{"id":4850,"date":"2020-01-09T14:11:44","date_gmt":"2020-01-09T14:11:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4850"},"modified":"2024-09-04T14:01:21","modified_gmt":"2024-09-04T14:01:21","slug":"resolucao-comentada-das-questoes-de-fisica-vestibulares-recentes-eletrodinamica-2018-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletrodinamica-2018-2019\/resolucao-comentada-das-questoes-de-fisica-vestibulares-recentes-eletrodinamica-2018-2019\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica \u2013 vestibulares recentes \u2013 Eletrodin\u00e2mica \u2013 2018\/2019"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica \u2013 vestibulares recentes \u2013 Eletrodin\u00e2mica \u2013 2018\/2019<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span><\/p>\n

<\/p>\n

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

<\/p>\n

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

C\u00e1lculo da pot\u00eancia da bateria\u00a0<\/b><\/span><\/span><\/span>\u00a0Po\u00a0= i.U = 2.5\u00a0\u00a0<\/b><\/span><\/span><\/span>Po\u00a0= 10 W.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>energia W utilizada no processo do carregamento foi de 90% de 105\u00a0j = 0,9.105\u00a0\u00a0W = 9.104\u00a0J.<\/b><\/span><\/span><\/span><\/p>\n

Po\u00a0= W\/t\u00a0\u00a010 = 9.104\/t\u00a0\u00a0t = 9.104\u00a0\/10\u00a0\u00a0<\/b><\/span><\/span><\/span>t\u00a0<\/b><\/span><\/span><\/span>= 9.103\u00a0= 9000s\/3,6\u00a0<\/b><\/span><\/span><\/span>= 2,5 h.<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/span><\/span><\/span><\/p>\n

Toda bateria possui um polo positivo e um polo negativo, caracter\u00edsticas semelhantes\u00a0<\/b><\/span><\/span><\/span>\u00e0s\u00a0<\/b><\/span><\/span><\/span>desse peixe, onde o polo positivo seria a cabe\u00e7a e o negativo a cauda.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Esses dois polos originam uma diferen\u00e7a de potencial\u00a0<\/b><\/span><\/span><\/span>(tens\u00e3o ou voltagem) que\u00a0<\/b><\/span><\/span><\/span>originam<\/b><\/span><\/span><\/span>\u00a0a movimenta\u00e7\u00e3o de\u00a0<\/b><\/span><\/span><\/span>corrente el\u00e9trica,\u00a0<\/b><\/span><\/span><\/span>que no caso do peixe, quando tocado, pode\u00a0<\/b><\/span><\/span><\/span>liberar uma forte descarga el\u00e9trica, imobilizando e at\u00e9 matando suas presas ou agressores.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

06-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se\u00a0<\/b><\/span><\/span><\/span>voc\u00ea dobrar o valor da resist\u00eancia a pot\u00eancia ficar\u00e1 2 vezes menor\u00a0<\/b><\/span><\/span><\/span>e se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>triplicar a ddp a pot\u00eancia ficar\u00e1 32\u00a0= 9 vezes maior\u00a0\u00a0P\u2019 = 9P\/2.<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/span><\/span><\/span><\/p>\n

S\u00e3o dados: P\u00a0<\/b><\/span><\/span><\/span>= 5,5.103\u00a0W;\u00a0<\/b><\/span><\/span><\/span>U\u00a0<\/b><\/span><\/span><\/span>= 220 V e\u00a0<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>\u00a0= ?<\/b><\/span><\/span><\/span><\/p>\n

P = i.U\u00a0<\/b><\/span><\/span><\/span>\u00a05,5.103\u00a0= i.220\u00a0\u00a0i = 5500\/220\u00a0\u00a0<\/b><\/span><\/span><\/span>i = 25 A.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Veja na\u00a0<\/b><\/span><\/span><\/span>tabela que a energia total consumida em 1 m\u00eas (30 dias) foi de Wt = 375 kWh e\u00a0<\/b><\/span><\/span><\/span>como a\u00a0<\/b><\/span><\/span><\/span>chaleira consome 2% dessa energia em 30 dias ela consumir\u00e1 W1<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0,02\u00d7375\u00a0<\/b><\/span><\/span><\/span>W1\u00a0= 7,5 kWh (em 30 dias).<\/b><\/span><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span><\/span>1 dia ela consumir\u00e1 W2\u00a0=<\/b><\/span><\/span><\/span>\u00a07,5\/30\u00a0<\/b><\/span><\/span><\/span>W2\u00a0= 0,25 kWh (energia gasta pela chaleira em 1 dia).<\/b><\/span><\/span><\/span><\/p>\n

Mas o\u00a0<\/b><\/span><\/span><\/span>tempo de funcionamento dela por dia \u00e9 t\u00a0<\/b><\/span><\/span><\/span>= 15 min = 15\/60 = 1\/4\u00a0<\/b><\/span><\/span><\/span>= 0,25 h.<\/b><\/span><\/span><\/span><\/p>\n

Po\u00a0= W2\/t\u00a0<\/b><\/span><\/span><\/span>= 0,25\/0,25\u00a0<\/b><\/span><\/span><\/span>Po\u00a0=<\/b><\/span><\/span><\/span>\u00a01 kW\u00a0<\/b><\/span><\/span><\/span>= 1000 W<\/b><\/span><\/span><\/span><\/p>\n

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

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

Chuveiro\u00a0\u00a0t\u00a0<\/b><\/span><\/span><\/span>= 30 diasx20\/60h\u00a0<\/b><\/span><\/span><\/span>= 10 h\u00a0\u00a0P<\/b><\/span><\/span><\/span>\u00a0= 7000 W\u00a0<\/b><\/span><\/span><\/span>= 7 kW\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energia\u00a0\u00a0W = P.t\u00a0<\/b><\/span><\/span><\/span>= 7.10\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W = 70 kWh\u00a0\u00a0pre\u00e7o\u00a0<\/b><\/span><\/span><\/span>\u00a0C$ = 70\u00d70,5\u00a0\u00a0<\/b><\/span><\/span><\/span>C$ = 35,00<\/b><\/span><\/span><\/span><\/p>\n

L\u00e2mpada\u00a0\u00a0t =<\/b><\/span><\/span><\/span>\u00a024\u00d730\u00a0<\/b><\/span><\/span><\/span>= 720 h\u00a0\u00a0P =<\/b><\/span><\/span><\/span>\u00a0100 W\u00a0<\/b><\/span><\/span><\/span>= 0,1 kW<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0energia\u00a0\u00a0W = P.t\u00a0<\/b><\/span><\/span><\/span>= 0,1.720\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W = 72 kWh\u00a0\u00a0pre\u00e7o\u00a0<\/b><\/span><\/span><\/span>\u00a0C$ = 72\u00d70,5\u00a0\u00a0<\/b><\/span><\/span><\/span>C$ = 36,00<\/b><\/span><\/span><\/span><\/p>\n

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

10-<\/span><\/span><\/span><\/p>\n

Veja o resumo te\u00f3rico abaixo:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

R \u2013 C<\/b><\/span><\/span><\/span><\/p>\n

11-<\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Nas condi\u00e7\u00f5es da figura A, a ddp U do gerador deve ser a mesma que a do resistor<\/b><\/span><\/span><\/span>\u00a0e na\u00a0<\/b><\/span><\/span><\/span>figura B<\/b><\/span><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span><\/span>o ponto onde a reta e a curva se interceptam ou seja, quando U = 10 V, i = 0,14 A.<\/b><\/span><\/span><\/span><\/p>\n

P = i.U\u00a0<\/b><\/span><\/span><\/span>= 0,14.10\u00a0<\/b><\/span><\/span><\/span>P = 1,4 W,<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

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

14-<\/span><\/span><\/span><\/p>\n

F\u00f3rmulas \u2013 Lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

R =\u00a0<\/b><\/span><\/span><\/span>22 \u03a9\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>U =<\/b><\/span><\/span><\/span>\u00a0220 V\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R =\u00a0<\/b><\/span><\/span><\/span>\u00a022 =\u00a0\u00a0i =\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>i = 10 A<\/b><\/span><\/span><\/span><\/p>\n

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

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

F\u00f3rmulas \u2013 Lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

S\u00e3o dados: U = 20 V, P = 300 W e t = 1 min = 60 s.<\/b><\/span><\/span><\/span><\/p>\n

P = i.U\u00a0<\/b><\/span><\/span><\/span>\u00a0300 = i.20\u00a0<\/b><\/span><\/span><\/span>\u00a0i\u00a0<\/b><\/span><\/span><\/span>= 300\/20\u00a0<\/b><\/span><\/span><\/span>= 15 A.<\/b><\/span><\/span><\/span><\/p>\n

Energia = pot\u00eanciaxtempo\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>W\u00a0<\/b><\/span><\/span><\/span>= 300\u00d760\u00a0<\/b><\/span><\/span><\/span>= 18000 J<\/b><\/span><\/span><\/span><\/p>\n

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

16-<\/span><\/span><\/span><\/p>\n

<\/p>\n

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

L\u00e2mpada fluorescente<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do\u00a0<\/b><\/span><\/span><\/span>custo total de uma l\u00e2mpada fluorescente\u00a0<\/b><\/span><\/span><\/span>de pot\u00eancia\u00a0<\/b><\/span><\/span><\/span>Pf\u00a0= 28 W = 28.10-3\u00a0kW\u00a0<\/b><\/span><\/span><\/span>durante sua\u00a0<\/b><\/span><\/span><\/span>vida \u00fatil fornecida (tf\u00a0= 10000 h = 104\u00a0h).<\/b><\/span><\/span><\/span><\/p>\n

Energia el\u00e9trica (Wf) consumida\u00a0<\/b><\/span><\/span><\/span>por essa l\u00e2mpada\u00a0<\/b><\/span><\/span><\/span>durante sua vida \u00fatil\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wf\u00a0= Pf.tf<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 28.10-3\u00a0kW. 104h\u00a0\u00a0<\/b><\/span><\/span><\/span>Wf\u00a0= 280 kWh.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Custo total de cada l\u00e2mpada fluorescente\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C$ = 70,00 + pre\u00e7o de uma l\u00e2mpada\u00a0<\/b><\/span><\/span><\/span>= 70,00 + 20,00\u00a0\u00a0<\/b><\/span><\/span><\/span>C$ = 90,00.<\/b><\/span><\/span><\/span><\/p>\n

L\u00e2mpada incandescente<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do\u00a0<\/b><\/span><\/span><\/span>custo total de uma l\u00e2mpada incandescente\u00a0<\/b><\/span><\/span><\/span>de pot\u00eancia\u00a0<\/b><\/span><\/span><\/span>Pf\u00a0= 100 W = 10-1\u00a0kW\u00a0<\/b><\/span><\/span><\/span>durante a mesma vida \u00fatil que a da fluorescente<\/b><\/span><\/span><\/span>\u00a0(tf\u00a0= 10 000 h = 104\u00a0h).<\/b><\/span><\/span><\/span><\/p>\n

Energia el\u00e9trica (Wf) consumida\u00a0<\/b><\/span><\/span><\/span>por essa l\u00e2mpada\u00a0<\/b><\/span><\/span><\/span>durante sua vida \u00fatil\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wf\u00a0= Pf.tf<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 10-1\u00a0kW. 104h\u00a0\u00a0<\/b><\/span><\/span><\/span>Wf\u00a0= 103\u00a0kWh = 1 000 kWh.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe que voc\u00ea\u00a0<\/b><\/span><\/span><\/span>necessita de 10 l\u00e2mpadas incandescentes\u00a0<\/b><\/span><\/span><\/span>(vida \u00fatil 10 000 h) para\u00a0<\/b><\/span><\/span><\/span>operar na mesma vida \u00fatil (mesmo per\u00edodo de tempo)<\/b><\/span><\/span><\/span>\u00a0que a\u00a0<\/b><\/span><\/span><\/span>das fluorescentes\u00a0<\/b><\/span><\/span><\/span>(vida \u00fatil 1 000 h).<\/b><\/span><\/span><\/span><\/p>\n

Assim, o\u00a0<\/b><\/span><\/span><\/span>custo total de\u00a0cada incandescente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>dever\u00e1 ser\u00a0<\/b><\/span><\/span><\/span>C$ = custo de uma incandescente + 10xpre\u00e7o de uma l\u00e2mpada incandescente = 250,00 + 10\u00d74,00\u00a0\u00a0C$ = 290,00\u00a0<\/b><\/span><\/span><\/span>(custo total de cada incandescente para a mesma vida \u00fatil de cada fluorescente)<\/b><\/span><\/span><\/span><\/p>\n

Assim, o\u00a0<\/b><\/span><\/span><\/span>valor poupado\u00a0<\/b><\/span><\/span><\/span>ser\u00e1\u00a0<\/b><\/span><\/span><\/span>C$ =\u00a0<\/b><\/span><\/span><\/span>290,00 \u2013 90,00\u00a0<\/b><\/span><\/span><\/span>= 200,00.<\/b><\/span><\/span><\/span><\/p>\n

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

18-<\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Antes da troca\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energia consumida\u00a0<\/b><\/span><\/span><\/span>pelas\u00a0<\/b><\/span><\/span><\/span>duas l\u00e2mpadas fluorescentes\u00a0<\/b><\/span><\/span><\/span>de\u00a0<\/b><\/span><\/span><\/span>20 W cada\u00a0<\/b><\/span><\/span><\/span>acesas\u00a0<\/b><\/span><\/span><\/span>durante \u2206t = 4 h\u00a0<\/b><\/span><\/span><\/span>por noite\u00a0\u00a0<\/b><\/span><\/span><\/span>W1\u00a0= 2.P. \u2206t\u00a0<\/b><\/span><\/span><\/span>= 2.20.4\u00a0<\/b><\/span><\/span><\/span>= 160 Wh\u00a0<\/b><\/span><\/span><\/span>+ dispositivos de\u00a0<\/b><\/span><\/span><\/span>200 W ligados por 4 h = 800 Wh.<\/b><\/span><\/span><\/span><\/p>\n

Wantes\u00a0= 160 + 800\u00a0\u00a0<\/b><\/span><\/span><\/span>Wantes\u00a0= 960 Wh.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Depois da troca\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energia consumida\u00a0<\/b><\/span><\/span><\/span>pela\u00a0<\/b><\/span><\/span><\/span>duas l\u00e2mpadas de LED\u00a0<\/b><\/span><\/span><\/span>de\u00a0<\/b><\/span><\/span><\/span>9 W cada\u00a0<\/b><\/span><\/span><\/span>acesas<\/b><\/span><\/span><\/span>\u00a0durante \u2206t = 6 h\u00a0<\/b><\/span><\/span><\/span>por noite\u00a0\u00a0<\/b><\/span><\/span><\/span>W2\u00a0= 2.P. \u2206t\u00a0<\/b><\/span><\/span><\/span>= 2.9.6\u00a0<\/b><\/span><\/span><\/span>= 108 Wh\u00a0<\/b><\/span><\/span><\/span>+ dispositivos de\u00a0<\/b><\/span><\/span><\/span>200 W ligados por 5 h = 1000 Wh.<\/b><\/span><\/span><\/span><\/p>\n

Wdepois\u00a0= 108 + 1000\u00a0\u00a0<\/b><\/span><\/span><\/span>Wdepois\u00a0= 1108 Wh.<\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a\u00a0\u00a0W\u2019 = Wdepois\u00a0\u2013 Wantes<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 1108 \u2013 960\u00a0\u00a0<\/b><\/span><\/span><\/span>W\u2019 = 148 Wh.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Como \u00e9 pedida a\u00a0<\/b><\/span><\/span><\/span>maior pot\u00eancia el\u00e9trica\u00a0<\/b><\/span><\/span><\/span>voc\u00ea deve selecionar a\u00a0<\/b><\/span><\/span><\/span>maior corrente el\u00e9trica fornecida pelo gr\u00e1fico, que \u00e9 de im\u00e1x\u00a0= 60 A<\/b><\/span><\/span><\/span>\u00a0e, se a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o \u00e9 constante U = 110 V<\/b><\/span><\/span><\/span>, basta voc\u00ea utilizar a\u00a0<\/b><\/span><\/span><\/span>express\u00e3o Pm\u00e1x\u00a0= im\u00e1x.U\u00a0<\/b><\/span><\/span><\/span>= 60.110\u00a0\u00a0<\/b><\/span><\/span><\/span>Pm\u00e1x\u00a0= 6600 W = 6,6 kW.<\/b><\/span><\/span><\/span><\/p>\n

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

20-<\/span><\/span><\/span><\/p>\n

Para esse exerc\u00edcio nem \u00e9 necess\u00e1rio utilizar conhecimentos de f\u00edsica, visto que\u00a0<\/b><\/span><\/span><\/span>a pot\u00eancia n\u00e3o muda, o importante mesmo \u00e9 o tempo utilizado<\/b><\/span><\/span><\/span>. Em\u00a0<\/b><\/span><\/span><\/span>6 dias<\/b><\/span><\/span><\/span>\u00a0por semana\u00a0<\/b><\/span><\/span><\/span>por 30 min (meia hora)<\/b><\/span><\/span><\/span>\u00a0n\u00f3s temos:<\/b><\/span><\/span><\/span><\/p>\n

6.0,5 horas = 3 horas<\/b><\/span><\/span><\/span><\/p>\n

No outro caso\u00a0<\/b><\/span><\/span><\/span>se utiliza o ferro 2 horas por semana<\/b><\/span><\/span><\/span>. Ent\u00e3o se economiza 3 \u2013 2 =\u00a0<\/b><\/span><\/span><\/span>1 hora por semana<\/b><\/span><\/span><\/span>. Se fizermos\u00a0<\/b><\/span><\/span><\/span>a divis\u00e3o 1\/3 = 0,33 = 33% temos o quanto economizamos por semana<\/b><\/span><\/span><\/span>\u00a0em rela\u00e7\u00e3o ao valor inicial. Alternativa B<\/b><\/span><\/span><\/span><\/p>\n

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

Energia el\u00e9trica W<\/b><\/span><\/span><\/span>\u00a0consumida por\u00a0<\/b><\/span><\/span><\/span>cada aparelho, por dia:<\/b><\/span><\/span><\/span><\/p>\n

Geladeira\u00a0\u00a0Wg\u00a0= Pg.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wg\u00a0= 0,25.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 6,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Computador\u00a0\u00a0Wc\u00a0= Pc.\u00a0\u00a0Wc\u00a0= 0,50.\u00a0=\u00a0<\/b><\/span><\/span><\/span>1,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Televis\u00e3o\u00a0\u00a0Wt\u00a0= Pt.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wt\u00a0= 0,10.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0,5 Wh .<\/b><\/span><\/span><\/span><\/p>\n

Ar condicionado\u00a0\u00a0Wa\u00a0= Pa.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wg\u00a0= 3,00.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 15 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Chuveiro\u00a0\u00a0Wc\u00a0= Pc.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wc\u00a0= 4,00.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 2,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Forno\u00a0\u00a0Wf\u00a0= Pf.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Wf\u00a0= 1,50.1\/3<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0,5 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Energia total consumida por dia\u00a0\u00a0Wt\u00a0= 25 kWh<\/b><\/span><\/span><\/span><\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>cada painel\u00a0<\/b><\/span><\/span><\/span>de 1,0 m2\u00a0de \u00e1rea\u00a0<\/b><\/span><\/span><\/span>gera, em m\u00e9dia, 0,5 kWh por dia,\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>n\u00famero m\u00ednimo\u00a0<\/b><\/span><\/span><\/span>de pain\u00e9is ser\u00e1 25\/0,5\u00a0<\/b><\/span><\/span><\/span>= 50 pain\u00e9is.<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 C<\/b><\/span><\/span><\/span><\/p>\n

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

Segunda lei de Ohm (resistividade)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

01. Correto<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 a express\u00e3o\u00a0<\/b><\/span><\/span><\/span>R =\u00a0l\/A indica que a resist\u00eancia el\u00e9trica R \u00e9 diretamente proporcional ao comprimento l do fio.<\/b><\/span><\/span><\/span><\/p>\n

02. Correto \u2013<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>express\u00e3o R =\u00a0l\/A indica que a resist\u00eancia el\u00e9trica R \u00e9 inversamente proporcional a\u00a0\u00e1rea A da se\u00e7\u00e3o transversal\u00a0<\/b><\/span><\/span><\/span>de um resistor.\u00a0Assim,\u00a0<\/b><\/span><\/span><\/span>quanto maior for a \u00e1rea A da se\u00e7\u00e3o transversal de um resistor, menor ser\u00e1 sua resist\u00eancia el\u00e9trica.<\/b><\/span><\/span><\/span><\/p>\n

04. Correto \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>se\u00a0<\/b><\/span><\/span><\/span>associados em s\u00e9rie voc\u00ea ter\u00e1 apenas um resistor de resist\u00eancia 2R\u00a0<\/b><\/span><\/span><\/span>e a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia oferecida por apenas um desses resistores com as mesmas caracter\u00edsticas, mas com o dobro do comprimento l tamb\u00e9m ser\u00e1 2R,\u00a0<\/b><\/span><\/span><\/span>pois\u00a0resist\u00eancia el\u00e9trica R \u00e9 diretamente proporcional ao comprimento l do fio.<\/b><\/span><\/span><\/span><\/p>\n

08. Correto \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>se\u00a0<\/b><\/span><\/span><\/span>associados em paralelo voc\u00ea ter\u00e1 apenas um resistor de resist\u00eancia R\/2 e a\u00a0resist\u00eancia oferecida por apenas um desses resistores com as mesmas caracter\u00edsticas, mas com o dobro da \u00e1rea de se\u00e7\u00e3o transversal tamb\u00e9m ser\u00e1 R\/2,\u00a0<\/b><\/span><\/span><\/span>pois\u00a0resist\u00eancia el\u00e9trica R \u00e9 inversamente proporcional \u00e0 \u00e1rea de se\u00e7\u00e3o transversal A do fio.<\/b><\/span><\/span><\/span><\/p>\n

16- Falso \u2013\u00a0<\/b><\/span><\/span><\/span>Dedu\u00e7\u00e3o da unidade da resistividade (\u03c1) no sistema internacional de unidades (SI)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

R- (01 + 02 + 04 + 08) = 15<\/b><\/span><\/span><\/span><\/p>\n

23-<\/span><\/span><\/span><\/p>\n

Efeito Joule<\/b><\/span><\/span><\/span><\/p>\n

Efeito Joule<\/b><\/span><\/span><\/span>\u00a0\u00a0 num\u00a0<\/b><\/span><\/span><\/span>condutor met\u00e1lico<\/b><\/span><\/span><\/span>, os\u00a0<\/b><\/span><\/span><\/span>el\u00e9trons livres da corrente el\u00e9trica,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>durante suas<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>movimenta\u00e7\u00f5es,<\/b><\/span><\/span><\/span>\u00a0sofrem continuamente\u00a0<\/b><\/span><\/span><\/span>colis\u00f5es\u00a0<\/b><\/span><\/span><\/span>com os\u00a0<\/b><\/span><\/span><\/span>\u00e1tomos da rede cristalina desse condutor,<\/b><\/span><\/span><\/span>\u00a0transferindo a eles\u00a0<\/b><\/span><\/span><\/span>parte\u00a0<\/b><\/span><\/span><\/span>de sua\u00a0<\/b><\/span><\/span><\/span>energia cin\u00e9tica e, como resultado, os \u00e1tomos do condutor,\u00a0<\/b><\/span><\/span><\/span>como um todo, passam a\u00a0<\/b><\/span><\/span><\/span>vibrar com uma energia maior.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Esse\u00a0<\/b><\/span><\/span><\/span>aumento do \u201cn\u00edvel de vibra\u00e7\u00e3o\u201d dos \u00e1tomos<\/b><\/span><\/span><\/span>\u00a0do condutor\u00a0<\/b><\/span><\/span><\/span>provocam um aumento<\/b><\/span><\/span><\/span>\u00a0de sua\u00a0<\/b><\/span><\/span><\/span>temperatura,<\/b><\/span><\/span><\/span>\u00a0fazendo-o\u00a0<\/b><\/span><\/span><\/span>liberar energia t\u00e9rmica (calor).<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esse fen\u00f4meno \u00e9 denominado efeito Joule.<\/b><\/span><\/span><\/span><\/p>\n

Sendo a\u00a0<\/b><\/span><\/span><\/span>pot\u00eancia da P da fonte (usina geradora) constante,\u00a0<\/b><\/span><\/span><\/span>pela express\u00e3o\u00a0<\/b><\/span><\/span><\/span>P = i.U,\u00a0<\/b><\/span><\/span><\/span>quanto\u00a0<\/b><\/span><\/span><\/span>maior a tens\u00e3o U\u00a0<\/b><\/span><\/span><\/span>no gerador,\u00a0<\/b><\/span><\/span><\/span>menor ser\u00e1 a corrente\u00a0<\/b><\/span><\/span><\/span>(menor perda por efeito Joule)\u00a0<\/b><\/span><\/span><\/span>na linha\u00a0<\/b><\/span><\/span><\/span>capaz de transmitir uma determinada pot\u00eancia ao consumidor.<\/b><\/span><\/span><\/span><\/p>\n

Essa\u00a0<\/b><\/span><\/span><\/span>perda de pot\u00eancia P\u00a0<\/b><\/span><\/span><\/span>(e consequentemente de energia el\u00e9trica) por\u00a0<\/b><\/span><\/span><\/span>efeito Joule\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecida pela\u00a0<\/b><\/span><\/span><\/span>express\u00e3o P = R.i2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>onde\u00a0<\/b><\/span><\/span><\/span>P \u00e9 a pot\u00eancia da fonte constante, R a resist\u00eancia equivalente da linha\u00a0<\/b><\/span><\/span><\/span>de transmiss\u00e3o e\u00a0<\/b><\/span><\/span><\/span>i a intensidade da corrente el\u00e9trica na mesma.\u00a0<\/b><\/span><\/span><\/span>Assim, como\u00a0<\/b><\/span><\/span><\/span>conv\u00e9m P dissipada menor, voc\u00ea deve diminuir R e i.<\/b><\/span><\/span><\/span><\/p>\n

Para\u00a0<\/b><\/span><\/span><\/span>diminuir<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia equivalente R\u00a0<\/b><\/span><\/span><\/span>da\u00a0<\/b><\/span><\/span><\/span>linha de transmiss\u00e3o,\u00a0<\/b><\/span><\/span><\/span>pela\u00a0<\/b><\/span><\/span><\/span>segunda lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

fornecida acima, mantendo L e S constantes,\u00a0<\/b><\/span><\/span><\/span>voc\u00ea deve\u00a0<\/b><\/span><\/span><\/span>diminuir o valor da resistividade\u00a0<\/b><\/span><\/span><\/span>(R e\u00a0\u00a0s\u00e3o diretamente proporcionais)<\/b><\/span><\/span><\/span><\/p>\n

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

24-<\/span><\/span><\/span><\/p>\n

Segunda lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

25-<\/span><\/span><\/span><\/p>\n

S\u00e3o\u00a0<\/b><\/span><\/span><\/span>dado<\/b><\/span><\/span><\/span>s:\u00a0<\/b><\/span><\/span><\/span>\u03c1 = 2,8.10\u20138\u00a0\u03a9; L = 1000 m = 103\u00a0m; A = 2 mm2\u00a0= 2.10-6\u00a0m2.<\/b><\/span><\/span><\/span><\/p>\n

R = \u03c1.L\/A\u00a0<\/b><\/span><\/span><\/span>= 2,8.10-8.103\/2.10-6\u00a0= 1,4.10-5\/10-6\u00a0= 1,4.10\u00a0<\/b><\/span><\/span><\/span>\u00a0R = 14 \u2126<\/b><\/span><\/span><\/span><\/p>\n

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

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

Primeira lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Segunda lei de Ohm (Resistividade)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>gr\u00e1fico Vxi fornecido refere-se \u00e0 primeira lei de Ohm\u00a0<\/b><\/span><\/span><\/span>(teoria acima) onde\u00a0<\/b><\/span><\/span><\/span>R \u00e9 constante para quaisquer valores de V e i\u00a0<\/b><\/span><\/span><\/span>e, por exemplo, escolhendo\u00a0<\/b><\/span><\/span><\/span>quando V = 50 V, i\u00a0<\/b><\/span><\/span><\/span>= 100 mA = 100.10-3\u00a0A =\u00a0<\/b><\/span><\/span><\/span>0,1 A\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R = v\/i\u00a0<\/b><\/span><\/span><\/span>\u00a0R = 50\/0,1\u00a0\u00a0<\/b><\/span><\/span><\/span>R = 500 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando agora a\u00a0<\/b><\/span><\/span><\/span>2a\u00a0lei de O<\/b><\/span><\/span><\/span>hm (veja teoria acima)\u00a0<\/b><\/span><\/span><\/span>sendo dados: R =\u00a0<\/b><\/span><\/span><\/span>500 \u2126;<\/b><\/span><\/span><\/span>\u00a0L =<\/b><\/span><\/span><\/span>\u00a02 m e\u00a01,75. 10-6<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\uf057<\/b><\/span><\/span><\/span>.m\u00a0<\/b><\/span><\/span><\/span>\u00a0R =\u00a0\/<\/b><\/span><\/span><\/span>S\u00a0\u00a0500 = 1,75.10-6.2\/S\u00a0\u00a0500S = 3,5.10-6\u00a0\u00a0S = 3,5.10-6\/5.102\u00a0= 0,7.10-8\u00a0m2\u00a0= 0,7.<\/b><\/span><\/span><\/span><\/p>\n

10-8.104\u00a0\u00a0<\/b><\/span><\/span><\/span>S = 0,7.10-4\u00a0cm2.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como tem\u00a0<\/b><\/span><\/span><\/span>curto circuito vamos refazer<\/b><\/span><\/span><\/span>\u00a0a associa\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>marcando todos os n\u00f3s que possuem o mesmo potencial com a mesma letra (A)<\/b><\/span><\/span><\/span>\u00a0e os\u00a0<\/b><\/span><\/span><\/span>outros, C e D<\/b><\/span><\/span><\/span>\u00a0e, em seguida vamos\u00a0<\/b><\/span><\/span><\/span>calcular a resist\u00eancia do resistor equivalente conforme a sequ\u00eancia a seguir:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Entre os pontos A e B\u00a0<\/b><\/span><\/span><\/span>a corrente circula\u00a0<\/b><\/span><\/span><\/span>apenas pelo ultimo resistor de 12 \u2126\u00a0\u00a0Req\u00a0= 12\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Ou ainda, veja na\u00a0<\/b><\/span><\/span><\/span>figura que se a corrente el\u00e9trica i entrar pelo ponto A ela se desviar\u00e1 conforme<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

I<\/b><\/span><\/span><\/span>ndicado<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>circular\u00e1 apenas pelo \u00faltimo resistor\u00a0\u00a0Req\u00a0= 12\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

Observe que os\u00a0<\/b><\/span><\/span><\/span>pontos P,Q, R e S\u00a0<\/b><\/span><\/span><\/span>possuem o\u00a0<\/b><\/span><\/span><\/span>mesmo potencial<\/b><\/span><\/span><\/span>\u00a0(curto circuito) ent\u00e3o voc\u00ea\u00a0<\/b><\/span><\/span><\/span>pode uni-los\u00a0<\/b><\/span><\/span><\/span>(primeira e segunda figuras)\u00a0<\/b><\/span><\/span><\/span>refazendo o circuito\u00a0<\/b><\/span><\/span><\/span>e, calculando a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia do resistor equivalente conforme a sequ\u00eancia de figuras.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

Formulas utilizadas na resolu\u00e7\u00e3o do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Rela\u00e7\u00f5es entre corrente el\u00e9trica (i), tens\u00e3o (U), resist\u00eancia el\u00e9trica (R) e<\/b><\/span><\/span><\/span><\/p>\n

potencia el\u00e9trica (Po)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Resolu\u00e7\u00e3o do exerc\u00edcio<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

30-<\/span><\/span><\/span><\/p>\n

Todos os dados j\u00e1 est\u00e3o no SI. \u00c9 poss\u00edvel notar que h\u00e1 uma chave aberta e uma chave fechada no circuito. O ramo que est\u00e1 com a chave aberta\u00a0<\/b><\/span><\/span><\/span>n\u00e3o permite a passagem de corrente<\/b><\/span><\/span><\/span>\u00a0e, portanto, pode ser\u00a0<\/b><\/span><\/span><\/span>desconsiderado junto com a resist\u00eancia<\/b><\/span><\/span><\/span>\u00a0que h\u00e1 nele, pois o circuito n\u00e3o o \u2018sente\u2019. Assim,\u00a0<\/b><\/span><\/span><\/span>a resist\u00eancia equivalente<\/b><\/span><\/span><\/span>\u00a0ser\u00e1 a opera\u00e7\u00e3o entre as\u00a0<\/b><\/span><\/span><\/span>duas resist\u00eancias em paralelo<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A resist\u00eancia equivalente \u00e9 a resist\u00eancia que produz os mesmos efeitos de\u00a0<\/b><\/span><\/span><\/span>corrente<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o<\/b><\/span><\/span><\/span>\u00a0quando substitui as duas outras resist\u00eancias no circuito, com isso e com a\u00a0<\/b><\/span><\/span><\/span>lei de Ohm<\/b><\/span><\/span><\/span>\u00a0\u00e9 poss\u00edvel calcular a corrente que flui pelo circuito, a lei de Ohm \u00e9 dada por:<\/b><\/span><\/span><\/span><\/p>\n

Onde U \u00e9 a tens\u00e3o imposta no resistor e I \u00e9 a corrente que passa por ele. Usando essa lei chegamos a:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Como as\u00a0<\/b><\/span><\/span><\/span>tr\u00eas pilhas est\u00e3o associadas em paralelo\u00a0<\/b><\/span><\/span><\/span>(mesma ddp) elas se\u00a0<\/b><\/span><\/span><\/span>comportam como se fossem uma \u00fanica pilha de ddp U = 25 V.<\/b><\/span><\/span><\/span><\/p>\n

Observe\u00a0<\/b><\/span><\/span><\/span>com aten\u00e7\u00e3o os\u00a0<\/b><\/span><\/span><\/span>resistores e veja que R1\u00a0est\u00e1 em s\u00e9rie com R2\u00a0e R3\u00a0que est\u00e3o em paralelo entre si<\/b><\/span><\/span><\/span>\u00a0e que o\u00a0<\/b><\/span><\/span><\/span>volt\u00edmetro est\u00e1 indicando apenas a ddp dessa associa\u00e7\u00e3o paralelo.<\/b><\/span><\/span><\/span><\/p>\n

Veja<\/b><\/span><\/span><\/span>\u00a0um\u00a0<\/b><\/span><\/span><\/span>esquema simb\u00f3lico dessa associa\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>na sequ\u00eancia das figuras o c\u00e1lculo da resist\u00eancia do resistor equivalente Req\u00a0e da corrente el\u00e9trica total i.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Quando<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>corrente total i = 2,5 A chega na associa\u00e7\u00e3o paralelo de resistores de mesmo valor (10 \u2126) ela se divide pela metade e cada resistor fica percorrido por i = 1,25 A\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>sob ddp de U\u00a0<\/b><\/span><\/span><\/span>= R.i = 10.1,25 =\u00a0<\/b><\/span><\/span><\/span>12,5 V,\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>\u00e9 a indica\u00e7\u00e3o do volt\u00edmetro colocado em paralelo em seus terminais.<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>amper\u00edmetro indica a corrente que passa por ele que vale i = 2,5 A.<\/b><\/span><\/span><\/span><\/p>\n

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

32-<\/span><\/span><\/span><\/p>\n

Se\u00a0<\/b><\/span><\/span><\/span>as l\u00e2mpadas s\u00e3o id\u00eanticas cada uma possui a mesma resist\u00eancia el\u00e9trica RL\u00a0<\/b><\/span><\/span><\/span>e s\u00e3o\u00a0<\/b><\/span><\/span><\/span>percorridas pela mesma corrente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>iL.<\/b><\/span><\/span><\/span><\/p>\n

Observe no\u00a0<\/b><\/span><\/span><\/span>esquema do circuito que, se o resistor R est\u00e1 sob ddp de U = 3V, cada uma das tr\u00eas l\u00e2mpadas ter\u00e3o a mesma ddp<\/b><\/span><\/span><\/span>\u00a0(paralelo a ddp \u00e9 a mesma) de\u00a0<\/b><\/span><\/span><\/span>U\u2019\u00a0<\/b><\/span><\/span><\/span>= (12 \u2013 3<\/b><\/span><\/span><\/span>) = 9 V.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>cada l\u00e2mpada est\u00e1 sob ddp de U\u2019 = 9 V pelo gr\u00e1fico voc\u00ea verifica que a corrente em cada l\u00e2mpada ser\u00e1 iL<\/b><\/span><\/span><\/span>\u00a0= 100 mA = 100.10-3\u00a0A =\u00a0<\/b><\/span><\/span><\/span>0,1 A<\/b><\/span><\/span><\/span>. Assim, a\u00a0<\/b><\/span><\/span><\/span>corrente total ser\u00e1 itotal<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 3iL\u00a0= 3.0,1<\/b><\/span><\/span><\/span>= 0,3 A.<\/b><\/span><\/span><\/span><\/p>\n

Essa\u00a0<\/b><\/span><\/span><\/span>corrente itotal\u00a0= 0,3 A \u00e9 a corrente que passa pelo resistor R\u00a0<\/b><\/span><\/span><\/span>que est\u00e1 sob\u00a0<\/b><\/span><\/span><\/span>ddp de U = 3V.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

33-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>aumenta a resist\u00eancia vari\u00e1vel R mantendo U constante\u00a0<\/b><\/span><\/span><\/span>voc\u00ea\u00a0<\/b><\/span><\/span><\/span>estar\u00e1 diminuindo i, pois na express\u00e3o R =U\/i, com U constante R e i s\u00e3o inversamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>diminui P voc\u00ea estar\u00e1 diminuindo i, pois na express\u00e3o P = U.i, com U constante, P e i s\u00e3o i s\u00e3o diretamente proporcionais<\/b><\/span><\/span><\/span>\u00a0e, ao\u00a0<\/b><\/span><\/span><\/span>mesmo tempo estar\u00e1 diminuindo a energia consumida E pois,<\/b><\/span><\/span><\/span><\/p>\n

P = E\/t, com t constante P e E s\u00e3o diretamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Na\u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o paralelo a resist\u00eancia do resistor equivalente pode ser calculada por Req\u00a0= produto\/soma<\/b><\/span><\/span><\/span>\u00a0= (40 x 10)\/(40 + 10) = 400\/50\u00a0\u00a0<\/b><\/span><\/span><\/span>Req\u00a0= 8 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o paralelo a tens\u00e3o U = 12 V \u00e9 a mesma para os dois resistores\u00a0\u00a0P40\u2126\u00a0= U2\/R\u00a0<\/b><\/span><\/span><\/span>= 122\/40 =<\/b><\/span><\/span><\/span><\/p>\n

144\/40\u00a0\u00a0<\/b><\/span><\/span><\/span>P40\u2126\u00a0= 3,6 W\u00a0\u00a0Req\u00a0= U\/i<\/b><\/span><\/span><\/span>\u00a0\u00a08 = 12\/i\u00a0\u00a0<\/b><\/span><\/span><\/span>i = 1,5 A<\/b><\/span><\/span><\/span><\/p>\n

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

35-<\/span><\/span><\/span><\/p>\n

Falsa\u00a0<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>pot\u00eancia \u00e9 fun\u00e7\u00e3o do valor da resist\u00eancia el\u00e9tr<\/b><\/span><\/span><\/span>ica e,\u00a0<\/b><\/span><\/span><\/span>como a corrente i \u00e9 a mesma<\/b><\/span><\/span><\/span>, para\u00a0<\/b><\/span><\/span><\/span>uma l\u00e2mpada a pot\u00eancia ser\u00e1 P1\u00a0= R.i2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>para as tr\u00eas P3\u00a0= 3Ri2<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>P3\u00a0> P1.<\/b><\/span><\/span><\/span><\/p>\n

B. Falsa\u00a0\u00a0veja (A).<\/b><\/span><\/span><\/span><\/p>\n

C. Falsa\u00a0\u00a0nas instala\u00e7\u00f5es residenciais a associa\u00e7\u00e3o utilizada \u00e9 a paralelo (mesma tens\u00e3o, ddp ou voltagem U).<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa\u00a0\u00a0uma das caracter\u00edsticas da associa\u00e7\u00e3o s\u00e9rie \u00e9 de que a corrente i \u00e9 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

E. Correta;<\/b><\/span><\/span><\/span><\/p>\n

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

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

Observe<\/b><\/span><\/span><\/span>\u00a0que a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia vari\u00e1vel Req\u00a0e o ferro el\u00e9trico de resist\u00eancia RF\u00a0est\u00e3o associados em s\u00e9rie<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sendo a\u00a0<\/b><\/span><\/span><\/span>corrente m\u00e1xima i = 10 A a mesma para os dois<\/b><\/span><\/span><\/span>\u00a0(caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie) e\u00a0<\/b><\/span><\/span><\/span>como a tens\u00e3o no ferro deve ser de VF\u00a0= 110 V, a tens\u00e3o em Req\u00a0tamb\u00e9m deve ser de Veq\u00a0= 110 V,<\/b><\/span><\/span><\/span>\u00a0pois\u00a0<\/b><\/span><\/span><\/span>Ventrada\u00a0= 220 V deve ser a soma das tens\u00f5es parciais ou seja, a tens\u00e3o em cada um deve ser de 110 V<\/b><\/span><\/span><\/span>\u00a0(caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/p>\n

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

A\u00a0<\/b><\/span><\/span><\/span>figura abaixo\u00a0<\/b><\/span><\/span><\/span>mostra um desses\u00a0<\/b><\/span><\/span><\/span>isolantes topol\u00f3gicos\u00a0<\/b><\/span><\/span><\/span>e, observe que a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o U \u00e9 a mesma<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

para\u00a0<\/b><\/span><\/span><\/span>cada resist\u00eancia (da superf\u00edcie e do interior) e que elas est\u00e3o associadas em paralelo.<\/b><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas da associa\u00e7\u00e3o paralelo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0Os\u00a0<\/b><\/span><\/span><\/span>resistores s\u00e3o associados pelos seus terminais<\/b><\/span><\/span><\/span>, ou seja,\u00a0<\/b><\/span><\/span><\/span>todos saem de um mesmo ponto e todos chegam a um mesmo ponto.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0A\u00a0<\/b><\/span><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o) U<\/b><\/span><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span><\/span>toda a associa\u00e7\u00e3o (entre A e B) \u00e9 a mesma\u00a0para todos os resistores<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0A\u00a0<\/b><\/span><\/span><\/span>corrente total i<\/b><\/span><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span><\/span>soma das correntes parciais<\/b><\/span><\/span><\/span>, ou seja,\u00a0i = i1\u00a0+ i2\u00a0+ i3<\/b><\/span><\/span><\/span><\/p>\n

F\u00f3rmulas v\u00e1lidas para a associa\u00e7\u00e3o paralelo<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

a) Claro que os\u00a0<\/b><\/span><\/span><\/span>aparelhos mencionados\u00a0<\/b><\/span><\/span><\/span>est\u00e3o associados em paralelo,\u00a0<\/b><\/span><\/span><\/span>pois a\u00a0<\/b><\/span><\/span><\/span>ddp (tens\u00e3o, voltagem) \u00e9 a mesma para todos (U = 110 V).<\/b><\/span><\/span><\/span><\/p>\n

Associa\u00e7\u00e3o paralelo<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

1. Correta\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Uma \u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o s\u00e9rie\u00a0<\/b><\/span><\/span><\/span>\u00e9 aquela em que os\u00a0<\/b><\/span><\/span><\/span>resistores est\u00e3o associados um em seguida do outro<\/b><\/span><\/span><\/span>\u00a0de modo que a\u00a0<\/b><\/span><\/span><\/span>corrente em cada um seja a mesma.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

C\u00e1lculo<\/b><\/span><\/span><\/span>\u00a0da resist\u00eancia\u00a0<\/b><\/span><\/span><\/span>(Req)<\/b><\/span><\/span><\/span>\u00a0do\u00a0<\/b><\/span><\/span><\/span>resistor equivalente\u00a0na associa\u00e7\u00e3o s\u00e9rie\u00a0<\/b><\/span><\/span><\/span>\u00a0a resist\u00eancia\u00a0do\u00a0<\/b><\/span><\/span><\/span>resistor<\/b><\/span><\/span><\/span>\u00a0equivalente entre A e B\u00a0\u00e9 fornecida pela\u00a0soma das resist\u00eancias de cada resistor.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Portanto, numa\u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o de resistores em s\u00e9rie,\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>resistor equivalente\u00a0<\/b><\/span><\/span><\/span>sempre tem\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia maior\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>qualquer uma das resist\u00eancias\u00a0<\/b><\/span><\/span><\/span>dos resistores que\u00a0<\/b><\/span><\/span><\/span>formam a associa\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

02. Falsa\u00a0\u00a0A tens\u00e3o U \u00e9 a mesma para todos os capacitores. Veja informa\u00e7\u00e3o abaixo:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

03. Correta\u00a0\u00a0Veja as informa\u00e7\u00f5es abaixo:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>processo de carga,<\/b><\/span><\/span><\/span>\u00a0quando os\u00a0<\/b><\/span><\/span><\/span>capacitores em s\u00e9rie\u00a0estiverem ligados ao\u00a0gerador de tens\u00e3o U<\/b><\/span><\/span><\/span>, as\u00a0<\/b><\/span><\/span><\/span>cargas\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>saem de uma armadura ser\u00e3o deslocadas para a seguinte\u00a0<\/b><\/span><\/span><\/span>at\u00e9 que as\u00a0<\/b><\/span><\/span><\/span>cargas de todas as armaduras tenham o mesmo m\u00f3dulo, ou seja, Q1\u00a0= Q2\u00a0= Q3\u00a0= Q, que \u00e9 igual \u00e0 carga do capacitor equivalente.<\/b><\/span><\/span><\/span><\/p>\n

Observe que,\u00a0se as\u00a0<\/b><\/span><\/span><\/span>cargas s\u00e3o iguais\u00a0e as\u00a0capacit\u00e2ncias s\u00e3o diferentes,\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o) tamb\u00e9m ser\u00e1 diferente.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Voc\u00ea tamb\u00e9m pode\u00a0<\/b><\/span><\/span><\/span>calcular o capacitor equivalente\u00a0<\/b><\/span><\/span><\/span>como\u00a0<\/b><\/span><\/span><\/span>Ceq\u00a0= produto\/soma tomados dois a dois.<\/b><\/span><\/span><\/span><\/p>\n

Lembre-se de que o\u00a0<\/b><\/span><\/span><\/span>Ceq\u00a0\u00e9 sempre menor que a capacit\u00e2ncia de cada um dos capacitores associados.<\/b><\/span><\/span><\/span><\/p>\n

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

40-<\/span><\/span><\/span><\/p>\n

Formulas utilizadas na resolu\u00e7\u00e3o do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Rela\u00e7\u00f5es entre corrente el\u00e9trica (i), tens\u00e3o (U), resist\u00eancia el\u00e9trica (R) e<\/b><\/span><\/span><\/span><\/p>\n

potencia el\u00e9trica (Po)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Resolu\u00e7\u00e3o do exerc\u00edcio<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

Todos os dados j\u00e1 est\u00e3o no SI. \u00c9 poss\u00edvel notar que h\u00e1 uma chave aberta e uma chave fechada no circuito. O ramo que est\u00e1 com a chave aberta\u00a0<\/b><\/span><\/span><\/span>n\u00e3o permite a passagem de corrente<\/b><\/span><\/span><\/span>\u00a0e, portanto, pode ser\u00a0<\/b><\/span><\/span><\/span>desconsiderado junto com a resist\u00eancia<\/b><\/span><\/span><\/span>\u00a0que h\u00e1 nele, pois o circuito n\u00e3o o \u2018sente\u2019. Assim,\u00a0<\/b><\/span><\/span><\/span>a resist\u00eancia equivalente<\/b><\/span><\/span><\/span>\u00a0ser\u00e1 a opera\u00e7\u00e3o entre as\u00a0<\/b><\/span><\/span><\/span>duas resist\u00eancias em paralelo<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A resist\u00eancia equivalente \u00e9 a resist\u00eancia que produz os mesmos efeitos de\u00a0<\/b><\/span><\/span><\/span>corrente<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o<\/b><\/span><\/span><\/span>\u00a0quando substitui as duas outras resist\u00eancias no circuito, com isso e com a\u00a0<\/b><\/span><\/span><\/span>lei de Ohm<\/b><\/span><\/span><\/span>\u00a0\u00e9 poss\u00edvel calcular a corrente que flui pelo circuito, a lei de Ohm \u00e9 dada por:<\/b><\/span><\/span><\/span><\/p>\n

Onde U \u00e9 a tens\u00e3o imposta no resistor e I \u00e9 a corrente que passa por ele. Usando essa lei chegamos a:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

42-<\/span><\/span><\/span><\/p>\n

A ideia principal nesse problema \u00e9 lembrar que\u00a0<\/b><\/span><\/span><\/span>o capacitor funciona como um circuito aberto em corrente cont\u00ednua<\/b><\/span><\/span><\/span>, assim\u00a0<\/b><\/span><\/span><\/span>n\u00e3o haver\u00e1 passagem de corrente\u00a0<\/b><\/span><\/span><\/span>na malha em que o capacitor se encontra,\u00a0<\/b><\/span><\/span><\/span>inutilizando o resistor que se encontra ali<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>A corrente I que flui pelo circuito \u00e9 ent\u00e3o apenas a que passa na malha em que se encontra a fonte U=30 V<\/b><\/span><\/span><\/span>\u00a0e a resist\u00eancia restante R, o circuito equivalente com essas modifica\u00e7\u00f5es \u00e9 exibido na imagem abaixo.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

Como as\u00a0<\/b><\/span><\/span><\/span>tr\u00eas pilhas est\u00e3o associadas em paralelo\u00a0<\/b><\/span><\/span><\/span>(mesma ddp) elas se\u00a0<\/b><\/span><\/span><\/span>comportam como se fossem uma \u00fanica pilha de ddp U = 25 V.<\/b><\/span><\/span><\/span><\/p>\n

Observe\u00a0<\/b><\/span><\/span><\/span>com aten\u00e7\u00e3o os\u00a0<\/b><\/span><\/span><\/span>resistores e veja que R1\u00a0est\u00e1 em s\u00e9rie com R2\u00a0e R3\u00a0que est\u00e3o em paralelo entre si<\/b><\/span><\/span><\/span>\u00a0e que o\u00a0<\/b><\/span><\/span><\/span>volt\u00edmetro est\u00e1 indicando apenas a ddp dessa associa\u00e7\u00e3o paralelo.<\/b><\/span><\/span><\/span><\/p>\n

Veja<\/b><\/span><\/span><\/span>\u00a0um\u00a0<\/b><\/span><\/span><\/span>esquema simb\u00f3lico dessa associa\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>na sequ\u00eancia das figuras o c\u00e1lculo da resist\u00eancia do resistor equivalente Req\u00a0e da corrente el\u00e9trica total i.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Quando<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>corrente total i = 2,5 A chega na associa\u00e7\u00e3o paralelo de resistores de mesmo valor (10 \u2126) ela se divide pela metade e cada resistor fica percorrido por i = 1,25 A\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>sob ddp de U\u00a0<\/b><\/span><\/span><\/span>= R.i = 10.1,25 =\u00a0<\/b><\/span><\/span><\/span>12,5 V,\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>\u00e9 a indica\u00e7\u00e3o do volt\u00edmetro colocado em paralelo em seus terminais.<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>amper\u00edmetro indica a corrente que passa por ele que vale i = 2,5 A.<\/b><\/span><\/span><\/span><\/p>\n

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

44-<\/span><\/span><\/span><\/p>\n

Relembrando alguns conceitos de\u00a0<\/b><\/span><\/span><\/span>plano inclinado<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

45-<\/span><\/span><\/span><\/p>\n

Se\u00a0<\/b><\/span><\/span><\/span>as l\u00e2mpadas s\u00e3o id\u00eanticas cada uma possui a mesma resist\u00eancia el\u00e9trica RL\u00a0<\/b><\/span><\/span><\/span>e s\u00e3o\u00a0<\/b><\/span><\/span><\/span>percorridas pela mesma corrente<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>iL.<\/b><\/span><\/span><\/span><\/p>\n

Observe no\u00a0<\/b><\/span><\/span><\/span>esquema do circuito que, se o resistor R est\u00e1 sob ddp de U = 3V, cada uma das tr\u00eas l\u00e2mpadas ter\u00e3o a mesma ddp<\/b><\/span><\/span><\/span>\u00a0(paralelo a ddp \u00e9 a mesma) de\u00a0<\/b><\/span><\/span><\/span>U\u2019\u00a0<\/b><\/span><\/span><\/span>= (12 \u2013 3<\/b><\/span><\/span><\/span>) = 9 V.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>cada l\u00e2mpada est\u00e1 sob ddp de U\u2019 = 9 V pelo gr\u00e1fico voc\u00ea verifica que a corrente em cada l\u00e2mpada ser\u00e1 iL<\/b><\/span><\/span><\/span>\u00a0= 100 mA = 100.10-3\u00a0A =\u00a0<\/b><\/span><\/span><\/span>0,1 A<\/b><\/span><\/span><\/span>. Assim, a\u00a0<\/b><\/span><\/span><\/span>corrente total ser\u00e1 itotal<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 3iL\u00a0= 3.0,1<\/b><\/span><\/span><\/span>= 0,3 A.<\/b><\/span><\/span><\/span><\/p>\n

Essa\u00a0<\/b><\/span><\/span><\/span>corrente itotal\u00a0= 0,3 A \u00e9 a corrente que passa pelo resistor R\u00a0<\/b><\/span><\/span><\/span>que est\u00e1 sob\u00a0<\/b><\/span><\/span><\/span>ddp de U = 3V.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

1 mA = 10-3\u00a0A\u00a0<\/b><\/span><\/span><\/span>\u00a0assim,\u00a0<\/b><\/span><\/span><\/span>10 000 mA\u00a0<\/b><\/span><\/span><\/span>= 10 000\u00d710-3\u00a0A\u00a0<\/b><\/span><\/span><\/span>= 10 A.<\/b><\/span><\/span><\/span><\/p>\n

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

47-<\/span><\/span><\/span><\/p>\n

Uma\u00a0<\/b><\/span><\/span><\/span>situa\u00e7\u00e3o pr\u00e1tica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>bastante comum<\/b><\/span><\/span><\/span>\u00a0nas\u00a0<\/b><\/span><\/span><\/span>resid\u00eancias<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>nos hot\u00e9is<\/b><\/span><\/span><\/span>\u00a0\u00e9 o chamado<\/b><\/span><\/span><\/span><\/p>\n

\u201cinterruptor paralelo\u201d,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>onde,\u00a0<\/b><\/span><\/span><\/span>na\u00a0realidade a liga\u00e7\u00e3o \u00e9 em s\u00e9rie<\/b><\/span><\/span><\/span>,\u00a0na qual\u00a0<\/b><\/span><\/span><\/span>\u00e9\u00a0poss\u00edvel ligar ou desligar uma determinada l\u00e2mpada, de forma independente, estando no ponto mais alto ou mais baixo de uma escada, da maneira indicada nas sequ\u00eancias abaixo.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

48-<\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>aumenta a resist\u00eancia vari\u00e1vel R mantendo U constante\u00a0<\/b><\/span><\/span><\/span>voc\u00ea\u00a0<\/b><\/span><\/span><\/span>estar\u00e1 diminuindo i, pois na express\u00e3o R =U\/i, com U constante R e i s\u00e3o inversamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>diminui P voc\u00ea estar\u00e1 diminuindo i, pois na express\u00e3o P = U.i, com U constante, P e i s\u00e3o i s\u00e3o diretamente proporcionais<\/b><\/span><\/span><\/span>\u00a0e, ao\u00a0<\/b><\/span><\/span><\/span>mesmo tempo estar\u00e1 diminuindo a energia consumida E pois,<\/b><\/span><\/span><\/span><\/p>\n

P = E\/t, com t constante P e E s\u00e3o diretamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Na\u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o paralelo a resist\u00eancia do resistor equivalente pode ser calculada por Req\u00a0= produto\/soma<\/b><\/span><\/span><\/span>\u00a0= (40 x 10)\/(40 + 10) = 400\/50\u00a0\u00a0<\/b><\/span><\/span><\/span>Req\u00a0= 8 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>associa\u00e7\u00e3o paralelo a tens\u00e3o U = 12 V \u00e9 a mesma para os dois resistores\u00a0\u00a0P40\u2126\u00a0= U2\/R\u00a0<\/b><\/span><\/span><\/span>= 122\/40 =<\/b><\/span><\/span><\/span><\/p>\n

144\/40\u00a0\u00a0<\/b><\/span><\/span><\/span>P40\u2126\u00a0= 3,6 W\u00a0\u00a0Req\u00a0= U\/i<\/b><\/span><\/span><\/span>\u00a0\u00a08 = 12\/i\u00a0\u00a0<\/b><\/span><\/span><\/span>i = 1,5 A<\/b><\/span><\/span><\/span><\/p>\n

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

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

Veja na sequ\u00eancia das figuras abaixo o c\u00e1lculo da resist\u00eancia do resistor equivalente:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Req\u00a0= U\/i\u00a0 <\/b><\/span><\/span><\/span>\u21d2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>3,5 = 12\/i\u00a0<\/b><\/span><\/span><\/span>\u21d2<\/b><\/span><\/span><\/span>\u00a0i = 3,2 A.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Falsa\u00a0<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>pot\u00eancia \u00e9 fun\u00e7\u00e3o do valor da resist\u00eancia el\u00e9tr<\/b><\/span><\/span><\/span>ica e,\u00a0<\/b><\/span><\/span><\/span>como a corrente i \u00e9 a mesma<\/b><\/span><\/span><\/span>, para\u00a0<\/b><\/span><\/span><\/span>uma l\u00e2mpada a pot\u00eancia ser\u00e1 P1\u00a0= R.i2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>para as tr\u00eas P3\u00a0= 3Ri2<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>P3\u00a0> P1.<\/b><\/span><\/span><\/span><\/p>\n

B. Falsa\u00a0\u00a0veja (A).<\/b><\/span><\/span><\/span><\/p>\n

C. Falsa\u00a0\u00a0nas instala\u00e7\u00f5es residenciais a associa\u00e7\u00e3o utilizada \u00e9 a paralelo (mesma tens\u00e3o, ddp ou voltagem U).<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa\u00a0\u00a0uma das caracter\u00edsticas da associa\u00e7\u00e3o s\u00e9rie \u00e9 de que a corrente i \u00e9 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

E. Correta;<\/b><\/span><\/span><\/span><\/p>\n

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

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

Primeiro circuito<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Segundo ciercuito<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Terceiro circuito<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Quarto circuito<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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


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

<\/p>\n

<\/p>\n

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

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

Observe<\/b><\/span><\/span><\/span>\u00a0que a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia vari\u00e1vel Req\u00a0e o ferro el\u00e9trico de resist\u00eancia RF\u00a0est\u00e3o associados em s\u00e9rie<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sendo a\u00a0<\/b><\/span><\/span><\/span>corrente m\u00e1xima i = 10 A a mesma para os dois<\/b><\/span><\/span><\/span>\u00a0(caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie) e\u00a0<\/b><\/span><\/span><\/span>como a tens\u00e3o no ferro deve ser de VF\u00a0= 110 V, a tens\u00e3o em Req\u00a0tamb\u00e9m deve ser de Veq\u00a0= 110 V,<\/b><\/span><\/span><\/span>\u00a0pois\u00a0<\/b><\/span><\/span><\/span>Ventrada\u00a0= 220 V deve ser a soma das tens\u00f5es parciais ou seja, a tens\u00e3o em cada um deve ser de 110 V<\/b><\/span><\/span><\/span>\u00a0(caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/p>\n

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

Vamos\u00a0<\/b><\/span><\/span><\/span>analisar apenas a parte inferior do circuito<\/b><\/span><\/span><\/span>\u00a0onde\u00a0<\/b><\/span><\/span><\/span>circulam as correntes i, i1\u00a0e i2,\u00a0<\/b><\/span><\/span><\/span>j\u00e1 que com<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

apenas esses dados \u00e9 poss\u00edvel calcular a tens\u00e3o U pedida.<\/b><\/span><\/span><\/span><\/p>\n

Observe que o\u00a0<\/b><\/span><\/span><\/span>trecho que possui o resistor de R2\u00a0= 120 \u2126 \u00e9 percorrido pela corrente i2\u00a0= 500 mA = 0,5 A (corrente nominal suportada pelo fus\u00edvel).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

<\/p>\n

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

Os valores nominais do LED s\u00e3o P = 8 W e U = 4 V fornecidos pelo enunciado (<\/b><\/span><\/span><\/span>LED (L) de 8 W,\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>opera<\/b><\/span><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span><\/span>4 V).<\/b><\/span><\/span><\/span><\/p>\n

Pelos valores nominais voc\u00ea pode calcular o valor da resist\u00eancia do LED atrav\u00e9s da express\u00e3o P =\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a08 =\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0R =\u00a0\u00a0\u00a0RL\u00a0= 2 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Para que o\u00a0<\/b><\/span><\/span><\/span>LED opere segundo sua tens\u00e3o nominal (U = 4 V)\u00a0<\/b><\/span><\/span><\/span>ele\u00a0<\/b><\/span><\/span><\/span>dever\u00e1 ser percorrido\u00a0<\/b><\/span><\/span><\/span>por uma\u00a0<\/b><\/span><\/span><\/span>corrente el\u00e9trica de R =\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a02 =\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>i = 2 A.<\/b><\/span><\/span><\/span><\/p>\n

Como\u00a0<\/b><\/span><\/span><\/span>R e RL\u00a0est\u00e3o em s\u00e9rie\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>corrente el\u00e9trica neles \u00e9 a mesma\u00a0<\/b><\/span><\/span><\/span>e a resist\u00eancia do\u00a0<\/b><\/span><\/span><\/span>resistor equivalente ser\u00e1 Req\u00a0=\u00a0<\/b><\/span><\/span><\/span>R + RL<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= R + 2.<\/b><\/span><\/span><\/span><\/p>\n

Esse\u00a0<\/b><\/span><\/span><\/span>Req\u00a0= R + 2\u00a0<\/b><\/span><\/span><\/span>est\u00e1 ligado na\u00a0<\/b><\/span><\/span><\/span>fonte de U = 6 V\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>lei de Ohm Req\u00a0=\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0R + 2 =\u00a0\u00a0\u00a0R = 3 \u2013 2\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R = 1 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

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

A\u00a0<\/b><\/span><\/span><\/span>figura abaixo\u00a0<\/b><\/span><\/span><\/span>mostra um desses\u00a0<\/b><\/span><\/span><\/span>isolantes topol\u00f3gicos\u00a0<\/b><\/span><\/span><\/span>e, observe que a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o U \u00e9 a mesma<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

para\u00a0<\/b><\/span><\/span><\/span>cada resist\u00eancia (da superf\u00edcie e do interior) e que elas est\u00e3o associadas em paralelo.<\/b><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas da associa\u00e7\u00e3o paralelo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0Os\u00a0<\/b><\/span><\/span><\/span>resistores s\u00e3o associados pelos seus terminais<\/b><\/span><\/span><\/span>, ou seja,\u00a0<\/b><\/span><\/span><\/span>todos saem de um mesmo ponto e todos chegam a um mesmo ponto.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0A\u00a0<\/b><\/span><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o) U<\/b><\/span><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span><\/span>toda a associa\u00e7\u00e3o (entre A e B) \u00e9 a mesma\u00a0para todos os resistores<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0A\u00a0<\/b><\/span><\/span><\/span>corrente total i<\/b><\/span><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span><\/span>soma das correntes parciais<\/b><\/span><\/span><\/span>, ou seja,\u00a0i = i1\u00a0+ i2\u00a0+ i3<\/b><\/span><\/span><\/span><\/p>\n

F\u00f3rmulas v\u00e1lidas para a associa\u00e7\u00e3o paralelo<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

60-<\/span><\/span><\/span><\/p>\n

Observe que os\u00a0<\/b><\/span><\/span><\/span>pontos P,Q, R e S\u00a0<\/b><\/span><\/span><\/span>possuem o\u00a0<\/b><\/span><\/span><\/span>mesmo potencial<\/b><\/span><\/span><\/span>\u00a0(curto circuito) ent\u00e3o voc\u00ea\u00a0<\/b><\/span><\/span><\/span>pode uni-los\u00a0<\/b><\/span><\/span><\/span>(primeira e segunda figuras)\u00a0<\/b><\/span><\/span><\/span>refazendo o circuito\u00a0<\/b><\/span><\/span><\/span>e, calculando a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia do resistor equivalente conforme a sequ\u00eancia de figuras.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

61-<\/span><\/span><\/span><\/p>\n

Primeiro, \u00e9 importante retomar \u00e0\u00a0<\/b><\/span><\/span><\/span>lei de Ohm<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Pela equa\u00e7\u00e3o, \u00e9 poss\u00edvel observar que para manter a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o U constante<\/b><\/span><\/span><\/span>, ao\u00a0<\/b><\/span><\/span><\/span>diminuir<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a corrente\u00a0<\/b><\/span><\/span><\/span>devemos\u00a0<\/b><\/span><\/span><\/span>aumentar<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a resist\u00eancia<\/b><\/span><\/span><\/span>, e ao\u00a0<\/b><\/span><\/span><\/span>aumentar<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a corrente\u00a0<\/b><\/span><\/span><\/span>devemos\u00a0<\/b><\/span><\/span><\/span>diminuir<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a resist\u00eancia.\u00a0<\/b><\/span><\/span><\/span>Portanto, como queremos a corrente de\u00a0<\/b><\/span><\/span><\/span>menor intensidade<\/b><\/span><\/span><\/span>, devemos buscar o ramo que oferece a\u00a0<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia<\/b><\/span><\/span><\/span>. Para isso, vamos calcular a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span><\/span>\u00a0de cada ramo para podermos comparar. Iremos utilizar as seguintes f\u00f3rmulas para associa\u00e7\u00f5es de resistores:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Caso I:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Como podemos ver o caso III tem a\u00a0<\/b><\/span><\/span><\/span>maior resist\u00eancia<\/b><\/span><\/span><\/span>, ent\u00e3o a alternativa correta \u00e9 ent\u00e3o a\u00a0<\/b><\/span><\/span><\/span>(C).<\/b><\/span><\/span><\/span><\/p>\n

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

<\/p>\n

<\/p>\n

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

A\u00a0<\/b><\/span><\/span><\/span>sequ\u00eancia abaixo<\/b><\/span><\/span><\/span>\u00a0mostra a\u00a0<\/b><\/span><\/span><\/span>representa\u00e7\u00e3o simb\u00f3lica\u00a0<\/b><\/span><\/span><\/span>do circuito e o\u00a0<\/b><\/span><\/span><\/span>c\u00e1lculo da resist\u00eancia equivalente Req\u00a0= 5,8 + r\u00a0<\/b><\/span><\/span><\/span>(soma porque (r e 5,8 \u2126) est\u00e3o associados em s\u00e9rie).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

Veja na\u00a0<\/b><\/span><\/span><\/span>figura e no esquema abaix<\/b><\/span><\/span><\/span>o onde\u00a0<\/b><\/span><\/span><\/span>E = 10 000 V\u00a0<\/b><\/span><\/span><\/span>\u00e9 a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a eletromotriz\u00a0<\/b><\/span><\/span><\/span>do gerador,\u00a0<\/b><\/span><\/span><\/span>r\u00a0<\/b><\/span><\/span><\/span>sua\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia interna, Rh\u00a0= 1 000<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2126\u00a0<\/b><\/span><\/span><\/span>\u00e9 a\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia corporal da pessoa e i = 0,01 A \u00e9 a m\u00e1xima corrente n\u00e3o letal atrav\u00e9s do corpo da pessoa.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

Curva caracter\u00edstica de um gerador<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

E = for\u00e7a eletromotriz<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>icc\u00a0= corrente de curto circuito<\/b><\/span><\/span><\/span><\/p>\n

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

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

C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>resist\u00eancia do resistor equivalente onde R = 100 \u2126 e r = 20 \u2126 est\u00e3o em s\u00e9rie\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Req\u00a0= R + r<\/b><\/span><\/span><\/span>\u00a0= 100 + 20\u00a0\u00a0<\/b><\/span><\/span><\/span>Req\u00a0= 120 \u2126<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>corrente el\u00e9trica i no circuito\u00a0\u00a0Req\u00a0= (U = E)\/i\u00a0<\/b><\/span><\/span><\/span>\u00a0120 = 120\/i\u00a0\u00a0i 120\/120\u00a0\u00a0<\/b><\/span><\/span><\/span>i = 1 A.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>pot\u00eancia dissipada no gerador \u00e9 devida a sua resist\u00eancia interna r = 20 \u2126<\/b><\/span><\/span><\/span>\u00a0e fornecida por\u00a0<\/b><\/span><\/span><\/span>P = r.i2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 20.12\u00a0= 20.1\u00a0\u00a0<\/b><\/span><\/span><\/span>P = 20 W<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

<\/p>\n

<\/p>\n

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

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

Representando<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>circuito\u00a0<\/b><\/span><\/span><\/span>por meio dos\u00a0<\/b><\/span><\/span><\/span>s\u00edmbolos convencionais:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como os\u00a0<\/b><\/span><\/span><\/span>aparelhos de medidas s\u00e3o ideais\u00a0<\/b><\/span><\/span><\/span>voc\u00ea pode\u00a0<\/b><\/span><\/span><\/span>\u201ccurto-circuitar\u201d o amper\u00edmetro\u00a0<\/b><\/span><\/span><\/span>(resist\u00eancia interna nula) e\u00a0<\/b><\/span><\/span><\/span>retirar o volt\u00edmetro\u00a0<\/b><\/span><\/span><\/span>(resist\u00eancia interna infinita).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe na\u00a0<\/b><\/span><\/span><\/span>figura acima\u00a0<\/b><\/span><\/span><\/span>que a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o (ddp ou voltagem) no resistor R\u00a0<\/b><\/span><\/span><\/span>vale\u00a0<\/b><\/span><\/span><\/span>UR<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 9,0 \u2013 5,7\u00a0<\/b><\/span><\/span><\/span>= 3,3 V\u00a0<\/b><\/span><\/span><\/span>e que ele\u00a0<\/b><\/span><\/span><\/span>\u00e9 percorrido por uma corrente el\u00e9trica de i = 0,15 A\u00a0<\/b><\/span><\/span><\/span>(indica\u00e7\u00e3o do amper\u00edmetro).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Quando o\u00a0<\/b><\/span><\/span><\/span>volt\u00edmetro \u00e9 colocado em contato com os polos da bateria<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>indicar 12 V\u00a0<\/b><\/span><\/span><\/span>ele\u00a0<\/b><\/span><\/span><\/span>estar\u00e1 fornecendo o valor da for\u00e7a eletromotriz E da bateria\u00a0\u00a0E = 12 V.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>ddp, tens\u00e3o ou voltagem U nos terminais do resistor de R = 1 k\u2126 = 1.103\u00a0\u2126, percorrido por i = 8 mA = 8.10-3\u00a0A\u00a0<\/b><\/span><\/span><\/span>(indica\u00e7\u00e3o do amper\u00edmetro<\/b><\/span><\/span><\/span>) ser\u00e1\u00a0\u00a0R = U\/i\u00a0<\/b><\/span><\/span><\/span>\u00a0103\u00a0= U\/8.10-3\u00a0<\/b><\/span><\/span><\/span>\u00a0U = 8 V.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o do gerador U = E \u2013 r.i\u00a0<\/b><\/span><\/span><\/span>\u00a08 = 12 \u2013 r.8.10-3\u00a0\u00a0r = 4\/8.10-3\u00a0\u00a0<\/b><\/span><\/span><\/span>r\u00a0<\/b><\/span><\/span><\/span>= 0,5 .103<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0,5 k\u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Vamos\u00a0<\/b><\/span><\/span><\/span>analisar apenas a parte inferior do circuito<\/b><\/span><\/span><\/span>\u00a0onde\u00a0<\/b><\/span><\/span><\/span>circulam as correntes i, i1\u00a0e i2,\u00a0<\/b><\/span><\/span><\/span>j\u00e1 que com<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

apenas esses dados \u00e9 poss\u00edvel calcular a tens\u00e3o U pedida.<\/b><\/span><\/span><\/span><\/p>\n

Observe que o\u00a0<\/b><\/span><\/span><\/span>trecho que possui o resistor de R2\u00a0= 120 \u2126 \u00e9 percorrido pela corrente i2\u00a0= 500 mA = 0,5 A (corrente nominal suportada pelo fus\u00edvel).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Como as\u00a0<\/b><\/span><\/span><\/span>tr\u00eas pilhas est\u00e3o associadas em paralelo\u00a0<\/b><\/span><\/span><\/span>(mesma ddp) elas se\u00a0<\/b><\/span><\/span><\/span>comportam como se fossem uma \u00fanica pilha de ddp U = 25 V.<\/b><\/span><\/span><\/span><\/p>\n

Observe\u00a0<\/b><\/span><\/span><\/span>com aten\u00e7\u00e3o os\u00a0<\/b><\/span><\/span><\/span>resistores e veja que R1\u00a0est\u00e1 em s\u00e9rie com R2\u00a0e R3\u00a0que est\u00e3o em paralelo entre si<\/b><\/span><\/span><\/span>\u00a0e que o\u00a0<\/b><\/span><\/span><\/span>volt\u00edmetro est\u00e1 indicando apenas a ddp dessa associa\u00e7\u00e3o paralelo.<\/b><\/span><\/span><\/span><\/p>\n

Veja<\/b><\/span><\/span><\/span>\u00a0um\u00a0<\/b><\/span><\/span><\/span>esquema simb\u00f3lico dessa associa\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>na sequ\u00eancia das figuras o c\u00e1lculo da resist\u00eancia do resistor equivalente Req\u00a0e da corrente el\u00e9trica total i.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Quando<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>corrente total i = 2,5 A chega na associa\u00e7\u00e3o paralelo de resistores de mesmo valor (10 \u2126) ela se divide pela metade e cada resistor fica percorrido por i = 1,25 A\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>sob ddp de U\u00a0<\/b><\/span><\/span><\/span>= R.i = 10.1,25 =\u00a0<\/b><\/span><\/span><\/span>12,5 V,\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>\u00e9 a indica\u00e7\u00e3o do volt\u00edmetro colocado em paralelo em seus terminais.<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>amper\u00edmetro indica a corrente que passa por ele que vale i = 2,5 A.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Considerando o galvan\u00f4metro ideal, podemos utilizar\u00a0<\/b><\/span><\/span><\/span>a f\u00f3rmula do divisor de corrente\u00a0<\/b><\/span><\/span><\/span>para calcular a raz\u00e3o pedida. O divisor de corrente diz que\u00a0<\/b><\/span><\/span><\/span>a corrente em um ramo IG<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ser\u00e1\u00a0<\/b><\/span><\/span><\/span>proporcional<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e0 corrente de entrada na malha I<\/b><\/span><\/span><\/span>\u00a0por um fator de resist\u00eancia dado por:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Todos os dados j\u00e1 est\u00e3o no SI. \u00c9 poss\u00edvel notar que h\u00e1 uma chave aberta e uma chave fechada no circuito. O ramo que est\u00e1 com a chave aberta\u00a0<\/b><\/span><\/span><\/span>n\u00e3o permite a passagem de corrente<\/b><\/span><\/span><\/span>\u00a0e, portanto, pode ser\u00a0<\/b><\/span><\/span><\/span>desconsiderado junto com a resist\u00eancia<\/b><\/span><\/span><\/span>\u00a0que h\u00e1 nele, pois o circuito n\u00e3o o \u2018sente\u2019. Assim,\u00a0<\/b><\/span><\/span><\/span>a resist\u00eancia equivalente<\/b><\/span><\/span><\/span>\u00a0ser\u00e1 a opera\u00e7\u00e3o entre as\u00a0<\/b><\/span><\/span><\/span>duas resist\u00eancias em paralelo<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A resist\u00eancia equivalente \u00e9 a resist\u00eancia que produz os mesmos efeitos de\u00a0<\/b><\/span><\/span><\/span>corrente<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o<\/b><\/span><\/span><\/span>\u00a0quando substitui as duas outras resist\u00eancias no circuito, com isso e com a\u00a0<\/b><\/span><\/span><\/span>lei de Ohm<\/b><\/span><\/span><\/span>\u00a0\u00e9 poss\u00edvel calcular a corrente que flui pelo circuito, a lei de Ohm \u00e9 dada por:<\/b><\/span><\/span><\/span><\/p>\n

Onde U \u00e9 a tens\u00e3o imposta no resistor e I \u00e9 a corrente que passa por ele. Usando essa lei chegamos a:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Primeira lei de Kirchhoff ou lei dos n\u00f3s:\u00a0\u201cA\u00a0<\/b><\/span><\/span><\/span>soma alg\u00e9brica\u00a0<\/b><\/span><\/span><\/span>das\u00a0<\/b><\/span><\/span><\/span>correntes que chegam a um n\u00f3\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0\u00a0<\/b><\/span><\/span><\/span>soma alg\u00e9brica das correntes que saem do mesmo n\u00f3\u201d.<\/b><\/span><\/span><\/span><\/p>\n

Exemplo num\u00e9rico\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0 determine a\u00a0<\/b><\/span><\/span><\/span>intensidade da corrente i3\u00a0do trecho de circuito abaixo:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

\u00a0<\/span>Segunda lei de Kirchhoff ou lei das malhas:\u00a0\u201c<\/b><\/span><\/span><\/span>Percorrendo uma malha num mesmo sentido<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>\u00e9 nula a soma alg\u00e9brica das tens\u00f5es encontradas em cada elemento do circuito\u201d<\/b><\/span><\/span><\/span><\/p>\n

Baseado na\u00a0<\/b><\/span><\/span><\/span>lei acima, escolhendo um sentido de percurso para cada malha<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>partindo de um ponto, retornando ao mesmo ponto e igualando esta soma alg\u00e9brica de tens\u00f5es a zero, voc\u00ea pode estabelecer as seguintes conven\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Conven\u00e7\u00e3o para o sinal de E:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

\u00a0<\/span>Conven\u00e7\u00e3o para o sinal de R.i<\/b><\/span><\/span><\/span><\/p>\n

Quando o\u00a0<\/b><\/span><\/span><\/span>sentido da corrente i coincidir com o sentido do percurso escolhido para a malha,<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>sinal de R.i ser\u00e1 positivo e quando o sentido da corrente i n\u00e3o coincidir com o sentido de percurso escolhido para a malha, o sinal R.i ser\u00e1 negativo.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe no\u00a0<\/b><\/span><\/span><\/span>ponto A que a corrente que chega i \u00e9 a soma das correntes que saem,<\/b><\/span><\/span><\/span>\u00a0ou seja<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 2 + 6\u00a0<\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>8 A.<\/b><\/span><\/span><\/span><\/p>\n

Percorrendo a malha I no sentido hor\u00e1rio a partir de A, retornando a A e igualando a zero\u00a0<\/b><\/span><\/span><\/span>\u00a0\u2013 8R \u2013 0,5.2 + 5 = 0\u00a0\u00a08R = 4\u00a0\u00a0<\/b><\/span><\/span><\/span>R = 0,5 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Percorrendo a malha II no sentido hor\u00e1rio a partir de A, retornando a A e igualando a zero\u00a0\u00a0<\/b><\/span><\/span><\/span>+ 6.3 \u2013U + 0,5.8 = 0\u00a0\u00a0<\/b><\/span><\/span><\/span>U = 22 V.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Uma\u00a0<\/b><\/span><\/span><\/span>situa\u00e7\u00e3o pr\u00e1tica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>bastante comum<\/b><\/span><\/span><\/span>\u00a0nas\u00a0<\/b><\/span><\/span><\/span>resid\u00eancias<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>nos hot\u00e9is<\/b><\/span><\/span><\/span>\u00a0\u00e9 o chamado<\/b><\/span><\/span><\/span><\/p>\n

\u201cinterruptor paralelo\u201d,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>onde,\u00a0<\/b><\/span><\/span><\/span>na\u00a0realidade a liga\u00e7\u00e3o \u00e9 em s\u00e9rie<\/b><\/span><\/span><\/span>,\u00a0na qual\u00a0<\/b><\/span><\/span><\/span>\u00e9\u00a0poss\u00edvel ligar ou desligar uma determinada l\u00e2mpada, de forma independente, estando no ponto mais alto ou mais baixo de uma escada, da maneira indicada nas sequ\u00eancias abaixo.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

A ideia principal nesse problema \u00e9 lembrar que\u00a0<\/b><\/span><\/span><\/span>o capacitor funciona como um circuito aberto em corrente cont\u00ednua<\/b><\/span><\/span><\/span>, assim\u00a0<\/b><\/span><\/span><\/span>n\u00e3o haver\u00e1 passagem de corrente\u00a0<\/b><\/span><\/span><\/span>na malha em que o capacitor se encontra,\u00a0<\/b><\/span><\/span><\/span>inutilizando o resistor que se encontra ali<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>A corrente I que flui pelo circuito \u00e9 ent\u00e3o apenas a que passa na malha em que se encontra a fonte U=30 V<\/b><\/span><\/span><\/span>\u00a0e a resist\u00eancia restante R, o circuito equivalente com essas modifica\u00e7\u00f5es \u00e9 exibido na imagem abaixo.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica \u2013 vestibulares recentes \u2013 Eletrodin\u00e2mica \u2013 2018\/2019 01- 02- 03- C\u00e1lculo da pot\u00eancia da bateria\u00a0\u00a0Po\u00a0= i.U = 2.5\u00a0\u00a0Po\u00a0= 10 W. A\u00a0energia W utilizada no processo do carregamento foi de 90% de 105\u00a0j = 0,9.105\u00a0\u00a0W = 9.104\u00a0J. Po\u00a0= W\/t\u00a0\u00a010 = 9.104\/t\u00a0\u00a0t = 9.104\u00a0\/10\u00a0\u00a0t\u00a0= 9.103\u00a0= 9000s\/3,6\u00a0= 2,5 h. R- B 04- Toda bateria possui um polo<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4848,"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-4850","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4850","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=4850"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4850\/revisions"}],"predecessor-version":[{"id":4851,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4850\/revisions\/4851"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4848"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}