das-associacoes-serie-paralelo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/eletrodinamica-caracteristicas-das-associacoes-serie-paralelo\/resolucao-comentada-caracteristicas-das-associacoes-serie-paralelo\/","title":{"rendered":"Caracter\u00edsticas das associa\u00e7\u00f5es S\u00e9rie Paralelo – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Caracter\u00edsticas das associa\u00e7\u00f5es S\u00e9rie Paralelo<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

C\u00e1lculo da corrente\u00a0 el\u00e9trica i de cada fus\u00edvel pelos valores <\/b><\/span><\/span><\/span>nominais\u00a0\u00a0P=iU\u00a0 —\u00a0 i=55\/36 A\u00a0\u00a0 —\u00a0 quando ligadas ao mesmo fus\u00edvel s\u00e3o percorridas pela nova corrente i\u2019, as duas l\u00e2mpadas, cada uma de <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

resist\u00eancia R, ficam associadas em paralelo e, nesse caso, o resistor equivalente vale Req=R\/2\u00a0 —\u00a0 se a resist\u00eancia equivalente cai pela metade, a nova corrente i\u2019 fica igual ao dobro da anterior , pois, R=U\/i com U constante\u00a0 —\u00a0 ou ainda, ambas as l\u00e2mpadas est\u00e3o funcionando dentro das especifica\u00e7\u00f5es e portanto percorridas pela corrente i=55\/36 A\u00a0 —\u00a0 i\u2019=2.(55\/36)\u00a0 —\u00a0 i\u2019=3,05 A\u00a0 — para suportar essa corrente el\u00e9trica, o menor valor do fus\u00edvel deve ser de 5 A, ou seja o laranja\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

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

\u00a0Resist\u00eancia equivalente dos m resistores de resist\u00eancia el\u00e9trica R1\u00a0 —\u00a0 Req1=mR1\u00a0 —\u00a0 idem\u00a0\u00a0 —\u00a0 Req2=nR2\u00a0 —\u00a0 Reqtotal=mR1\u00a0+ nR2\u00a0 —\u00a0 a corrente \u00e9 a mesma em todos os resistores (associa\u00e7\u00e3o s\u00e9rie) e vale\u00a0 —\u00a0 Reqtotal=U\/i\u00a0 —\u00a0 (mR1\u00a0+ nR2)=U\/i\u00a0 —\u00a0\u00a0\u00a0 i=U\/(mR1\u00a0+nR2)\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

03-(CEFET-RJ)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0A tens\u00e3o total \u00e9 a soma das tens\u00f5es de cada l\u00e2mpada\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=n.U<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 117=n.15\u00a0 —\u00a0 n=117\/15\u00a0 —\u00a0 n=7,8\u00a0 —\u00a0 como a tens\u00e3o nos terminais de cada l\u00e2mpada n\u00e3o pode ultrapassar 15V, deve-se selecionar 8 l\u00e2mpadas\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0\u00a0Fechando a chave C, provoca-se um curto circuito nos terminais da l\u00e2mpada A, a corrente se desvia e ela se apaga. Assim, como a resist\u00eancia total diminui, a corrente aumenta na l\u00e2mpada B aumentando seu brilho.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

Pot\u00eancia fornecida pela bateria\u00a0 —\u00a0 Pb=i.U=330.10-3.6\u00a0 —\u00a0 Pb=1,98W\u00a0 —\u00a0 como o LED consome\u00a0 uma pot\u00eancia de 1,0W, sobra para o resistor R uma pot\u00eancia de 1,98 \u2013 1,0\u00a0 —\u00a0 PR=0,98W\u00a0 —\u00a0 PR=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —0,98=R.(330.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=98.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\/(33.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=9,0\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

06-(PUC-SP)<\/b><\/span><\/span><\/span><\/p>\n

Resposta correta\u00a0 —\u00a0 circuito da figura III onde voc\u00ea consegue ligar a l\u00e2mpada, independentemente, em qualquer um dos interruptores\u00a0 —\u00a0 observe a sequ\u00eancia abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

I- Falsa\u00a0 —\u00a0 quando uma das l\u00e2mpadas queima, no circuito em s\u00e9rie, a corrente el\u00e9trica \u00e9 interrompido e a outra l\u00e2mpada apaga.<\/b><\/span><\/span><\/span><\/p>\n

II- Falsa\u00a0 —\u00a0 como as l\u00e2mpadas s\u00e3o id\u00eanticas, cada uma ficar\u00e1 sujeita a uma tens\u00e3o de 110V e, a pot\u00eancia em cada uma delas ficar\u00e1 4 vezes menor, ou seja, de 25W.<\/b><\/span><\/span><\/span><\/p>\n

III- Falsa\u00a0 —\u00a0\u00a0 como as l\u00e2mpadas s\u00e3o id\u00eanticas e cada uma delas ficar\u00e1 sujeita a uma tens\u00e3o de 110V, elas estar\u00e3o dentro das especifica\u00e7\u00f5es, funcionando normalmente.<\/b><\/span><\/span><\/span><\/p>\n

IV- Correta\u00a0 —\u00a0 como s\u00e3o id\u00eanticas, cada uma delas ficar\u00e1 com metade da tens\u00e3o total que \u00e9 de 220V, ou seja, cada uma ficar\u00e1 sujeita \u00e0 tens\u00e3o de 110V.<\/b><\/span><\/span><\/span><\/p>\n

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

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

S\u00e9rie\u00a0 —\u00a0 R1\u00a0+ R2=6\u00a0 —\u00a0 R1=6 \u2013 R2\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R1.R2\/(R1\u00a0+ R2)=4\/3\u00a0 —\u00a0 (6 \u2013 R2).R2=(6 \u2013 R2\u00a0+ R2)\u00a0 —\u00a0 R2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0– 6R2\u00a0+ 8=0\u00a0 —\u00a0 \u221a\u0394=2\u00a0 —\u00a0 R2=(6 \u00b1 2)\/2\u00a0 —\u00a0 R\u20192=4\u2126\u00a0 —\u00a0 R\u2019\u2019=2\u2126\u00a0 —\u00a0 quando um \u00e9 2\u2126, o outro \u00e9 4\u2126 e vice versa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

R e a l\u00e2mpada est\u00e3o em paralelo e ambas sob ddp de 12V\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 6=il.12\u00a0 —\u00a0 il=0,5A\u00a0 —\u00a0 Rl=U\/i=12\/0,5\u00a0\u00a0—\u00a0 Rl=24\u2126\u00a0 —\u00a0 Req=24R\/(24 + R)\u00a0 —\u00a0 Req=U\/i\u00a0 —\u00a0 24R\/(24 + R)=12\/3\u00a0 —\u00a0 20R=96\u00a0 —\u00a0 R=4,8\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

10-(UCS-RS)<\/b><\/span><\/span><\/span><\/p>\n

O fio superior corresponde \u00e0 um dos polos (fases) da fonte e o inferior ao outro\u00a0 —\u00a0 observe atentamente as figuras\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

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

Corrente na l\u00e2mpada\u00a0 —\u00a0 Pl=il.U\u00a0 —\u00a0 150=il.110\u00a0 —\u00a0 il=214\/15A\u00a0 —\u00a0 corrente no ferro\u00a0 —\u00a0 como l\u00e2mpada e ferro est\u00e3o associados em paralelo\u00a0 —\u00a0 idisjuntor=il\u00a0+ if\u00a0 —\u00a0 if=(15 \u2013 214\/15)\u00a0 —\u00a0 if=214\/15A\u00a0 —\u00a0 Pf=if.U\u00a0 — Pf=214\/15.110=1.570W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

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

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

Resolu\u00e7\u00e3o Comentada Caracter\u00edsticas das associa\u00e7\u00f5es S\u00e9rie Paralelo \u00a0 \u00a0 01-(ENEM-MEC) C\u00e1lculo da corrente\u00a0 el\u00e9trica i de cada fus\u00edvel pelos valores nominais\u00a0\u00a0P=iU\u00a0 —\u00a0 i=55\/36 A\u00a0\u00a0 —\u00a0 quando ligadas ao mesmo fus\u00edvel s\u00e3o percorridas pela nova corrente i\u2019, as duas l\u00e2mpadas, cada uma de resist\u00eancia R, ficam associadas em paralelo e, nesse caso, o resistor equivalente vale Req=R\/2\u00a0 —\u00a0 se a resist\u00eancia<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3798,"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-3800","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3800","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=3800"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3800\/revisions"}],"predecessor-version":[{"id":10632,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3800\/revisions\/10632"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3798"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3800"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}