{"id":1786,"date":"2016-01-11T22:20:46","date_gmt":"2016-01-11T22:20:46","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1786"},"modified":"2024-12-02T20:49:02","modified_gmt":"2024-12-02T20:49:02","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-caracteristicas-das-associacoes-serie-e-paralelo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/serie-paralelo-caracteristicas\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-caracteristicas-das-associacoes-serie-e-paralelo\/","title":{"rendered":"Caracter\u00edsticas das associa\u00e7\u00f5es s\u00e9rie e paralelo – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/b><\/p>\n

Caracter\u00edsticas das associa\u00e7\u00f5es s\u00e9rie e paralelo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0Como eles s\u00e3o id\u00eanticos, a tens\u00e3o nos terminais de cada um deles \u00e9 igual \u00e0 metade da tens\u00e3o<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

total\u00a0 —\u00a0 U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U\u2019=252\/2\u00a0 —\u00a0U\u2019=126V\u00a0 —\u00a0 R=U\u2019\/i=126\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=63\u2126<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=100 + 100 + 200\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=400\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/I\u00a0 —\u00a0 400=U\/0,5\u00a0 —\u00a0 U=200V\u00a0 —\u00a0 P=i.U=0,5.200\u00a0 —\u00a0 P=100W\u00a0 —\u00a0\u00a0 ou\u00a0 —\u00a0 P=R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0400.0,25=100W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> C\u00e1lculo da corrente i, que \u00e9 a mesma nos dois resistores\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 110=220\/i\u00a0 —\u00a0 i=2\u00aa\u00a0 —\u00a0 no resistor de 10\u2126\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 10=U\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=20V<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Resist\u00eancia equivalente dos m resistores de resist\u00eancia el\u00e9trica R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eq1<\/b><\/span><\/span><\/sub>=mR<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 idem\u00a0\u00a0 —\u00a0 R<\/b><\/span><\/span>eq2<\/b><\/span><\/span><\/sub>=nR<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub>=mR<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ nR<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a corrente \u00e9 a mesma em todos os resistores (associa\u00e7\u00e3o s\u00e9rie) e vale\u00a0 —\u00a0 R<\/b><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 (mR<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ nR<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=U\/i\u00a0 —\u00a0\u00a0\u00a0 i=U\/(mR<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+nR<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0A tens\u00e3o total \u00e9 a soma das tens\u00f5es de cada l\u00e2mpada\u00a0 —\u00a0 U<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=n.U<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>\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>\u00a0R- D<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\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>\u00a0R- A<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0l\u00e2mpada\u00a0 —\u00a0 P=i.U\u00a0 —\u00a0 2,25=4,5.i\u00a0 —\u00a0 i=0,5A\u00a0 —\u00a0 como est\u00e3o em s\u00e9rie a corrente i\u00a0 tamb\u00e9m \u00e9 0,5A no resistor\u00a0 —\u00a0\u00a0 no resistor\u00a0 —\u00a0 U=12 \u2013 4,5=7,5V\u00a0 —\u00a0 R=U\/i=7,5\/0,5\u00a0 —\u00a0 R=15 \u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>As l\u00e2mpadas ter\u00e3o correntes el\u00e9tricas diferentes, pois possuem valores nominais diferentes\u00a0 —\u00a0 corrente em cada l\u00e2mpada\u00a0 —\u00a0 i<\/b><\/span><\/span>100W<\/b><\/span><\/span><\/sub>=P\/U=100\/110=0,91A —\u00a0 i<\/b><\/span><\/span>60W<\/b><\/span><\/span><\/sub>=60\/220=0,28A\u00a0 — A corrente nessa associa\u00e7\u00e3o s\u00e9rie n\u00e3o pode exceder 0,28A, pois ultrapassaria a pot\u00eancia da l\u00e2mpada de menor pot\u00eancia (60W)\u00a0 —\u00a0 quando i=0,28A\u00a0\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 100=0.28U\u00a0 —\u00a0 U=360V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Corrente na l\u00e2mpada\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 100=i.120\u00a0 —\u00a0 i=5\/6A\u00a0 —\u00a0 resistor\u00a0 —\u00a0 U=240 \u2013 120=120V\u00a0 —\u00a0 P = iU=5\/6.120\u00a0 —\u00a0 100W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0U em cada l\u00e2mpada\u00a0 —\u00a0 U=110\/20=5,5V\u00a0 —\u00a0 i em cada l\u00e2mpada\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 5=i.5,5\u00a0 —\u00a0 i=1\/1,1A\u00a0 —\u00a0 R=U\/i=5,5\/(1\/1,1)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- 6,05\u2126<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=(R + 100)=U\/i\u00a0 — R + 100=110\/0,5\u00a0 —\u00a0 R=120\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0U<\/b><\/span><\/span>rede<\/b><\/span><\/span><\/sub>=U\u00a0 —\u00a0 U<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,1U\u00a0 —\u00a0 U<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=0,9U\u00a0 —\u00a0 i<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=0,9U\/X\u00a0 —\u00a0 i<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,1U\/R\u00a0 —\u00a0 s\u00e9rie\u00a0 —\u00a0 i<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0.9U\/X=0,1U\/R\u00a0 —\u00a0 X=9R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0I- Correta \u2013 9 l\u00e2mpadas \u2013 90W + 1 l\u00e2mpada de 5W\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=95W\u00a0 —\u00a0 P de cada l\u00e2mpada\u00a0 =95\/10=9,5W<\/b><\/span><\/span><\/p>\n

II- Correta –\u00a0 9 l\u00e2mpadas \u2013 90W + 1 l\u00e2mpada de 12W\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=102W\u00a0 —\u00a0 P de cada l\u00e2mpada\u00a0 =102\/10=10,2W<\/b><\/span><\/span><\/p>\n

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

14-<\/b><\/span><\/span><\/span>\u00a0Pot\u00eancia fornecida pela bateria\u00a0 —\u00a0 P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=i.U=330.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.6\u00a0 —\u00a0 P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=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 P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,98W\u00a0 —\u00a0 P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 0,98=R.(330.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=98.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/(33.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=9,0\u2126\u00a0 —<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Resposta 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 seq\u00fc\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"Associa\u00e7\u00f5es<\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0R- C\u00a0 —\u00a0 veja exerc\u00edcio anterior<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da corrente el\u00e9trica i no circuito que est\u00e1 em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2 + 5 + 8 + 5\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=20\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 20=100\/i\u00a0 —<\/b><\/span><\/span><\/p>\n

i=5A\u00a0 —\u00a0 pot\u00eancia dissipada por R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P=R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=8.(5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=200W\u00a0 —\u00a0 energia liberada pelo resistor imerso\u00a0 —\u00a0 W=P.\u0394t\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

W=200.\u0394t (I)\u00a0 —\u00a0 energia necess\u00e1ria para transformar 200g de gelo a 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C at\u00e9 \u00e1gua a 20<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 Q=mL + mc\u0394\u03b8=200.80 + 200.1.(20 \u2013 0)=16.000 + 4.000\u00a0 —\u00a0 Q=20.000cal\u00a0 —\u00a0 1cal=4J\u00a0 —\u00a0 Q=W=4.20.000\u00a0 —\u00a0 W=80.000J (II)\u00a0 —\u00a0 igualando II com I\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

200\u0394t=80.000\u00a0 —\u00a0 \u0394t=400s —\u00a0 \u0394t=400s\/60=6,7\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394t=6,7 minutos<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Observe que R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0e R<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em s\u00e9rie\u00a0 —\u00a0 o volt\u00edmetro ideal indica U<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=10V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

esquematizando o circuito\u00a0 —\u00a0 como est\u00e3o\u00a0 todos em s\u00e9rie, a corrente i \u00e9 a mesma e vale\u00a0 —\u00a0 U<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 10=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.i\u00a0 —\u00a0 i=10\/10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 i=10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>A\u00a0\u00a0 —\u00a0 U<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 U<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 U<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=10V\u00a0 —\u00a0 U<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>P\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 30=10 + U<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0+ 10\u00a0 —\u00a0 U<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=10V\u00a0 — \u00a0U<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 10=R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=10\/10\u20115<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=1M\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

19-<\/b><\/span><\/span><\/span>\u00a0I- 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><\/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><\/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><\/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><\/p>\n

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

20-<\/b><\/span><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

b) entre R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R\u2019<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=240\/3=80\u2126\u00a0 —\u00a0 entre R\u2019<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=80\u2126 e 80\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=80\/2\u00a0 —<\/b><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=40\u2126<\/b><\/span><\/span>\u00a0 — R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 40=120\/i\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=3A\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

c) R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U\/i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 240=120\/i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,5A\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U\/i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 120=120\/i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1,0A —\u00a0 R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=U\/i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 80=120\/i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=1,5A\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=8\/3\u2126 \u00a0—\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 8\/3=U\/3\u00a0 —\u00a0 U=8V\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=8\/4=<\/b><\/span><\/span>2A<\/b><\/span><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=8\/8=<\/b><\/span><\/span>1A<\/b><\/span><\/span>\u00a0 —<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja resolu\u00e7\u00e3o anterior<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da ddp U no resistor de 20\u2126\u00a0 — U=R.i=20.4\u00a0 —\u00a0 U=80V\u00a0 —\u00a0 c\u00e1lculo de i no resistor de 10\u2126\u00a0 —\u00a0 i=U\/R=80\/10=<\/b><\/span><\/span>8A<\/b><\/span><\/span>\u00a0 —\u00a0 c\u00e1lculo de R\u00a0 —\u00a0 R=U\/i=80\/16=<\/b><\/span><\/span>5<\/b><\/span><\/span>\u2126\u00a0 —<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0S\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=6\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=4\/3\u00a0 —\u00a0 (6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>).R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0– 6R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 8=0\u00a0 —\u00a0 \u221a\u0394=2\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(6 \u00b1 2)\/2\u00a0 —\u00a0 R\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=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>R- C<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0a) Sim, \u00e9 \u00f4hmico\u00a0 —\u00a0 a resist\u00eancia \u00e9 constante\u00a0 —\u00a0 R=U\/i=0,6\/0,2=1,8\/0,6=3,0\/1,0=<\/b><\/span><\/span>3,0\u2126<\/b><\/span><\/span><\/p>\n

b) P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=(9)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0 P=27W\u00a0 —\u00a0 W=P.\u0394t=27.300\u00a0 —\u00a0 W=8.100J<\/b><\/span><\/span><\/p>\n

c) R<\/b><\/span><\/span>eq1<\/b><\/span><\/span><\/sub>=R\/3\u00a0 —\u00a0 R<\/b><\/span><\/span>eq2<\/b><\/span><\/span><\/sub>=3R\u00a0 —\u00a0 como P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0\u00e9 inversamente proporcional a P, pois U \u00e9 a mesma\u00a0 —\u00a0 aquele que aquece mais rapidamente a \u00e1gua \u00e9 o que tem maior pot\u00eancia e consequentemente menor R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0associa\u00e7\u00e3o I<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0a) Os aparelhos est\u00e3o associados em paralelo e submetidos \u00e0 mesma ddp de U=110V\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 i=P\/U\u00a0 —\u00a0 i<\/b><\/span><\/span>aquec<\/b><\/span><\/span><\/sub>=2.200\/110=20A\u00a0 —\u00a0 i<\/b><\/span><\/span>ferro<\/b><\/span><\/span><\/sub>=770\/110=7A\u00a0 —\u00a0 i<\/b><\/span><\/span>l\u00e2mpada<\/b><\/span><\/span><\/sub>=100\/110=0,91A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>ferro e l\u00e2mpada<\/b><\/span><\/span><\/p>\n

b) n=15\/0.91=16,5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a016 l\u00e2mpadas<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq1<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq1<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=12\/1\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) (I)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eq2<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12\/0.4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=30\u2126<\/b><\/span><\/span>\u00a0(II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.30=12(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 30)\u00a0 —\u00a0 18R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=360\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=20\u2126<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0Todos os dispositivos el\u00e9tricos de uma resid\u00eancia s\u00e3o associados em paralelo para que fiquem sujeitos \u00e0 mesma diferen\u00e7a de potencial\u00a0 —\u00a0 aquecedor el\u00e9trico\u00a0 —\u00a0 W=P.\u0394t=200.1=200Wh\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 W=P.\u0394t=60.24=1.440W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0S\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eqs<\/b><\/span><\/span><\/sub>=10R\u00a0 —\u00a0 R<\/b><\/span><\/span>eqs<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 10R=120\/0,05\u00a0 —\u00a0 R=240\u2126\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span>eqp<\/b><\/span><\/span><\/sub>=R\/10\u00a0 —\u00a0 R\/10=U\/i\u00a0 —\u00a0 240\/10=120\/i\u00a0 —\u00a0 i=5A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0Se a ddp em seus terminais \u00e9 constante as l\u00e2mpadas est\u00e3o ligadas em paralelo\u00a0 —\u00a0 c\u00e1lculo da corrente em cada l\u00e2mpada\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 60=i.120\u00a0 —\u00a0 i=0,5A\u00a0 —\u00a0 n=15\/0,5=<\/b><\/span><\/span>30 l\u00e2mpadas\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0R 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=i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.12\u00a0 —\u00a0 i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=0,5A\u00a0 —\u00a0 R<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=U\/i=12\/0,5\u00a0\u00a0<\/b><\/span><\/span>—<\/b><\/span><\/span>\u00a0 R<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=24\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=24R\/(24 + R)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 24R\/(24 + R)=12\/3\u00a0 —\u00a0 20R=96\u00a0 —\u00a0 R=4,8\u2126\u00a0 —\u00a0 R- E<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0Corrente em cada l\u00e2mpada\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 330=220\/i\u00a0 —\u00a0 i=22\/33 A\u00a0 —\u00a0 n=50\/(22\/33)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0n=75 l\u00e2mpadas<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0a) P=iU=6\/11.110\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=60W<\/b><\/span><\/span><\/p>\n

b) n=14\/(6\/11)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>n=25 l\u00e2mpadas<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O fio superior corresponde \u00e0 um dos p\u00f3los da fonte e o inferior ao outro\u00a0 —\u00a0 observe atentamente as figuras\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0Observe que os pontos M,N,O,P,Q,R,S e T est\u00e3o em curto circuito, ent\u00e3o os 4 resistores est\u00e3o em<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0R\/4=U\/i\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=4U\/R<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0corrente na l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.U\u00a0 —\u00a0 150=i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.110\u00a0 —\u00a0 i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=214\/15A\u00a0 —\u00a0 corrente no ferro\u00a0 —\u00a0 i<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=(15 \u2013 214\/15)\u00a0 —\u00a0 i<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=214\/15A\u00a0 —\u00a0 P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>.U=214\/15.110=1.570W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0Chuveiro\u00a0 —\u00a0 i<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=P\/U=3.000\/110=27,27A\u00a0 —\u00a0 geladeira\u00a0 —\u00a0 i<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>=400\/110=3,63A\u00a0 —\u00a0 TV\u00a0 —\u00a0 i<\/b><\/span><\/span>TV<\/b><\/span><\/span><\/sub>=150\/110=1,36A\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 i<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=60\/110=0,54A\u00a0 —\u00a0 faltam 30 \u2013 27,7=2,3A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- L\u00e2mpada e TV<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(2R\/3)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=3U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2R<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0Corrente total i no disjuntor\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>t.<\/b><\/span><\/span><\/sub>U\u00a0 —\u00a0 (1.400 + 920)=i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>.110\u00a0 —\u00a0 i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=21,1A\u00a0 —\u00a0 para que o disjuntor n\u00e3o desligue, a corrente el\u00e9trica m\u00ednima deve ser de 25 A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0L\u00e2mpada Q\u00a0 —\u00a0 P<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>=(U\/2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

41<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> Dados\u00a0 —\u00a0 U = 12,4 V\u00a0 —\u00a0 i = 0,8 A\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 0,2 \u03a9\u00a0 —\u00a0 r = 0,3 \u03a9\u00a0 —\u00a0 \u00a0resist\u00eancia equivalente da associa\u00e7\u00e3o\u00a0 —\u00a0 U = R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 12,4 = R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a00,8\u00a0 —\u00a0 Req =12,4\/0,8\u00a0 —\u00a0 Req = 15,5 \u03a9\u00a0 —\u00a0 os valores das resist\u00eancias est\u00e3o em progress\u00e3o aritm\u00e9tica (P.A.)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0= a<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ (n \u2013 1) r\u00a0 —\u00a0 R<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0= R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ (n \u2013 1) r\u00a0 —\u00a0 R<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0= 0,2 + (n \u2013 1) 0,3\u00a0 —\u00a0 R<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0= 0,3 n \u2013 0,1\u00a0 — como os resistores est\u00e3o associados em s\u00e9rie, a resist\u00eancia equivalente \u00e9 a soma das resist\u00eancias\u00a0 —\u00a0 soma dos n primeiros termos de uma P.A.\u00a0 —\u00a0\u00a0 S<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0=(a<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>).n\/2\u00a0 —\u00a0 \u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>).n\/2\u00a0 —\u00a0 15,5={0,2 + (0,3n \u2013 1).n}\/2\u00a0 —\u00a0 31=0,3n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 0,1n\u00a0 —\u00a0 3n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ n \u2013 310=0\u00a0 —\u00a0 resolvendo a equa\u00e7\u00e3o do 2\u00ba grau\u00a0 —\u00a0 n= (-1 \u00b1\u221a3.721)\/6\u00a0 —\u00a0 desprezando a resposta negativa\u00a0 —\u00a0 n= (-1 + 61)\/6\u00a0 —\u00a0 n=10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0(01) Correta\u00a0 —\u00a0 a pot\u00eancia el\u00e9trica \u00e9 dado por: P = U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R, sendo\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>\u00a0a resist\u00eancia el\u00e9trica\u00a0 —\u00a0 como a tens\u00e3o \u00e9 a mesma para as duas l\u00e2mpadas, a de maior pot\u00eancia apresenta menor resist\u00eancia, ou seja, R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0<<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 segunda lei de Ohm\u00a0 —\u00a0 \u00a0R = \u03c1L\/S\u00a0 —\u00a0 nessa express\u00e3o,\u00a0<\/b><\/span><\/span>\u03c1<\/b><\/span><\/span>\u00a0\u00e9 a resistividade do material;\u00a0<\/b><\/span><\/span>S<\/b><\/span><\/span>\u00a0\u00e9 a \u00e1rea da sec\u00e7\u00e3o transversal do filamento e\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>\u00a0\u00e9 o seu comprimento\u00a0 —\u00a0 se ambas forem do mesmo material e de mesmo di\u00e2metro, a resist\u00eancia ser\u00e1 diretamente proporcional ao comprimento\u00a0 —\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0<<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0e<\/b><\/span><\/span><\/p>\n

L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0<<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 a intensidade luminosa \u00e9 diretamente proporcional a pot\u00eancia luminosa emitida\u00a0 —\u00a0\u00a0 pot\u00eancia luminosa de A\u00a0 —\u00a0\u00a0 P\u2019<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= 30% de 100 = 0,3(100) = 30 W\u00a0 —\u00a0 pot\u00eancia luminosa de B P\u2019<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= 30% de 40 = 0,3(40) = 12 W\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u2019\/P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u2019=30\/12=2,5
\n(04) Correta\u00a0 —\u00a0 a resistividade do filamento aumenta com a temperatura\u00a0 —\u00a0 o brilho de uma l\u00e2mpada \u00e9 devido ao aquecimento de seu filamento a altas temperaturas\u00a0 —\u00a0 assim, quando desligadas, as l\u00e2mpadas apresentam menor resistividade, consequentemente, menor resist\u00eancia.<\/b><\/span><\/span><\/p>\n

(08) Falsa\u00a0 —\u00a0 quando est\u00e3o associadas em s\u00e9rie, ambas s\u00e3o percorridas pela mesma corrente el\u00e9trica\u00a0 —\u00a0 \u00a0pot\u00eancia el\u00e9trica\u00a0 — \u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0= R<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como j\u00e1 conclu\u00eddo, R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0<<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 ent\u00e3o P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0< P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a intensidade luminosa de<\/b><\/span><\/span>A<\/b><\/i><\/span><\/span>\u00a0\u00e9 menor que a de\u00a0<\/b><\/span><\/span>B<\/b><\/i><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

(16) Falsa\u00a0 —\u00a0 Como j\u00e1 justificado, R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0<<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

R- (01+ 02 + 04) = 07<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Correta\u00a0 —\u00a0 em toda associa\u00e7\u00e3o s\u00e9rie a intensidade da corrente el\u00e9trica \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

II. C\u00e1lculo da resist\u00eancia, suposta constante, de cada l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 60=14.400\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=240\u03a9\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

100=14.400\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=144\u03a9\u00a0 —\u00a0 resistor equivalente\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=240 + 144\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=384\u03a9\u00a0 —\u00a0 corrente el\u00e9trica em cada resistor\u00a0 —<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 384=120\/i\u00a0 —\u00a0 i=120\/384\u00a0 —\u00a0 i=0,3 A\u00a0 —\u00a0 c\u00e1lculo da pot\u00eancia dissipada em cada l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=240.0,9\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=216W\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=144.0,9\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=129,6W\u00a0 — a l\u00e2mpada 1 brilha mais, pois P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>>P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 veja justificativa anterior<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A afirmativa errada \u00e9 a C\u00a0 —\u00a0 veja seq\u00fc\u00eancia abaixo\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,72 + 1,44 + 1,44=3,6\u03a9\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 3,6=6\/i\u00a0 —\u00a0 i=1,66 A\u00a0<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

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

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Falsa – Em uma liga\u00e7\u00e3o em s\u00e9rie, a corrente el\u00e9trica que percorre as l\u00e2mpadas \u00e9 a mesma. Assim, quando uma l\u00e2mpada queima, abre-se o circuito.<\/b><\/span><\/span><\/p>\n

II. Falsa – Em uma liga\u00e7\u00e3o em s\u00e9rie, a d.d.p. total \u00e9 dividida para as duas l\u00e2mpadas. Assim, elas ter\u00e3o d.d.p de 110V, n\u00e3o funcionando como especificado no problema.<\/b><\/span><\/span><\/p>\n

III. Falsa – A d.d.p. total \u00e9 de 220V. Se colocarmos duas l\u00e2mpadas de 110V em s\u00e9rie, funcionar\u00e3o corretamente.<\/b><\/span><\/span><\/p>\n

IV. Verdadeira – Como foi dito na afirma\u00e7\u00e3o II, cada l\u00e2mpada ser\u00e1 submetida \u00e0 d.d.p. de 110V.<\/b><\/span><\/span><\/p>\n

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

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= R\u00a0 e\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 3R\u00a0 —\u00a0\u00a0 resist\u00eancia equivalente em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0= R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= R + 3 R\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0= 4 R\u00a0 —\u00a0 sendo E a for\u00e7a<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

eletromotriz da bateria ideal, a pot\u00eancia por ela fornecida nessa associa\u00e7\u00e3o \u00e9\u00a0 —\u00a0 Ps=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4R (I)\u00a0 —\u00a0 resist\u00eancia equivalente em paralelo\u00a0 —\u00a0 quando um \u00e9 o triplo do outro, divide o maior por 4\u00a0 —\u00a0 R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=3R\/4\u00a0 —\u00a0 pot\u00eancia fornecida\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(3R\/4)\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=4E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/3R (II)\u00a0 —\u00a0 dividindo membro a membro (I) por (II) \u00a0—\u00a0 Ps\/P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4Rx3R\/4E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Ps\/P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=3\/16<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Os tr\u00eas resistores est\u00e3o em paralelo, portanto e ddp \u00e9 a mesma para o tr\u00eas, ou seja, U = 24 V\u00a0 —\u00a0\u00a0 resistor R<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=24<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/32\u00a0 —\u00a0 R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=18\u03a9\u00a0 —\u00a0 c\u00e1lculo da R<\/b><\/span><\/span>rq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1\/3 + 1\/9 + 1\/18\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2\u03a9\u00a0 —<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2.000\/2\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1.000\u03a9\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=100<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1.000\u00a0 —\u00a0 P=10W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P=iU\u00a0 —\u00a0 i=55\/36 A\u00a0 —\u00a0 quando ligados a um mesmo fus\u00edvel, a corrente dobra\u00a0 —\u00a0 I=2.(55\/36)\u00a0 —\u00a0 I=3,05 A\u00a0 —\u00a0 para suportar essa corrente el\u00e9trica, o menor valor do fus\u00edvel deve ser de 5 A, ou seja o laranja\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A invers\u00e3o de polaridade da pilha apenas inverte o sentido da corrente n\u00e3o alterando as demais caracter\u00edsticas do circuito\u00a0 —\u00a0 observe na express\u00e3o P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R que a pot\u00eancia \u00e9 inversamente proporcional \u00e0 resist\u00eancia da l\u00e2mpada\u00a0 —\u00a0 assim, quanto menor a resist\u00eancia do filamento, maior ser\u00e1 a pot\u00eancia e consequentemente maior ser\u00e1 o brilho\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Os dois resistores est\u00e3o em paralelo\u00a0 —\u00a0 Req=100\/3\u03a9\u00a0 —\u00a0 Req=U\/i=E\/i\u00a0 —\u00a0 100\/3=E\/1\u00a0 —\u00a0 E=100\/3V\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

54-<\/b><\/span><\/span><\/span> R- C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

 <\/p>\n

55)<\/b><\/span><\/span><\/span>– C\u00e1lculo, pelos valores nominais, da corrente que deve circular por cada l\u00e2mpada\u00a0 —\u00a0 P=i.U\u00a0 —\u00a0 6=i.10\u00a0 —\u00a0 i=0,6 A\u00a0 —\u00a0 na associa\u00e7\u00e3o paralelo a corrente total do fus\u00edvel (atualmente disjuntor) de 6 A,<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

se divide, passando 0,6 A por cada l\u00e2mpada\u00a0 —\u00a0\u00a0 n=6\/0,6=10 l\u00e2mpadas\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

56-<\/b><\/span><\/span><\/span> Observe na figura que, como a bateria de 12V est\u00e1 ligada nos pontos A e B, os resistores de 5k\u2126<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

e de 7k\u2126 est\u00e3o em s\u00e9rie sendo substitu\u00eddos por um \u00fanico resistor de 12k\u2126\u00a0 —\u00a0 agora os resistores de 12k\u2126 e de 3k\u2126\u00a0 est\u00e3o em paralelo e submetidos \u00e0 mesma ddp da bateria de U=12V\u00a0\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U\/R=12\/12k\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1mA\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U\/R=12\/3k\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4mA\u00a0 —\u00a0 i=i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1 + 4=5mA\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\u00a057-<\/b><\/span><\/span><\/span> Todas devem estar associadas em paralelo, pois devem estar sob a mesma ddp de U=120V\u00a0 —\u00a0 c\u00e1lculo de cada resist\u00eancia\u00a0 —\u00a0 3 l\u00e2mpadas\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 300=14.400\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=14.400\/300\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=48\u2126 —\u00a0 chapa de sandu\u00edche\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2.000=14.400\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=14.400\/2.000\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=7,2\u2126\u00a0 —\u00a0 forno de micro-ondas\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.500=14.400\/R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=14.400\/1.500\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9,6\u2126\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

c\u00e1lculo da resist\u00eancia equivalente desses tr\u00eas resistores que est\u00e3o associados em paralelo\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

1\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1\/48 + 1\/7,2 + 1\/9,6\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>= (69,12 + 460,8 + 345,6)\/3 317,76\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u22483,8\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3,8=120\/i<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=31,6 A\u00a0 —\u00a0 qualquer valor menor que 31,6 A desliga o disjuntor\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

58-<\/b><\/span><\/span><\/span> Veja na figura como o circuito deve ser montado numa associa\u00e7\u00e3o s\u00e9rie\u00a0 —\u00a0 nessa associa\u00e7\u00e3o, para que as l\u00e2mpadas<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

funcionem de maneira plena e eficiente, elas devem ser ligadas dentro das especifica\u00e7\u00f5es (3V-9W)\u00a0 —\u00a0 assim, cada l\u00e2mpada deve ser percorrida por uma corrente de P=i.U\u00a0 —\u00a0 9=3.i\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>i=3 A<\/b><\/span><\/span>\u00a0 —\u00a0 a tens\u00e3o total U=12V deve ser igual \u00e0 soma das tens\u00f5es de cada l\u00e2mpada, ent\u00e3o voc\u00ea ter\u00e1\u00a0<\/b><\/span><\/span>n=12V\/3V=4 l\u00e2mpadas<\/b><\/span><\/span>\u00a0 —\u00a0 resist\u00eancia de cada l\u00e2mpada\u00a0 —\u00a0 R=U\/i=3\/3=1\u2126\u00a0 —\u00a0 resist\u00eancia equivalente\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=4×1=4\u2126\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 pot\u00eancia total do circuito\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>.i=12.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=36W\u00a0<\/b><\/span><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

59-<\/b><\/span><\/span><\/span> Como esses resistores tem intensidades diferentes e s\u00e3o percorridos pela mesma corrente el\u00e9trica, eles est\u00e3o<\/b><\/span><\/span><\/p>\n

\"Associa\u00e7\u00f5es<\/p>\n

associados em s\u00e9rie\u00a0 —\u00a0 resist\u00eancia equivalente\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=4 + 2=6\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 6 = 12\/i\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Caracter\u00edsticas das associa\u00e7\u00f5es s\u00e9rie e paralelo \u00a001-\u00a0Como eles s\u00e3o id\u00eanticos, a tens\u00e3o nos terminais de cada um deles \u00e9 igual \u00e0 metade da tens\u00e3o total\u00a0 —\u00a0 U1=U2=U\u2019=252\/2\u00a0 —\u00a0U\u2019=126V\u00a0 —\u00a0 R=U\u2019\/i=126\/2\u00a0 —\u00a0\u00a0R=63\u2126 02-\u00a0Req=100 + 100 + 200\u00a0 —\u00a0 Req=400\u2126\u00a0 —\u00a0 Req=U\/I\u00a0 —\u00a0 400=U\/0,5\u00a0 —\u00a0 U=200V\u00a0 —\u00a0 P=i.U=0,5.200\u00a0 —\u00a0 P=100W\u00a0 —\u00a0\u00a0 ou\u00a0 —\u00a0 P=Req.i2\u00a0400.0,25=100W\u00a0 —\u00a0\u00a0\u00a0R-<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1783,"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-1786","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1786","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=1786"}],"version-history":[{"count":5,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1786\/revisions"}],"predecessor-version":[{"id":11950,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1786\/revisions\/11950"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1783"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1786"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}