{"id":1776,"date":"2016-01-11T22:15:50","date_gmt":"2016-01-11T22:15:50","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1776"},"modified":"2024-08-19T16:37:50","modified_gmt":"2024-08-19T16:37:50","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-resistor-equivalente","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/resistor-equivalente\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-resistor-equivalente\/","title":{"rendered":"Resistor equivalente – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Resistor equivalente<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0a) Veja figura abaixo, onde os resistores de 2 \u2126, 4 \u2126 e 4 \u2126 est\u00e3o associados em paralelo\u00a0 —\u00a0 4 com 4=4\/2=2 \u2126\u00a0 —\u00a0 2 com<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

2=2\/2=1 \u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1 \u2126<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/p>\n

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c)<\/b><\/span><\/span><\/p>\n

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\"\"<\/p>\n

 <\/p>\n

d)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

f)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

g)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

h)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

i)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

j)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n

l)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0 —\u00a0 R\/5<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Correta\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0paralelo com R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 — R\u2019=20\/5=4\u2126\u00a0 —\u00a0 R\u2019 s\u00e9rie com R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eq1<\/b><\/span><\/span><\/sub>=20 + 4=24\u2126\u00a0 —\u00a0 errada\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0paralelo com R<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sub>=20\/2=10\u2126 —\u00a0 10\u2126 em s\u00e9rie com R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>eq2<\/b><\/span><\/span><\/sub>=10 + 5=15\u2126\u00a0 —\u00a0 diferen\u00e7a\u00a0 —\u00a0 R=24 \u2013 15=9\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R-B<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Observe as resolu\u00e7\u00f5es nas figuras abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=R\/1 x 4\/3R=4\/3\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

b) Como o n\u00famero de resistores n\u00e3o foi especificado, existem v\u00e1rias associa\u00e7\u00f5es cuja resist\u00eancia equivalente \u00e9 1,5\u2126\u00a0 —\u00a0 uma delas est\u00e1 no esquema abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0A) Sim, \u00e9 \u00f4hmico, pois o gr\u00e1fico Uxi \u00e9 uma reta inclinada a partir da origem, ou seja , U \u00e9\u00a0 proporcional a i de modo que R seja constante\u00a0 —\u00a0 R=U\/i=0,6\/0,2=1.2\/0,4=3,0\/1,0=<\/b><\/span><\/span>3\u2126<\/b><\/span><\/span><\/p>\n

B) R=U\/i\u00a0 —\u00a0 3=9\/i\u00a0 —\u00a0 i=3A\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=i.U=3.9=27W\u00a0 —\u00a0 W=P<\/b><\/span><\/span>o.<\/b><\/span><\/span><\/sub>.\u0394t=27.300\u00a0 —\u00a0<\/b><\/span><\/span>W=8.100J=8,1kJ<\/b><\/span><\/span><\/p>\n

C)<\/b><\/span><\/span>\u00a0\u00c8 a associa\u00e7\u00e3o I onde a resist\u00eancia equivalente \u00e9 menor (R\/2), dissipando assim, maior pot\u00eancia.<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0(5 + R) paralelo com 4\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=produto pela soma\u00a0 —\u00a0 3=(5 + R).4\/5 + R + 4\u00a0 —\u00a0 3=(20 + 4R)\/9 + R\u00a0 —\u00a0 27 + 3R=20 + 4R\u00a0 —\u00a0 R=7\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/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 6=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>1<\/b><\/span><\/span><\/sub>=6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/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 4\/3=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 3R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 4R<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>(II)\u00a0 — (I) em (II)\u00a0 —\u00a0 3.(6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>).R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4(6 \u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) + 4R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 3R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 18R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 24=0\u00a0 —\u00a0 resolvendo essa equa\u00e7\u00e3o, obt\u00e9m-se 2\u2126 e 4\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

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

12-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=30\/3=10 \u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Observe nas figuras abaixo a seq\u00fc\u00eancia de opera\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

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

Os dois que est\u00e3o em paralelo\u00a0 —\u00a0 R\u2019=R\/2\u00a0 —\u00a0 R\u2019 fica em s\u00e9rie com R e R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=R + R\/2 + R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=5R\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0O sistema que aquece mais rapidamente a \u00e1gua \u00e9 aquele que possui\u00a0<\/b><\/span><\/span>menor<\/b><\/span><\/span>\u00a0resist\u00eancia equivalente\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,5\u2126\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0a) chave fechada\u00a0 —\u00a0 todos os resistores s\u00e3o percorridos por corrente e participam do circuito\u00a0 —\u00a0 R paralelo com R \u2013 R\/2\u00a0 —\u00a0 R\/2 em s\u00e9rie com R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3R\/2=1,5R\u00a0 — R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 1,5R=U\/i\u00a0 —\u00a0 i=U\/1,5R\u00a0 —\u00a0 chave aberta\u00a0 —\u00a0 passa corrente apenas pelos dois resistores que est\u00e3o em s\u00e9rie\u00a0 —\u00a0 R\u2019<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2R\u00a0 —\u00a0 R\u2019<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u2019\u00a0 —\u00a0 2R=U\/i\u2019\u00a0 —\u00a0 i\u2019=U\/2R\u00a0 —\u00a0 i\u2019\/i=U\/2R x 1,5R\/U=1,5\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>i\u2019\/i=0,75.<\/b><\/span><\/span><\/p>\n

b) Com a chave fechada onde a resist\u00eancia equivalente \u00e9 menor, dissipando maior pot\u00eancia e aquecendo mais a \u00e1gua.<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0Como o resistor maior \u00e9 o A e ele vai ser somado com o R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0dos outros dois, a alternativa correta \u00e9 a D\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=R\/3<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00e9 a menor resist\u00eancia, tem o maior\u00a0\u00a0 —\u00a0 s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R (I)\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>)=3R\/16\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/R=3R\/16\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/16 (II)\u00a0 —\u00a0 resolvendo o sistema composto por (I) e (II)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R\/4 e R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3R\/4<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0Para que a pot\u00eancia dissipada seja a menor poss\u00edvel, a resist\u00eancia equivalente deve ser a maior poss\u00edvel, o que ocorrer\u00e1 quando eles forem ligados em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1 + 2=3\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3 \u2126\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=4\/3W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Maior corrente implica em maior pot\u00eancia e consequentemente em menor resist\u00eancia\u00a0 —<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span> a) menor pot\u00eancia implica em maior resist\u00eancia equivalente, no caso, em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2 + 6=8\u2126\u00a0 —\u00a0 pot\u00eancia dissipada\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=144\/8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=18W<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Sendo P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00e9 diretamente proporcional a P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 para que a pot\u00eancia fique 4 vezes maior, a tens\u00e3o U deve ser dobrada\u00a0 —\u00a0 U=2.12\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0U=24V<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Refazendo o circuito e chamando de P e Q na figura abaixo os pontos de interse\u00e7\u00e3o de fios:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como R<\/b><\/span><\/span>eqAB<\/b><\/span><\/span><\/sub>=2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 3R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/9\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/8\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3\/8\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0Chamando de r a resist\u00eancia equivalente dos infinitos resistores que est\u00e3o a direita dos pontos P e Q (regi\u00e3o delimitada em<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

vermelho)\u00a0 —\u00a0 lembre-se que,se a seq\u00fc\u00eancia \u00e9 infinita, a resist\u00eancia da seq\u00fc\u00eancia situada \u00e0 direita dos pontos P e Q tamb\u00e9m \u00e9 r, e ent\u00e3o pode-se fazer —\u00a0 r = r paralelo com R (r\u2019) + r\u2019 em s\u00e9rie com 2R\u00a0 —\u00a0 r= r.R\/(r + R) + 2R\u00a0 —\u00a0 r(r + R) = rR + (r + R).2R\u00a0 —\u00a0 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ rR = rR + 2Rr + 2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2Rr \u2013 2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0 \u0394=B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 4.A.C=4R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 4.1.(-2R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u0394=4R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 8R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=12R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u221a\u0394=2\u221a3R\u00a0 —\u00a0 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= (2R\u00a0 + 2\u221a3R)\/2\u00a0 —\u00a0 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R(1 + \u221a3) e r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R(1 – \u221a3)\u00a0 —\u00a0 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0n\u00e3o satisfaz por ser negativa\u00a0 —<\/b><\/span><\/span><\/p>\n

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

25-<\/b><\/span><\/span><\/span>\u00a0lado oeste\u00a0 —\u00a0\u00a0 cada km tem 13\u2126 de resist\u00eancia\u00a0 —\u00a0 13x + R + 13x=60\u00a0 —\u00a0 26x + R=60\u00a0 —\u00a0 R=60 \u2013 26x (I)\u00a0 —\u00a0 lado leste\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

13.(10 \u2013 x) + R + 13.(10 \u2013 x)=216\u00a0 —\u00a0 130 \u2013 13x + R + 130 \u2013 13x=216\u00a0 —\u00a0 -26x + R=-44 (II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0 -26x + (60 \u2013 26x)=-44\u00a0 —\u00a0 -52x=-104\u00a0 —\u00a0 x=2km\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que os dois resistores est\u00e3o entre os pontos P e Q\u00a0 —\u00a0 paralelos\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=20\/3=6,66\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da resist\u00eancia de cada l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 40=(120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=360\u2126\u00a0 —\u00a0 c\u00e1lculo do valor da resist\u00eancia equivalente das 4 l\u00e2mpadas associadas em paralelo \u00a0—\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=360\/4\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=90\u2126\u00a0 —\u00a0 c\u00e1lculo da pot\u00eancia total fornecida por essas 4 l\u00e2mpadas em paralelo (uma chocadeira)\u00a0 —\u00a0 P<\/b><\/span><\/span>ot<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=(120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/90\u00a0 —\u00a0 P<\/b><\/span><\/span>ot<\/b><\/span><\/span><\/sub>=160W\u00a0 —\u00a0 c\u00e1lculo da energia consumida por essa chocadeira em um dia\u00a0 —\u00a0 W=P<\/b><\/span><\/span>ot<\/b><\/span><\/span><\/sub>.\u0394t=160W.1dia\u00a0 —\u00a0 W=160W por dia ou W=0,16kW por dia\u00a0 —\u00a0 hidrel\u00e9trica fornece 24kW por dia\u00a0 —\u00a0 chocadeira consome 0,16kW por dia\u00a0 —\u00a0 n\u00famero de chocadeiras\u00a0 —\u00a0 n=24\/0,16\u00a0 —\u00a0 n=150 chocadeiras por dia\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span> a)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

30-<\/b><\/span><\/span><\/span>\u00a0S\u00f3 s\u00e3o iluminados os trechos percorridos por corrente\u00a0 —\u00a0 os trechos que est\u00e3o em curo circuito foram eliminados nas figuras abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

 <\/p>\n

\"\"<\/p>\n

32<\/b><\/span><\/span><\/span>– <\/b><\/span><\/span><\/span>Observe na figura abaixo que todos os pontos B est\u00e3o em curto circuito, ou seja, s\u00e3o coincidentes e o mesmo ocorre com todos<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

os pontos A\u00a0 —\u00a0 ent\u00e3o os 4 resistores est\u00e3o em paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=R\/4\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R\/4=U\/i\u00a0 —\u00a0 i=4U\/R\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Marcando os n\u00f3s e colocando letras:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

34-<\/b><\/span><\/span><\/span> a)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

b)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

 <\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0Observe que, por simetria, os tr\u00eas pontos da reta vertical tem o mesmo potencial\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe que existe um curto circuito e que a corrente passar\u00e1 somente pelos dois resistores de baixo\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=R + R=2R<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2R<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

42-<\/b><\/span><\/span><\/span>\u00a0Resist\u00eancia equivalente R dos tr\u00eas resistores de resist\u00eancias a, b e c, associados em paralelo\u00a0 —\u00a0 1\/R=1\/a + 1\/b + 1\/c\u00a0 —\u00a0 esses valores est\u00e3o em progress\u00e3o geom\u00e9trica\u00a0 —\u00a0 b\/a=c\/b=1\/2\u00a0 —\u00a0 a=2b\u00a0 —\u00a0 c=b\/2\u00a0 — do enunciado\u00a0 —\u00a0 R=2\u03a9\u00a0 —\u00a0 \u00bd=1\/2b + 1\/b +1\/(b\/2)\u00a0 —\u00a0 1\/2=1\/b(1\/2 + 1 + 2)=7\/2b\u00a0 —\u00a0 b=7\u03a9\u00a0 —\u00a0 a=2b=2.7=14\u03a9\u00a0 —\u00a0 c=b\/2=7\/2=3,5\u03a9\u00a0 —\u00a0 (a + b + c)=(7 + 14 + 3,5)=24,5\u03a9
\n<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja a seq\u00fc\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A resist\u00eancia entre A e B corresponde \u00e0 soma de R com a associa\u00e7\u00e3o em paralelo entre R e R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0(restante do circuito)\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>= R + R,R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\/(R + R<\/b><\/span><\/span>eq)<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0j\u00e1 que o padr\u00e3o do circuito se repete ao infinito e os dois s\u00e3o t\u00e3o grandes que podem ser igualados\u00a0 —\u00a0 R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>= R + R. R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\/(R + R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>AB<\/b><\/span><\/span><\/sub>\u00a0\u2013 R. R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0\u2013 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0 R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=(R + R\u221a5)\/2\u00a0 —\u00a0 R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=(R – R\u221a5)\/2 (exclu\u00eddo, pois R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0deve ser positivo)\u00a0 —\u00a0 R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R( 1 + \u221a5)\/2\u00a0 —<\/b><\/span><\/span><\/p>\n

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

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Refazendo o circuito:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a\u00a0 sequ\u00eancia de figuras abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a seq\u00fc\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- B<\/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 Resistor equivalente \u00a001-\u00a0a) Veja figura abaixo, onde os resistores de 2 \u2126, 4 \u2126 e 4 \u2126 est\u00e3o associados em paralelo\u00a0 —\u00a0 4 com 4=4\/2=2 \u2126\u00a0 —\u00a0 2 com 2=2\/2=1 \u2126\u00a0 —\u00a0 Req=1 \u2126 b) c)   d) e) f) g) h) i) j)     l)   02- 03- R- B 04-\u00a0R-<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1774,"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-1776","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1776","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=1776"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1776\/revisions"}],"predecessor-version":[{"id":10547,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1776\/revisions\/10547"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1774"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1776"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}