{"id":1815,"date":"2016-01-25T00:24:04","date_gmt":"2016-01-25T00:24:04","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1815"},"modified":"2024-08-19T15:17:24","modified_gmt":"2024-08-19T15:17:24","slug":"resolucao-comentada-dos-exercicios-de-vestibulares","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/calculo-de-correntes-eletricas-tensoes-e-potencias-eletricas-em-circuitos-mistos\/resolucao-comentada-dos-exercicios-de-vestibulares\/","title":{"rendered":"Resolu\u00e7\u00e3o- exerc\u00edcios de vestibulares"},"content":{"rendered":"

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

C\u00e1lculo de correntes el\u00e9tricas, tens\u00f5es e pot\u00eancias el\u00e9tricas em circuitos mistos<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0a) Observe a figura abaixo\u00a0 —\u00a0 i=2i\u2019\u00a0 —\u00a0 R=U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i\u2019\u00a0 —\u00a0 R=U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>\/i\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i\u2019=U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>\/i\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i\u2019=U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>\/2i\u2019\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>\/U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=2<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) P<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 1,2=R.(2i\u2019)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 i\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=1,2\/4R\u00a0 —\u00a0 i\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0,3\/R\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=R.i\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R.0,3\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=0,3W<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 a pot\u00eancia \u00e9 a mesma nos dois resistores em paralelo e vale\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=1,2W<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0Como as l\u00e2mpadas devem ter brilho normal, devem funcionar dentro das especifica\u00e7\u00f5es\u00a0 —\u00a0 corrente em L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>L1<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>L1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=80\/20\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4A\u00a0 —\u00a0 corrente em L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>L2<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>L2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=36\/12\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3A\u00a0 —\u00a0 observe no esquema\u00a0 que a corrente que deve<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

passar por R deve ser de 1A\u00a0 —\u00a0 como L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e R est\u00e3o em paralelo eles tem a mesma ddp U=12V\u00a0 —\u00a0 R=U\/i=12\/1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R=12\u2126<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da tens\u00e3o U\u2019 no resistor de 12\u2126\u00a0 —\u00a0 R=U\u2019\/I\u00a0 —\u00a0 U\u2019=12.2\u00a0 —\u00a0 U\u2019=24V\u00a0 —\u00a0 como a corrente<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

total \u00e9 4A, pelos resistores em s\u00e9rie (3 + R) passam 2A\u00a0 —\u00a0 (3 + R)=U\u2019\/I\u00a0 —\u00a0 3 + R=24\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R=9\u2126\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 tens\u00e3o U\u2019\u2019 em cada resistor de 6\u2126\u00a0 —\u00a0 R=U\u2019\u2019\/I\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

6=U\u2019\u2019\/4\u00a0 —\u00a0 U\u2019\u2019=24V\u00a0 —\u00a0 U=U\u2019\u2019 + U\u2019 + U\u2019\u2019=24 + 24 + 24\u00a0 —\u00a0 U=72V\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que existe curto circuito\u00a0 —\u00a0\u00a0 marcando os n\u00f3s e colocando letras nos mesmos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

R\/3=V\/i\u00a0 —\u00a0 i=3V\/R\u00a0 —\u00a0mas a corrente pedida \u00e9 no resistor R2 e sua intensidade vale i2 — \u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

No resistor de 9\u2126\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=9.(0,8)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P=5,76W<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0a) Chuveiro\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 4.400=11.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 i=20A<\/b><\/span><\/span><\/span><\/p>\n

b) Pot\u00eancia total\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=((335 + 100 + 55 + 110 + 4400)=5.000W=5kW\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>.\u0394t=5.15\/60\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=1,25kWh<\/b><\/span><\/span><\/span><\/p>\n

c) Observe que, se o fio neutro queimar em A, geladeira, l\u00e2mpada e chuveiro continuar\u00e3o dentro das especifica\u00e7\u00f5es (geladeira 110V, l\u00e2mpada 110V e chuveiro 220V)\u00a0 —\u00a0 apenas ventilador e TV ser\u00e3o afetados e ficar\u00e3o em s\u00e9rie, submetidos a ddp de 220V\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 P<\/b><\/span><\/span><\/span>ventilador<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=220.(0,7)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=107,8W, que \u00e9 maior que 60,5W\u00a0 —\u00a0 ventilador queima \u00a0—\u00a0 P<\/b><\/span><\/span><\/span>TV<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=110.(0,7)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=53,9W, que \u00e9 menor que 121W\u00a0 —\u00a0 a TV n\u00e3o queima\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>quem queima \u00e9 s\u00f3 o ventilador<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b5 =U\u00a0 —\u00a0 K aberta\u00a0 —\u00a0 a corrente i passa apenas pelas 3 l\u00e2mpadas de cima\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=3R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 3R=U\/i\u00a0 —\u00a0 i=U\/3R\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

K fechada\u00a0 —\u00a0 passa corrente por todas as l\u00e2mpadas\u00a0 —\u00a0 3R em paralelo com 2R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=2R.3R\/(2R + 3R)=6R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/5R\u00a0 — \u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=6R\/5\u00a0 —\u00a0 i\u2019=U\/R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/(6R\/5)\u00a0 —\u00a0 i\u2019=5U\/6R\u00a0 —\u00a0 i\/i\u2019=U\/3R x 6R\/5U\u00a0 —\u00a0 i\u2019=15\/6i\u00a0 —\u00a0 i\u2019=2,5i\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0Observe que existe curto circuito\u00a0 —\u00a0 marcando os n\u00f3s, colocando letras nos mesmos e refazendo o circuito\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

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

\"\"<\/p>\n

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

13-<\/b><\/span><\/span><\/span>\u00a0Refazendo o circuito:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

14-<\/b><\/span><\/span><\/span>\u00a0Observe o esquema abaixo\u00a0 —\u00a0 i=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U\u2019\u2019\/3\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=U\u2019\u2019\/6\u00a0 —\u00a0 i=U\u2019\u2019\/3 + U\u2019\u2019\/6\u00a0 —\u00a0 6= U\u2019\u2019\/3 +<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

U\u2019\u2019\/6\u00a0 —\u00a0 6=3U\u2019\u2019\/6\u00a0 —\u00a0U\u2019\u2019=12V\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=U\u2019\u2019\/6=12\/6=2A\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>6\u2126<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=6.(2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>6\u2126<\/b><\/span><\/span><\/sub><\/span>=24W\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Chave aberta\u00a0 —\u00a0 como as l\u00e2mpadas s\u00e3o id\u00eanticas, a tens\u00e3o total de 2V fica distribu\u00edda para cada uma e elas ficam submetidas \u00e0 mesma tens\u00e3o V\u00a0 —\u00a0 chave fechada\u00a0 —\u00a0 a tens\u00e3o em L<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>(V)<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>n\u00e3o se altera e ent\u00e3o a tens\u00e3o em L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(V) tamb\u00e9m permanece a mesma.<\/b><\/span><\/span><\/span><\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0Se V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, a tens\u00e3o nos terminais de R e 90\u2126 \u00e9 a mesma (U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) e a tens\u00e3o nos terminais de 60\u2126 e 120\u2126 tamb\u00e9m \u00e9 a mesma (U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=60i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=120i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 60i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=120i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=90i<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u00a0U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=Ri<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 90i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=Ri<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 90i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R.2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R=45\u2126<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\u00a0R=10\u2126 e U<\/b><\/span><\/span><\/span>ab<\/b><\/span><\/span><\/sub><\/span>=15V\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

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

U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u03b5\/3\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2\u03b5\/R\u00a0 —U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) \u2013 (V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)=\u03b5\/3 – 2\u03b5\/3\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= – \u03b5\/3\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= -\u03b5\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u03b5\/3<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0Observe a figura a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A pot\u00eancia total \u00e9 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=40 + 40 +40=120\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=120W.<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0Se cada conjunto tem 5.000 c\u00e9lulas de 150mV=0,15V cada, a tens\u00e3o desse conjunto vale\u00a0 —\u00a0 U=5.000×0,15\u00a0 —\u00a0 U=750V\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Como a corrente total \u00e9 de 0,5A e existem 5 desses conjuntos em paralelo, essa corrente em cada<\/b><\/span><\/span><\/span><\/p>\n

conjunto \u00e9 de i=5\/5=1A\u00a0 —\u00a0 pot\u00eancia de um conjunto\u00a0 —\u00a0 P=i.U=0,1.750\u00a0 —\u00a0 P=75W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

A corrente \u00e9 a mesma em R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e R<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=3,0mA<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Em R<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 U\u2019=9,6V\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u2126\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=9,6\/4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=2,4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>A ou\u00a0 i<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=2,4mA<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

A tens\u00e3o nos terminais das 5 l\u00e2mpadas id\u00eanticas de resist\u00eancia R \u00e9 de U\u2019=107,8V e a corrente total \u00e9 i=2,2A\u00a0 —\u00a0 cada l\u00e2mpada ser\u00e1 percorrida por uma corrente i\u2019=2,2\/5=0,44A\u00a0 —\u00a0 R=U\u2019\/i\u2019=107,8\/0,44\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R=245\u2126<\/b><\/span><\/span><\/span><\/p>\n

b) fios de liga\u00e7\u00e3o\u00a0 —\u00a0 P=i.U=2,2.2,2=4,84W\u00a0 —\u00a0 W=P.\u0394t=4,84.30.5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0W=726Wh<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0a) Observe no esquema abaixo que todas as l\u00e2mpadas est\u00e3o sob suas especifica\u00e7\u00f5es normais<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=12 \u2126 paralelo com 12 \u2126 (6 \u2126) + 6 \u2126 em s\u00e9rie com 6 \u2126 e em s\u00e9rie com 6 \u2126 ( 18 \u2126)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=18 \u2126\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 18=12\/i\u00a0 —\u00a0 i=0,7A<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) resist\u00eancia do aquecedor P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 1.100=(110)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 R=11\u2126\u00a0 —\u00a0 observe que existe curto circuito\u00a0 —\u00a0 marcando os n\u00f3s,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

colocando letras nos mesmos e refazendo o circuito\u00a0 —\u00a0 observe a seq\u00fc\u00eancia\u00a0 —\u00a0 a corrente no aquecedor \u00e9<\/b><\/span><\/span><\/span>\u00a0i\u2019=2,5A<\/b><\/span><\/span><\/span><\/p>\n

b) P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>=(44\/3).(7,5)<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span>=825W<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 5=1,5\/i\u00a0 —\u00a0 i=0,3 A\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.(0,3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P=0,18W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

29-<\/b><\/span><\/span><\/span>\u00a0a) Estimando a intensidade da for\u00e7a que o dedo exerce sobre a tela como sendo F=0,8N\u00a0 —\u00a0 A=0,25cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0,24.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P=F\/A\u00a0 —\u00a0 P=0,8\/0,25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span>3,2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(Pa)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a fonte a mesma a ddp U aplicada aos arranjos \u00e9 a mesma\u00a0 — arranjo I\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=R\/3\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(R\/3)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=3U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 arranjo II\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=3R\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/3R\u00a0 —\u00a0 arranjo III\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=R + R\/2=3R\/2\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(3R\/2)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>=2U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/3R\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>R1<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 100=R.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(I)\u00a0 —\u00a0 em R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(II) —\u00a0 em R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R=(U\/2)\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R=U\/2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(III)\u00a0 — \u00a0igualando II com III\u00a0 —\u00a0 U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U\/2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0\u00a0 —\u00a0 pot\u00eancia em R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 P\u2019=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R.(i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P\u2019=R.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4 (VI) —\u00a0 substituindo I em IV\u00a0 —\u00a0\u00a0 P\u2019=100\/4\u00a0 —\u00a0 P\u2019=25W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B\u00a0<\/b><\/span><\/span><\/span>.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na figura abaixo que todos os resistores est\u00e3o em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=10k + 20k + 30k + 40k\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=100.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u03a9\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/—\u00a0 i=9\/100.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 i=9.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0A\u00a0 —\u00a0 P=i.U=9.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>.9\u00a0 —\u00a0 P=81.10<\/b><\/span><\/span><\/span>-5\u00a0<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P=8,1.10<\/b><\/span><\/span><\/span>-4\u00a0<\/b><\/span><\/span><\/sup><\/span>W<\/b><\/span><\/span><\/span><\/p>\n

b) Observe na figura abaixo que o circuito est\u00e1 em aberto, n\u00e3o est\u00e1 fechado e n\u00e3o passa corrente<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

pelo mesmo\u00a0 —\u00a0 a pot\u00eancia el\u00e9trica dissipada \u00e9 nula.<\/b><\/span><\/span><\/span><\/p>\n

c) Observe na figura abaixo que os tr\u00eas resistores est\u00e3o em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=20k + 40k + 30k\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=90.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u03a9\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a090.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=9\/i\u00a0 —\u00a0 i=1,0.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0A\u00a0 —\u00a0 P=iU=10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.9\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P=9,0.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>W<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Corrente el\u00e9trica em cada l\u00e2mpada\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 0,5=i.6\u00a0 —\u00a0 i=1\/12 A\u00a0 —\u00a0 corrente total no circuito\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=3.(1\/12)\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=0,25A\u00a0 —\u00a0 a tens\u00e3o no resistor somada \u00e0 tens\u00e3o nas l\u00e2mpadas deve ser igual a tens\u00e3o da fonte\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0+ U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 6 + U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= 12\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= 6 V\u00a0 —\u00a0\u00a0 1\u00aa lei de Ohm no resistor\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= R.i<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0 6 = R (0,25)\u00a0 —\u00a0 R = 24 \u03a9\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

34-\u00a0<\/b><\/span><\/span><\/span>Observe a resolu\u00e7\u00e3o na seq\u00fc\u00eancia abaixo, na ida e na volta:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 trata-se de uma associa\u00e7\u00e3o s\u00e9rie e as l\u00e2mpadas apresentam o mesmo brilho<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 est\u00e3o em s\u00e9rie<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 a corrente em vada l\u00e2mpada \u00e9 a mesma\u00a0 —\u00a0 brilham por igual.<\/b><\/span><\/span><\/span><\/p>\n

08- Correta\u00a0 —\u00a0 a l\u00e2mpada A tem um resistor associado em s\u00e9rie e \u00e9 percorrida por menor corrente e brilha menos que a B.<\/b><\/span><\/span><\/span><\/p>\n

16- Falsa\u00a0 —\u00a0 a l\u00e2mpada B fica em curto circuito e apaga.<\/b><\/span><\/span><\/span><\/p>\n

R- (02 + 08) = 10<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O disjuntor D, que limita a corrente que entra no circuito at\u00e9 certo limite evitando curto circuito deve ser ligado no fio fase, que \u00e9 por onde entra a corrente el\u00e9trica (op\u00e7\u00f5es c e d invalidadas)\u00a0 —\u00a0 na op\u00e7\u00e3o a, com o interruptor I ligado ou desligado, sempre passar\u00e1 corrente pela l\u00e2mpada, permanecendo sempre acesa\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 Observa\u00e7\u00f5es\u00a0 —\u00a0 01. Note na figura b que o fio Fase (por onde entra a corrente) \u00e9 quem deve ser interrompido pela chave de luz e n\u00e3o o Neutro \u00a0pois \u00a0assim \u00a0quando \u00a0voc\u00ea ou algu\u00e9m for trocar a l\u00e2mpada ou simplesmente limp\u00e1-la, basta ter certeza de que desligou a chave de luz para que n\u00e3o haja tens\u00e3o no recept\u00e1culo e conseq\u00fcentemente n\u00e3o ocorram choques\u00a0 —\u00a0 02. Para voc\u00ea saber qual fio \u00e9 o fase e qual \u00e9<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

o neutro, use uma l\u00e2mpada de teste, com fios isolados, ligue uma das pontas no fio a testar; encoste a outra ponta na parede ou em um cano de ferro. Se for o fio neutro, a l\u00e2mpada n\u00e3o acender\u00e1. Se for o fio fase (positivo), a l\u00e2mpada acender\u00e1. Muito cuidado para somente segurar a l\u00e2mpada teste nas partes isoladas.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Refazendo o circuito ele ter\u00e1 o formato da figura abaixo\u00a0 —\u00a0 corrente total no circuito\u00a0 —\u00a0 P=i.U\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 100=12.i\u00a0 —\u00a0 i=100\/12 A\u00a0 —A corrente em cada l\u00e2mpada vale\u00a0 —\u00a0 i\u2019=(100\/12)\/5=100\/60\u00a0 —\u00a0 i\u2019=1,666 A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre C\u00e1lculo de correntes el\u00e9tricas, tens\u00f5es e pot\u00eancias el\u00e9tricas em circuitos mistos     01-\u00a0a) Observe a figura abaixo\u00a0 —\u00a0 i=2i\u2019\u00a0 —\u00a0 R=UAB\/i\u2019\u00a0 —\u00a0 R=UBC\/i\u00a0 —\u00a0 UAB\/i\u2019=UBC\/i\u00a0 —\u00a0 UAB\/i\u2019=UBC\/2i\u2019\u00a0 —\u00a0\u00a0UBC\/UAB=2 b) PBC=R.i2\u00a0 —\u00a0 1,2=R.(2i\u2019)2\u00a0 —\u00a0 i\u20192=1,2\/4R\u00a0 —\u00a0 i\u20192=0,3\/R\u00a0 —\u00a0 PAB=R.i\u20192=R.0,3\/R\u00a0 —\u00a0\u00a0PAB=0,3W\u00a0 —\u00a0 a pot\u00eancia \u00e9 a mesma nos dois resistores em paralelo e vale\u00a0PBC=1,2W<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1812,"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-1815","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1815","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=1815"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1815\/revisions"}],"predecessor-version":[{"id":10531,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1815\/revisions\/10531"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1812"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}