{"id":1826,"date":"2016-01-25T00:34:51","date_gmt":"2016-01-25T00:34:51","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1826"},"modified":"2024-12-02T19:23:12","modified_gmt":"2024-12-02T19:23:12","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-geradores-associacao-de-geradores","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/geradores-associacao-de-geradores\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-geradores-associacao-de-geradores\/","title":{"rendered":"Associa\u00e7\u00e3o de geradores – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Geradores \u2013 Associa\u00e7\u00e3o de geradores<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando i=50 A\u00a0 —\u00a0 P=0,6kW=600W\u00a0 —\u00a0 P=E.i —\u00a0 600=E.50\u00a0 —\u00a0 E=12V\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

04<\/b><\/span><\/span>– Veja a figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

05<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Do gr\u00e1fico\u00a0 —\u00a0 E=12V\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 4=12\/r\u00a0 —\u00a0 r=3 \u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>E=12V\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 4=12\/r\u00a0 —\u00a0 r=3\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>E=10V\u00a0 —\u00a0 pelo gr\u00e1fico\u00a0 —\u00a0 i=2A\u00a0 —\u00a0 U=9V\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 9=10 \u2013 r.2\u00a0 —\u00a0 r=0,5\u2126\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,5 + 3,5=4,0\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i=E\/i\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 4=10\/i \u00a0—\u00a0 i=2,5A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>tamanho\u00a0<\/b><\/span><\/span>da pilha ou da bateria est\u00e1\u00a0<\/b><\/span><\/span>diretamente relacionado com a intensidade de corrente el\u00e9trica<\/b><\/span><\/span>\u00a0que ela deve<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecer a um circuito\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>i=Q\/\u0394t\u00a0 —\u00a0 Q=i.\u0394t=30A.1h\u00a0 —\u00a0 Q=30Ah\u00a0 —\u00a0 Q=30A.3.600s\u00a0 —\u00a0 Q=108.000C\u00a0 —\u00a0<\/b><\/span><\/span>Q=108kC<\/b><\/span><\/span>\u00a0 —\u00a0 W=Q.U\u00a0 —\u00a0 W=108.000.12\u00a0 —\u00a0 W=1.296.000J=(W.s)\u00a0 —\u00a0 W=1.296.000\/3.600=360Wh\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=0,36kWh\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/p>\n

11<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>E=1V (circuito em aberto)\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 20A=1\/r\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>r=0,05\u2126<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>a) Do gr\u00e1fico\u00a0 —\u00a0 fem\u00a0 —\u00a0 \u00e9 a tens\u00e3o U quando i=0\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 U=E \u2013 r.0\u00a0 —\u00a0 U= E=15V\u00a0 — corrente de curto circuito\u00a0 —\u00a0 ocorre quando U=0\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 0=E \u2013 r.i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 60=15\/r\u00a0 —\u00a0 r=0,25\u2126<\/b><\/span><\/span><\/p>\n

b) P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=E.i \u2013 r.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 esta express\u00e3o fornece o gr\u00e1fico abaixo, com suas principais caracter\u00edsticas\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=E.i<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>\u00a0\u2013 r.i<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=15.30 \u2013 0,25.(30)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=450 \u2013 225\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=225W<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>a) Quando P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=50A, i<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=5A\u00a0 —\u00a0 P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 50=U<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>.5\u00a0 —\u00a0 U<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=10V\u00a0 —\u00a0 U<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=E \u2013 r.i<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10=E \u2013 r.5\u00a0 —<\/b><\/span><\/span><\/p>\n

r=(E \u2013 10)\/5\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 10=E\/(E \u2013 10)\/5\u00a0 —\u00a0 5E=100\u00a0 —\u00a0 E=20V\u00a0 —\u00a0 verdadeira<\/b><\/span><\/span><\/p>\n

b) Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

c) i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 10=20\/r\u00a0 —\u00a0 r=2\u2126\u00a0 —\u00a0 verdadeira<\/b><\/span><\/span><\/p>\n

d) correta\u00a0 —\u00a0 quando P \u00e9 m\u00e1xima\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=r=2\u2126<\/b><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 a pot\u00eancia \u00e9 m\u00e1xima quando i=5A<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/span>R-= E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=E.i\u00a0 — 300=E.20\u00a0 —<\/b><\/span><\/span>\u00a0 E=15V<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Corrente el\u00e9trica que ela fornece ao circuito externo\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=U.i\u00a0 —\u00a0 1,8=3,6.i\u00a0 —\u00a0 i=0,5A\u00a0 —\u00a0 i=Q\/\u0394t\u00a0 —\u00a0 0,5 A=0,6 A.h\/\u0394t\u00a0 —\u00a0 \u0394t=0,6A.h\/0,5A\u00a0 —\u00a0 \u0394t=1,2h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como no gerador a corrente el\u00e9trica entra pelo p\u00f3lo negativo e sai pelo positivo, sobram as alternativas a), b) e e)\u00a0 —\u00a0\u00a0 a maior transfer\u00eancia ocorre com a alternativa que fornecer maior pot\u00eancia \u00fatil\u00a0 —\u00a0 a) U<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>= E \u2013 r.i=15 \u2013 0.1=15V \u2013 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=U.i=15.1=15W\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

b) U<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=E \u2013 r.i=12 \u2013 0.1=12V\u00a0 — P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=U.i=12.1=12W\u00a0 —\u00a0 e) U<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=E \u2013 r.i=10 \u2013 1.2=8V\u00a0 —\u00a0 P<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=U.i=8.2=16W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I- Correta\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

II- Quando r.i=E\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —U=E \u2013 E\u00a0 —\u00a0 U=0\u00a0 —\u00a0 ele n\u00e3o submete nenhuma d.d.p. ao circuito externo\u00a0 —\u00a0 se U=0\u00a0 —\u00a0 i=0\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

III- Correta\u00a0 —\u00a0 real tem resist\u00eancia externa, ou seja, tem r.i\u00a0 —\u00a0 U=E \u2013 r.i=E \u2013 U\u2019\u00a0 —\u00a0 U \u00e9 menor que E.<\/b><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Situa\u00e7\u00e3o II\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>\u03b5 \u2013\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>gerador<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>\u00a0\u03b5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0situa\u00e7\u00e3o II\u00a0 —\u00a0 i<\/b><\/span><\/span>geradoe<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>l\u00e2mpada<\/b><\/span><\/span><\/sub>=i\u00a0 —\u00a0 na l\u00e2mpada\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4.i\u00a0 —\u00a0 i=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/4=(<\/b><\/span><\/span>\u00a0\u03b5<\/b><\/span><\/span>\/1,2)\/4\u00a0\u00a0 —\u00a0 i=<\/b><\/span><\/span>\u00a0\u03b5<\/b><\/span><\/span>\/4,8\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>\u00a0\u03b5 –\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b5<\/b><\/span><\/span>\/1,2=<\/b><\/span><\/span>\u00a0\u03b5 \u2013\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/sub>(<\/b><\/span><\/span>\u03b5\/<\/b><\/span><\/span>4,8)\u00a0 —\u00a0 4=4,8 \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,8\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Do gr\u00e1fico\u00a0 —\u00a0 E=1,5V\u00a0 —\u00a0 quando i=1A\u00a0 —\u00a0 U=1,2V\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 1,2=1,5 \u2013 r.1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0r=0,3\u2126<\/b><\/span><\/span><\/p>\n

b) R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,3 + 1,7\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=E\/i\u00a0 —\u00a0 2=1,5\/i\u00a0 —\u00a0 i=0,75A<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- i=0,75A<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando R=5,8\u2126\u00a0 —\u00a0 U\u2019=11,6V\u00a0 —\u00a0 R=U\u2019\/i\u2019\u00a0 —\u00a0 5,8=11,6\/i\u2019\u00a0 —\u00a0 i\u2019=11,6\/5,8\u00a0 —\u00a0 i\u2019=2A\u00a0 —\u00a0 quando R=3,8\u2126\u00a0 —\u00a0 U\u2019\u2019=11,4V\u00a0 —\u00a0 R=U\u2019\u2019\/i\u2019\u2019\u00a0 —\u00a0 3,8=11,4\/i\u2019\u2019\u00a0 —\u00a0 i\u2019\u2019=3A\u00a0 —\u00a0 U\u2019=E \u2013 r.i\u2019\u00a0 —\u00a0 11,6=E \u2013 r.2 (I)\u00a0 —\u00a0 U\u2019\u2019=E \u2013 r.i\u2019\u2019\u00a0 —\u00a0 11,4=E \u2013 r.3 (II)\u00a0 —\u00a0 resolvendo I com II\u00a0 —\u00a0 r=0,2\u2126 e E=12V\u00a0 —\u00a0 acrescentando o resistor de 11,8\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i=E\/i\u00a0 —\u00a0 (11,8 + 0,2)=12\/i\u00a0 —\u00a0 i=1 A\u00a0 — pot\u00eancia el\u00e9trica dissipada no resistor de 11,8\u2126\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a011,8.(1)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=11,8W\u00a0 —\u00a0 energia dissipada em 10s\u00a0 —\u00a0 W=P.\u0394t=11,8.10\u00a0 —\u00a0<\/b><\/span><\/span>W=118J<\/b><\/span><\/span>\"\"<\/p>\n

21<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>A transfer\u00eancia de energia ser\u00e1 m\u00e1xima quando r=50\u2126=R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0 —\u00a0 observe que as pilhas est\u00e3o associadas em oposi\u00e7\u00e3o de fase.<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja figuras abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5E\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5r\u00a0 —\u00a0 com R=10\u2126 e com i=5A\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 10 + 5r=5E\/5\u00a0 —\u00a0 5E=25r + 50 ( I )\u00a0 —\u00a0 com R=28\u2126 e com i=2A\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 28 + 5r=5E\/2\u00a0 —\u00a0 10r + 56=5E ( II )\u00a0 —\u00a0 igualando I com II\u00a0 —\u00a0 25r + 50=10r + 56\u00a0 —\u00a0 r=6\/15\u00a0 —\u00a0 r=0,4\u2126\u00a0 —\u00a0 E=12V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>tamanho\u00a0<\/b><\/span><\/span>da pilha ou da bateria est\u00e1\u00a0<\/b><\/span><\/span>diretamente relacionado com a intensidade de corrente el\u00e9trica<\/b><\/span><\/span>\u00a0que ela deve<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecer a um circuito\u00a0 —\u00a0 observe na figura acima que as l\u00e2mpadas possuem a mesma for\u00e7a eletromotriz e, assim, como a maior fornece maior corrente, consequentemente ela deve fornecer maior pot\u00eancia (P=U.i) e possuir menor resist\u00eancia interna\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que as pilhas est\u00e3o em paralelo e a ddp fornecida a R=30\u2126 \u00e9 de E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U=1,5V\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 30=1,5\/i\u00a0 —\u00a0 i=0.05A\u00a0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=6,0V<\/b><\/span><\/span><\/p>\n

29<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da resist\u00eancia da l\u00e2mpada\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 9=12<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=16\u2126\u00a0 —\u00a0 corrente na l\u00e2mpada que est\u00e1 submetida \u00e0 tens\u00e3o fornecida pelas 4 pilhas em s\u00e9rie e que \u00e9 de E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U=1,5.4=6V\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 16=6\/i\u00a0 —\u00a0 i=0,375 A\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=16.(0,375)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=16.0,140625\u00a0 —\u00a0 P=2,25W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

31<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Observe que est\u00e3o em paralelo\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1,2V e r=1,2\/3=0,4\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

32<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Quanto maior a corrente atrav\u00e9s de cada l\u00e2mpada, maior ser\u00e1 seu brilho\u00a0 —\u00a0 resist\u00eancia da l\u00e2mpada R\u00a0 —\u00a0 figura 1\u00a0 — R=E\/i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=E\/R\u00a0 —\u00a0 a) R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3E\/i<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=3E\/2R=1,5E\/R\u00a0 —\u00a0 b) R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3E\/i<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0 — i<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=3E\/3R=E\/R\u00a0 —\u00a0 c) R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2E\/i<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>=2R\/2R=E\/R\u00a0 —\u00a0 d) R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2E\/i<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=2E\/3R\/2=4E\/3R=1,3E\/R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3E\/i<\/b><\/span><\/span>\u2019U<\/b><\/span><\/span><\/sub>—\u00a0 i\u2019<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=3E\/(R\/3)=9E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=i\u2019<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 i\u2019<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=3i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

3i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=9E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=3E\/R\u00a0 —\u00a0 maiores que E\/R s\u00e3o P,T e U\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span>\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

33<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Duas pilhas est\u00e3o em oposi\u00e7\u00e3o de fase e sua for\u00e7a eletromotrizes se anulam\u00a0 —\u00a0 E<\/b><\/span><\/span>equ<\/b><\/span><\/span><\/sub>=1,5V\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3.r=3.2\/3=2\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 2 + 3=1,5\/i\u00a0 —\u00a0 i=1,5\/5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=0,3A<\/b><\/span><\/span><\/p>\n

b) na l\u00e2mpada\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.(0,3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.0,09\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=0,27W<\/b><\/span><\/span><\/p>\n

c) Pilha invertida\u00a0 —\u00a0 P=0,27W \u00a0—\u00a0 pilhas corretas\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u2019\u00a0 —\u00a0 2 + 3=4,5\/i\u2019\u00a0 —\u00a0 i\u2019=0,9 A\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=R.(i\u2019)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.(0,9)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2,43W\u00a0 —\u00a0 F=P\/P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,27\/2,43\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=1\/9<\/b><\/span><\/span><\/p>\n

34<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>Maior corrente i\u00a0 —\u00a0 maior brilho\u00a0 —\u00a0 i<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=3E\/2R=1,5E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=3E\/3R=E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=E\/2R=0,5E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>=2E\/(3R\/2)=4E\/3R=1,3E\/r\u00a0 —\u00a0 i<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>\/2=0,7E\/R\u00a0 —\u00a0 i\u2019<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=3E\/(R\/3)=9E\/R\u00a0 —\u00a0 i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=i\u2019<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 i<\/b><\/span><\/span>U<\/b><\/span><\/span><\/sub>=3E\/R\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

 <\/p>\n

37<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span>\u00a0Como est\u00e3o associadas em paralelo a diferen\u00e7a de potencial U \u00e9 a mesma para cada ramo e para o aparelho e vale U=2V\u00a0 — no aparelho\u00a0 — R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R=2V\/i\u00a0 —\u00a0 i=2V\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo do gerador equivalente\u00a0 —\u00a0 s\u00e9rie\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5.000×0,15\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=750V\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5.000×0,25\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1.250\u2126\u00a0 —\u00a0 t\u00eam-se 140 geradores de E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=750V e r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1.250\u2126 associados em<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0paralelo onde a fem equivalente \u00e9 E\u2019=750V e a resist\u00eancia interna equivalente \u00e9 r\u2019=1250\/140\u00a0 —\u00a0 r\u2019=8,92\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 (8,92 + 800)=750\/i\u00a0 —\u00a0 i=750\/808,92\u00a0 —\u00a0 i=0,927A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) calculando a for\u00e7a eletromotriz e a resist\u00eancia interna do gerador equivalente\u00a0 —\u00a0 as tr\u00eas pilhas em paralelo\u00a0 —<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>\u00a0E\u2019=1,5V e r\u2019=0,3\/3\u00a0 — \u00a0r\u2019=0,1\u2126 \u00a0—\u00a0 essa pilha equivalente est\u00e1 em s\u00e9rie<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

com a quarta pilha\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1,5 + 1,5\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3,0V\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,1 + 0,3\u00a0 — \u00a0r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,4\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 (0,4 + 13,6)=3\/i\u00a0 —\u00a0 i=3\/14\u00a0 —\u00a0 i=0,21A\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=13,6.(0,21)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0 P=0,6W<\/b><\/span><\/span><\/p>\n

b) i<\/b><\/span><\/span><\/span>cc<\/b><\/span><\/span><\/span><\/sub>=E<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/span><\/sub>\/r<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/span><\/sub>=3\/0,4\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 i<\/b><\/span><\/span><\/span>cc<\/b><\/span><\/span><\/span><\/sub>=7,5A<\/b><\/span><\/span><\/span><\/p>\n

40<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span>\u00a0Para que a l\u00e2mpada brilhe normalmente (dentro das especifica\u00e7\u00f5es), a corrente i atrav\u00e9s dela deve ser de\u00a0 —\u00a0 P=i.U\u00a0 —\u00a0 6=18.i\u00a0 —\u00a0 i=1\/3=0,33A\u00a0 — resist\u00eancia da l\u00e2mpada\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 6=(18)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=54\u2126\u00a0 —\u00a0\u00a0 na seq\u00fc\u00eancia abaixo est\u00e1 todo processo do c\u00e1lculo do gerador equivalente\u00a0 —\u00a0 E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=18V\u00a0 —\u00a0 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,55\u2126\u00a0 —\u00a0 c\u00e1lculo da corrente i\u2019 no circuito\u00a0 — R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u2019\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

54 + 0,55=18\/i\u2019\u00a0 —\u00a0 i\u2019=18\/54,55\u00a0 —\u00a0 i\u2019=0,33A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=i\u2019=0,33 A\u00a0 —\u00a0 a l\u00e2mpada brilhar\u00e1 normalmente<\/b><\/span><\/span><\/p>\n

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

42-<\/b><\/span><\/span><\/span> <\/b>R- E<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A bateria \u00e9 um gerador eletroqu\u00edmico. A bobina acoplada ao motor produz corrente alternada,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0que \u00e9, no alternador,transformada em corrente cont\u00ednua, fornecendo energia para recarregar a bateria que, por sua vez, fornece energia para acender os far\u00f3is, alimentar o r\u00e1dio, as l\u00e2mpadas do painel etc.\u00a0<\/b><\/span><\/span><\/p>\n

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

44-<\/b><\/span><\/span><\/span> <\/b>R- D<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

 <\/p>\n

45-<\/b><\/span><\/span><\/span> <\/b>Curva caracter\u00edstica do gerador\u00a0\u2013 \u00e9 representada por um gr\u00e1fico que relaciona a intensidade de corrente el\u00e9trica i no gerador com a diferen\u00e7a de potencial (tens\u00e3o) U em seus terminais.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0Se\u00a0i=0,\u00a0ou seja, se o gerador estiver em\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\u00a0Se\u00a0U=0, ou seja, se o gerador estiver em curto circuito, tem-se<\/b><\/span><\/span><\/p>\n

circuito aberto, tem-se:<\/b><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

\u00a0Os pares de valores U=E e icc=E\/r determinam dois pontos no gr\u00e1fico Uxi, que unidos por um segmento de reta (fun\u00e7\u00e3o do primeiro grau) fornece a curva caracter\u00edstica de um gerador\u00a0 —\u00a0 como as grandezas s\u00e3o proporcionais, completando o gr\u00e1fico\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=24 A\u00a0 —\u00a0 E=12V\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=24 A\u00a0 —\u00a0 E=12V —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 24= 12\/r\u00a0 —\u00a0 r=0,5\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\u00a0a) Per\u00edodo da roda (tempo que ela demora para efetuar uma volta completa)\u00a0 —\u00a0 T<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,5s\u00a0 —\u00a0 velocidade angular da roda de raio R=50cm=0,5m\u00a0 —\u00a0 W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2.\u03c0\/T<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2.3\/0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=12 rad\/s\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 como a roda de raio R=50 cm e o d\u00ednamo de raio r=0,8 cm est\u00e3o acoplados, todos os pontos da superf\u00edcie externa de cada uma tem a mesma velocidade linear (escalar)\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>D\u00ad\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>.R = W<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>.r\u00a0 —\u00a0 12.0,5 =<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>.0,8\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>= 750rad\/s<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/p>\n

b) per\u00edodo T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0do d\u00ednamo\u00a0 —\u00a0 W<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=2.\u03c0\/T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 750=2.3\/T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=8,0.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>s<\/b><\/span><\/span><\/p>\n

c) C\u00e1lculo da for\u00e7a eletromotriz \u03b5 =U\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 24=\u03b5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/6\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b5=12V<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span> <\/b>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 polos da l\u00e2mpada s\u00e3o a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

parte inferior da rosca (ponto I) e a parte lateral da l\u00e2mpada (ponto L)\u00a0 —\u00a0 os dois p\u00f3los da pilha est\u00e3o indicados na figura\u00a0 —\u00a0 para que a l\u00e2mpada acenda o ponto L deve estar ligados a um dos polos da pilha e o ponto I da l\u00e2mpada ao outro p\u00f3lo da pilha, o que ocorre nas situa\u00e7\u00f5es 1, 3 e 7\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Geradores \u2013 Associa\u00e7\u00e3o de geradores   01-\u00a0R- D\u00a0 —\u00a0 veja teoria 02-\u00a0Quando i=50 A\u00a0 —\u00a0 P=0,6kW=600W\u00a0 —\u00a0 P=E.i —\u00a0 600=E.50\u00a0 —\u00a0 E=12V\u00a0\u00a0\u00a0 03– R- C\u00a0 —\u00a0 veja teoria 04– Veja a figura abaixo R-\u00a0 A 05– Do gr\u00e1fico\u00a0 —\u00a0 E=12V\u00a0 —\u00a0 icc=E\/r\u00a0 —\u00a0 4=12\/r\u00a0 —\u00a0 r=3 \u2126\u00a0 —\u00a0\u00a0R- C 06-\u00a0E=12V\u00a0 —\u00a0 icc=E\/r\u00a0 —\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1823,"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-1826","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1826","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=1826"}],"version-history":[{"count":8,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1826\/revisions"}],"predecessor-version":[{"id":11937,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1826\/revisions\/11937"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1823"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}