{"id":3304,"date":"2017-10-01T13:33:49","date_gmt":"2017-10-01T13:33:49","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3304"},"modified":"2024-09-04T14:00:21","modified_gmt":"2024-09-04T14:00:21","slug":"resolucao-comentada-eletrodinamica-2017-2016-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletrodinamica-2017-2016-2015\/resolucao-comentada-eletrodinamica-2017-2016-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada Eletrodin\u00e2mica – 2017 \u2013 2016 \u2013 2015"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada <\/span><\/span><\/span><\/p>\n

Eletrodin\u00e2mica – 2017 \u2013 2016 \u2013 2015.<\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

\u00a0<\/span>Assim como no c\u00e1lculo de\u00a0<\/b><\/span><\/span><\/span>\u0394S em um gr\u00e1fico V x \u0394t<\/b><\/span><\/span><\/span>, nossa carga <\/b><\/span><\/span><\/span>tamb\u00e9m pode ser calculada pela \u00e1rea do gr\u00e1fico<\/b><\/span><\/span><\/span>, pois como\u00a0<\/b><\/span><\/span><\/span>a varia\u00e7\u00e3o da dist\u00e2ncia \u00e9 a multiplica\u00e7\u00e3o entre tempo e velocidade, a carga \u00e9 a multiplica\u00e7\u00e3o entre intensidade da corrente e o tempo<\/b><\/span><\/span><\/span>, logo:<\/b><\/span><\/span><\/span>\"\"<\/p>\n

\"\"<\/p>\n

N\u00e3o podemos esquecer dos <\/b><\/span><\/span><\/span>coeficientes que multiplicam as unidades<\/b><\/span><\/span><\/span>, na hora de substituir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora que temos a carga,\u00a0<\/b><\/span><\/span><\/span>podemos fazer uma regra de tr\u00eas e achar quantidade de \u00edons<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

1 \u00edon -> 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>\u00a0C<\/b><\/span><\/span><\/span><\/p>\n

x \u00edons -> 4.10<\/b><\/span><\/span><\/span>-16<\/b><\/span><\/span><\/sup><\/span>\u00a0C<\/b><\/span><\/span><\/span><\/p>\n

Multiplicando em cruz:<\/b><\/span><\/span><\/span><\/p>\n

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

Isolando o x:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"Como, pelo\u00a0<\/b><\/span><\/span>enunciado, X estava eletrizada com carga de 5,0.10<\/b><\/span><\/span>-12<\/b><\/span>\u00a0C<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ela transferiu<\/b><\/span><\/span>\u00a0toda sua carga\u00a0<\/b><\/span><\/span>para Y,<\/b><\/span><\/span>\u00a0ficando\u00a0<\/b><\/span><\/span>X neutra (carga zero).<\/b><\/span><\/span><\/p>\n

Acontece que\u00a0<\/b><\/span><\/span>todo corpo ligado \u00e0 Terra fica totalmente descarregado <\/b><\/span><\/span>transferindo integralmente\u00a0<\/b><\/span><\/span>toda sua carga (5,0.10<\/b><\/span><\/span>-12<\/b><\/sup><\/span>\u00a0C) para a Terra. <\/b><\/span><\/span>Assim\u00a0<\/b><\/span><\/span>Y tamb\u00e9m ficar\u00e1 neutro com carga zero.<\/b><\/span><\/span><\/p>\n

Portanto, no\u00a0<\/b><\/span><\/span>final desse processo,<\/b><\/span><\/span>\u00a0as\u00a0<\/b><\/span><\/span>cargas el\u00e9tricas totais dos objetos X e Y ser\u00e3o, respectivamente, zero e zero.<\/b><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Os f\u00edsicos,\u00a0<\/b><\/span><\/span>Murray Gell-Mann e George\u00a0Zweig\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>desenvolveram, independentemente, um modelo para classificar as\u00a0<\/b><\/span><\/span>Part\u00edculas Elementares<\/b><\/span><\/span>.<\/b><\/span><\/span>\u00a0Segundo esse modelo (<\/b><\/span><\/span>hoje conhecido como\u00a0Modelo dos<\/b><\/span><\/span>Quarks<\/b><\/span><\/span>), os\u00a0<\/b><\/span><\/span>h\u00e1drons\u00a0<\/b><\/span><\/span>(b\u00e1rions,\u00a0m\u00e9sons e outras part\u00edculas subat\u00f4micas)<\/b><\/span><\/span>formavam os\u00a0<\/b><\/span><\/span>quarks,<\/b><\/span><\/span>\u00a0que por sua vez eram componentes dos\u00a0<\/b><\/span><\/span>pr\u00f3tons e n\u00eautrons.<\/b><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Veja a\u00a0<\/b><\/span><\/span>express\u00e3o abaixo\u00a0<\/b><\/span><\/span>que relaciona\u00a0<\/b><\/span><\/span>pot\u00eancia el\u00e9trica (Po) com energia el\u00e9trica (W):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Usa-se tamb\u00e9m como\u00a0<\/b><\/span><\/span>unidade de medida o kWh\u00a0<\/b><\/span><\/span>onde a\u00a0<\/b><\/span><\/span>pot\u00eancia deve estar em kW = 1000W<\/b><\/span><\/span>\u00a0e o\u00a0<\/b><\/span><\/span>tempo em horas (h).<\/b><\/span><\/span><\/p>\n

Lembrando que\u00a0\"\"<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>ano<\/b><\/span><\/span>\u00a0tem\u00a0<\/b><\/span><\/span>365 dias\u00a0<\/b><\/span><\/span>e que s\u00e3o\u00a0<\/b><\/span><\/span>10 pain\u00e9is que funcionam durante 6 horas por dia, temos:<\/b><\/span><\/span><\/p>\n

Por dia\u00a0<\/b><\/span><\/span>e por\u00a0<\/b><\/span><\/span>cada painel,\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Por dia<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0(10 pain\u00e9is)<\/b><\/span><\/span><\/p>\n

No ano, com 365 dias,\u00a0<\/b><\/span><\/span>temos, por fim\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span><\/p>\n

Como o\u00a0<\/b><\/span><\/span>exerc\u00edcio<\/b><\/span><\/span>\u00a0pede em\u00a0<\/b><\/span><\/span>kWh\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

Diferen\u00e7a di\u00e1ria \"\" <\/b><\/span><\/span>d = (0,21 \u2013 0,0189)\u00a0\"\" <\/b><\/span><\/span>d = C$ 0,1911\u00a0\"\" <\/b><\/span><\/span>descontando dos <\/b><\/span><\/span>C$ 60,00 os 5,00\u00a0<\/b><\/span><\/span>que ele\u00a0<\/b><\/span><\/span>pagou pela l\u00e2mpada incandescente, sobram C$ 55,00<\/b><\/span><\/span>, que\u00a0<\/b><\/span><\/span>divididos<\/b><\/span><\/span>\u00a0pela\u00a0<\/b><\/span><\/span>diferen\u00e7a di\u00e1ria de C$0,1011 fornecem 55\/0,1911= 287,8 \u2248 288 dias consecutivos.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

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

Pot\u00eancia =\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>W = Po.\u2206t\u00a0<\/b><\/span><\/span>= 20x24hx3600s\u00a0\"\"\u00a0<\/b><\/span><\/span>Po<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 1 728 000 =\u00a0<\/b><\/span><\/span>1,728.106\u00a0J.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

10- <\/span><\/span><\/span>Para\u00a0<\/b><\/span><\/span><\/span>calcular<\/b><\/span><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span><\/span>pot\u00eancia<\/b><\/span><\/span><\/span>\u00a0m\u00e1xima de\u00a0<\/b><\/span><\/span><\/span>sa\u00edda que o carregador pode fornecer<\/b><\/span><\/span><\/span>\u00a0voc\u00ea deve utilizar a\u00a0<\/b><\/span><\/span><\/span>tens\u00e3o de sa\u00edda U = 5 V e a corrente de sa\u00edda i = 1,3 A,\u00a0<\/b><\/span><\/span><\/span>fornecidas nas\u00a0<\/b><\/span><\/span><\/span>especifica\u00e7\u00f5es utilizando a express\u00e3o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P\u00a0= i.U<\/b><\/span><\/span><\/span>\u00a0= 1,3.5\u00a0<\/b><\/span><\/span><\/span>= 6,5 W.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>carga m\u00e1xima armazenada na bateria<\/b><\/span><\/span><\/span>\u00a0tamb\u00e9m \u00e9\u00a0<\/b><\/span><\/span><\/span>fornecida<\/b><\/span><\/span><\/span>\u00a0pelas<\/b><\/span><\/span><\/span>especifica\u00e7\u00f5es da bateria Q = 1650<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mAh\u00a0<\/b><\/span><\/span><\/span>que, transformado em\u00a0<\/b><\/span><\/span><\/span>coulomb (C)\u00a0<\/b><\/span><\/span><\/span>fornece Q = 1650.<\/b><\/span><\/span><\/span>10-3.\"\".3600s\u00a0\"\"\u00a0Q = 5940 C.<\/b><\/span><\/span><\/span><\/p>\n

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

11- <\/span><\/span><\/span>Observe na\u00a0<\/b><\/span><\/span>express\u00e3o P=U2\/R<\/b><\/span><\/span>\u00a0que, sendo a\u00a0<\/b><\/span><\/span>resist\u00eancia el\u00e9trica R constante<\/b><\/span><\/span>, a\u00a0<\/b><\/span><\/span>pot\u00eancia \u00e9 diretamente<\/b><\/span><\/span>\u00a0proporcional ao\u00a0<\/b><\/span><\/span>quadrado da voltagem U<\/b><\/span><\/span>. Assim, se ele\u00a0<\/b><\/span><\/span>for ligad<\/b><\/span><\/span>o a uma\u00a0<\/b><\/span><\/span>voltagem duas vezes menor (220 V para 110 V), a pot\u00eancia ficar\u00e1 22\u00a0= 4 vezes menor.<\/b><\/span><\/span><\/p>\n

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

12- <\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P = W\/\u2206t<\/b><\/span><\/span>\u2014\u00a0<\/b><\/span><\/span>l\u00e2mpada<\/b><\/span><\/span>\u00a0\u2014 60 = Wl\/3600 \u2014\u00a0<\/b><\/span><\/span>Wl\u00a0= 216 000 J<\/b><\/span><\/span><\/p>\n

P = W\/\u2206t<\/b><\/span><\/span>\u2014\u00a0<\/b><\/span><\/span>televis\u00e3o<\/b><\/span><\/span>\u00a0\u2014 90 = Wt\/3600 \u2014\u00a0<\/b><\/span><\/span>Wt\u00a0= 324 000 J<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>televis\u00e3o<\/b><\/span><\/span>\u00a0consome\u00a0<\/b><\/span><\/span>mais energia el\u00e9trica,<\/b><\/span><\/span>\u00a0mas a\u00a0<\/b><\/span><\/span>tens\u00e3o\u00a0<\/b><\/span><\/span>em cada aparelho\u00a0<\/b><\/span><\/span>\u00e9 a mesma<\/b><\/span><\/span>, pois est\u00e3o\u00a0<\/b><\/span><\/span>associados em paralelo<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

13-<\/b><\/span><\/span><\/span> Veja abaixo a express\u00e3o matem\u00e1tica da segunda lei de Ohm:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A resist\u00eancia el\u00e9trica\u00a0<\/b><\/span><\/span>R\u00a0<\/b><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span>diretamente proporcional ao comprimento l do fio e \u00e0 resistividade \u03c1<\/b><\/span><\/span>\u00a0(caracter\u00edstica do material que constitui o fio) e\u00a0<\/b><\/span><\/span>inversamente proporcional \u00e0 \u00e1rea se se\u00e7\u00e3o reta transversal (S<\/b><\/span><\/span>) \u201cgrossura do fio\u201d.<\/b><\/span><\/span><\/p>\n

Assim, se voc\u00ea\u00a0<\/b><\/span><\/span>dobrar o comprimento do fio estar\u00e1 dobrando o valor da resist\u00eancia el\u00e9trica R<\/b><\/span><\/span>\u00a0do mesmo.<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0= R e R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0= 2R<\/b><\/span><\/span><\/p>\n

Corrente i<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0no fio de\u00a0<\/b><\/span><\/span>comprimento L <\/b><\/span><\/span>submetido \u00e0 voltagem\u00a0<\/b><\/span><\/span>U <\/b><\/span><\/span>\u2014 R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0= U\/i<\/b><\/span><\/span>1 <\/b><\/span><\/sub><\/span>\u00a0\u2014 <\/b><\/span><\/span>i<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0= U\/R. (I)<\/b><\/span><\/span><\/p>\n

Corrente i<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0no fio de\u00a0<\/b><\/span><\/span>comprimento 2L <\/b><\/span><\/span>submetido \u00e0 voltagem\u00a0<\/b><\/span><\/span>U <\/b><\/span><\/span>\u2014 R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0= U\/i<\/b><\/span><\/span>2 <\/b><\/span><\/sub><\/span>\u2014 \u00a0<\/b><\/span><\/span>i<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0= U\/2R. (II)<\/b><\/span><\/span><\/p>\n

Comparando (I) com (II) \u2014\u00a0<\/b><\/span><\/span>i<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0=2i<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/p>\n

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

14- <\/span><\/span><\/span>Segunda lei de Ohm\u00a0\"\"\u00a0R = \u03c1L\/A.<\/b><\/span><\/span><\/p>\n

Resistor 1\u00a0\"\"\u00a0R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0= \u03c12L\/A\u00a0\"\"\u00a0resistor 2 R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0= \u03c1L\/2A.<\/b><\/span><\/span><\/p>\n

Resistor equivalente Req\u00a0= R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0= \u03c12L\/A + \u03c1L\/2A = 5\u03c1L\/2A.<\/b><\/span><\/span><\/p>\n

Req\u00a0= V\/i \"\" i = V\/Req\u00a0= V\/5\u03c1L\/2A \"\"i = 2AV\/5\u03c1L.<\/b><\/span><\/span><\/p>\n

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

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

a)<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span>, em\u00a0<\/b><\/span><\/span>\u03a9<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observe a\u00a0<\/b><\/span><\/span>sequ\u00eancia abaixo<\/b><\/span><\/span>\u00a0para o\u00a0<\/b><\/span><\/span>c\u00e1lculo da resist\u00eancia equivalente Req\u00a0no circuito da figura 2, onde os tr\u00eas resistores est\u00e3o em paralelo<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>pot\u00eancia total dissipada<\/b><\/span><\/span>, em\u00a0<\/b><\/span><\/span>W.<\/b><\/span><\/span><\/p>\n

Voc\u00ea pode calcular a\u00a0<\/b><\/span><\/span>pot\u00eancia total dissipada<\/b><\/span><\/span>\u00a0utilizando a\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span>\u00a0que est\u00e1 submetida \u00e0 uma\u00a0<\/b><\/span><\/span>tens\u00e3o de U = 12 V.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

16- <\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>Com o\u00a0<\/b><\/span><\/span>ohm\u00edmetro conectado entre A e B <\/b><\/span><\/span>voc\u00ea deve calcular a <\/b><\/span><\/span>resist\u00eancia<\/b><\/span><\/span>\u00a0do resistor\u00a0<\/b><\/span><\/span>equivalente R<\/b><\/span><\/span>AB<\/b><\/span><\/sub><\/span> entre esses dois pontos.<\/b><\/span><\/span><\/p>\n

Assim, o circuito\u00a0<\/b><\/span><\/span>ficar\u00e1 conforme a figura (1),<\/b><\/span><\/span>\u00a0onde os\u00a0<\/b><\/span><\/span>dois resistores de 10 k\u2126 est\u00e3o em s\u00e9rie <\/b><\/span><\/span>podendo ser\u00a0<\/b><\/span><\/span>substituidos por um \u00fanico resistor de<\/b><\/span><\/span>(10 k\u2126 + 10 k\u2126)\u00a0<\/b><\/span><\/span>= 20 k\u2126.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Com o\u00a0<\/b><\/span><\/span>ohm\u00edmetro conectado entre B e C<\/b><\/span><\/span>\u00a0voc\u00ea deve calcular a <\/b><\/span><\/span>resist\u00eancia<\/b><\/span><\/span>\u00a0do resistor\u00a0<\/b><\/span><\/span>equivalente R<\/b><\/span><\/span>BC<\/b><\/span><\/sub><\/span> entre esses dois pontos.<\/b><\/span><\/span><\/p>\n

Observe na sequ\u00eancia abaixo o c\u00e1lculo da resist\u00eancia entre B e C, R<\/b><\/span><\/span>BC<\/b><\/span><\/sub><\/span> .<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

17- <\/span><\/span><\/span>Calculando a\u00a0<\/b><\/span><\/span>resist\u00eancia\u00a0<\/b><\/span><\/span>do resistor\u00a0<\/b><\/span><\/span>equivalente\u00a0<\/b><\/span><\/span>entre os pontos\u00a0<\/b><\/span><\/span>A e B\u00a0<\/b><\/span><\/span>da figura que mostra os\u00a0<\/b><\/span><\/span>trechos pelos quais n\u00e3o circula corrente el\u00e9trica\u00a0<\/b><\/span><\/span>(circuito em aberto):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, a\u00a0<\/b><\/span><\/span>intensidade de corrente el\u00e9trica\u00a0<\/b><\/span><\/span>que ir\u00e1\u00a0<\/b><\/span><\/span>aquecer a \u00e1gua do aqu\u00e1rio \u00e9 de i = 10 A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"Agora vamos calcular a quantidade de\u00a0<\/b><\/span><\/span>energia t\u00e9rmica (calor)W\u00a0<\/b><\/span><\/span>que ir\u00e1 aquecer a massa de \u00e1gua do aqu\u00e1rio<\/b><\/span><\/span>m = 120000 g<\/b><\/span><\/span>, de\u00a0<\/b><\/span><\/span>calor espec\u00edfico c= 1g.cm-3.oC-<\/b><\/span><\/span>1\u00a0de modo que sofra uma <\/b><\/span><\/span>varia\u00e7\u00e3o de temperatura \u2206\u019f<\/b><\/span><\/span>\u00a0= 36oF\u00a0<\/b><\/span><\/span>= 20oC.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

18- <\/b><\/span><\/span><\/span>C\u00e1lculo da\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente e da fem do gerador com a chave C aberta:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa fem \u03b5 = 15R \u00e9\u00a0<\/b><\/span><\/span>constante<\/b><\/span><\/span>, pois o gerador \u00e9 o mesmo:<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>sequ\u00eancia\u00a0<\/b><\/span><\/span>de figuras abaixo mostra o\u00a0<\/b><\/span><\/span>c\u00e1lculo da resist\u00eancia do resistor equivalente<\/b><\/span><\/span>\u00a0com a chave\u00a0<\/b><\/span><\/span>C fechada<\/b><\/span><\/span>\u00a0e retirando o amper\u00edmetro ideal (resist\u00eancia interna nula):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observe\u00a0<\/b><\/span><\/span>atentamente na figura abaixo<\/b><\/span><\/span>\u00a0que a corrente el\u00e9trica\u00a0<\/b><\/span><\/span>indicada pelo amper\u00edmetro \u00e9 i<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0= 3 A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

19- <\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Para que as\u00a0<\/b><\/span><\/span>tomadas,\u00a0<\/b><\/span><\/span>como pede o enunciado<\/b><\/span><\/span>\u00a0estejam submetidas \u00e0 mesma diferen\u00e7a de potencial<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(ddp), tens\u00e3o ou voltagem U,<\/b><\/span><\/span>\u00a0essas duas tomadas\u00a0<\/b><\/span><\/span>devem estar ligadas em paralelo j\u00e1 que nesse tipo de associa\u00e7\u00e3o a tens\u00e3o \u00e9 a mesma que a tens\u00e3o nominal da rede el\u00e9trica (U).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Mas, o\u00a0<\/b><\/span><\/span>interruptor deve estar ligado em s\u00e9rie com a l\u00e2mpada<\/b><\/span><\/span>\u00a0e, quando<\/b><\/span><\/span>fechado ele acende a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>l\u00e2mpada<\/b><\/span><\/span>\u00a0que fica em\u00a0<\/b><\/span><\/span>paralelo com as tomadas\u00a0<\/b><\/span><\/span>e, portanto,\u00a0<\/b><\/span><\/span>submetida \u00e0 tens\u00e3o nominal (U) da rede que as tomadas.<\/b><\/span><\/span><\/p>\n

Desligando o interruptor<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>desliga-se a l\u00e2mpada,<\/b><\/span><\/span>\u00a0sem alterar a\u00a0<\/b><\/span><\/span>liga\u00e7\u00e3o paralelo das tomadas<\/b><\/span><\/span>\u00a0com a rede, que\u00a0<\/b><\/span><\/span>continuam sujeitas \u00e0 mesma ddp nominal U, da rede.<\/b><\/span><\/span><\/p>\n

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

20- <\/span><\/span><\/span>No<\/b><\/span><\/span><\/span>\u00a0chuveiro el\u00e9trico, podemos alterar a resist\u00eancia el\u00e9trica equivalente\u00a0<\/b><\/span><\/span><\/span>atrav\u00e9s da<\/b><\/span><\/span><\/span>\u00a0varia\u00e7\u00e3o do comprimento do resistor<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

21- <\/span><\/span><\/span>a) C\u00e1lculo da\u00a0<\/b><\/span><\/span>intensidade de corrente el\u00e9trica i\u00a0<\/b><\/span><\/span>utilizando os\u00a0<\/b><\/span><\/span>valores nominais\u00a0<\/b><\/span><\/span>fornecidos:\u00a0<\/b><\/span><\/span>P = 25 W\u00a0<\/b><\/span><\/span>e\u00a0<\/b><\/span><\/span>U = 40 V\u00a0\"\"\u00a0P = i.U\u00a0\"\"<\/b><\/span><\/span>\u00a025 = i.40\u00a0\"\"<\/b><\/span><\/span>\u00a0i\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>= 0,625 A.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>carga el\u00e9trica \u2206Q\u00a0<\/b><\/span><\/span>no intervalo de tempo\u00a0<\/b><\/span><\/span>\u2206t =\u00a0<\/b><\/span><\/span>20 minx60\u00a0<\/b><\/span><\/span>= 1200 s\u00a0<\/b><\/span><\/span>pode ser determinada por\u00a0<\/b><\/span><\/span>i =\u00a0\"\"\"\"\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>0,625 =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

\"\"R\u2019 + 16 = 96<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>R\u2019 = 80 \u2126.<\/b><\/span><\/span><\/p>\n

Ent\u00e3o, voc\u00ea\u00a0<\/b><\/span><\/span>deve associar em paralelo n resistores de 320\u2126 de maneira a obter um \u00fanico resistor de 80 \u2126\u00a0\"\"<\/b><\/span><\/span>\u00a0n =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>n = 4.<\/b><\/span><\/span><\/span><\/p>\n

22- <\/span><\/span><\/span>Trata-se de uma\u00a0<\/b><\/span><\/span>associa\u00e7\u00e3o de resistores<\/b><\/span><\/span>\u00a0(l\u00e2mpadas)\u00a0<\/b><\/span><\/span>associados em paralelo <\/b><\/span><\/span>e, cujas<\/b><\/span><\/span>\u00a0caracter\u00edsticas s\u00e3o\u00a0<\/b><\/span><\/span>fornecidas a seguir:<\/b><\/span><\/span><\/p>\n

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

\u00a0\"\"\u00a0\u00a0<\/span>Os\u00a0resistores s\u00e3o associados pelos seus terminais, ou seja,\u00a0todos saem de um mesmo ponto e todos chegam a um mesmo ponto.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o) U<\/b><\/span><\/span>\u00a0de\u00a0toda a associa\u00e7\u00e3o<\/b><\/span><\/span>\u00a0(entre A e B) \u00e9 a mesma para todos os resistores<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/p>\n

Observe que\u00a0<\/b><\/span><\/span>a intensidade de corrente i<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> <\/b>(corrente total) que\u00a0<\/b><\/span><\/span>sai do polo positivo<\/b><\/span><\/span>\u00a0da bateria deve ser a\u00a0<\/b><\/span><\/span>mesma que i<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span>\u00a0que chega ao polo negativo <\/b><\/span><\/span>da mesma bateria.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como, pelo\u00a0<\/b><\/span><\/span>enunciado, as l\u00e2mpadas s\u00e3o id\u00eanticas<\/b><\/span><\/span>\u00a0elas possuem a\u00a0<\/b><\/span><\/span>mesma resist\u00eancia<\/b><\/span><\/span>\u00a0e, consequentemente\u00a0<\/b><\/span><\/span>s\u00e3o percorridas pela mesma corrente el\u00e9trica, ou seja, i<\/b><\/span><\/span>C<\/b><\/span><\/sub><\/span>\u00a0= i<\/b><\/span><\/span>D<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

Observe que a alternativa correta \u00e9 a C, cuja an\u00e1lise \u00e9 feita a seguir:<\/b><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>sequ\u00eancia abaixo\u00a0<\/b><\/span><\/span>\u00e9 fornecido o\u00a0<\/b><\/span><\/span>c\u00e1lculo da corrente el\u00e9trica total i.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Observe que o\u00a0<\/b><\/span><\/span>circuito de n\u00famero 2 possui menor resist\u00eancia el\u00e9trica equivalente<\/b><\/span><\/span>, portanto ser\u00e1\u00a0<\/b><\/span><\/span>percorrido por maior corrente el\u00e9trica.<\/b><\/span><\/span><\/p>\n

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

26- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>C\u00e1lculo da resist\u00eancia equivalente R<\/b><\/span><\/span>eqA<\/b><\/span>\u00a0quando a chave est\u00e1\u00a0<\/b><\/span><\/span>ligada <\/b><\/span><\/span>no\u00a0<\/b><\/span><\/span>ponto A<\/b><\/span><\/span>\u00a0e a corrente<\/b><\/span><\/span><\/p>\n

\"\"el\u00e9trica passando apenas pelos dois resistores que est\u00e3o em paralelo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b)\u00a0<\/b><\/span><\/span>Se a<\/b><\/span><\/span>\u00a0densidade da \u00e1gua \u00e9 d = 1 kg\/L,\u00a0<\/b><\/span><\/span>isso significa que o\u00a0<\/b><\/span><\/span>volume de 1 L de \u00e1gua possui massa de m = 1 kg.<\/b><\/span><\/span><\/p>\n

Assim, o<\/b><\/span><\/span>\u00a0volume de 4L\u00a0<\/b><\/span><\/span>de \u00e1gua equivale \u00e0<\/b><\/span><\/span>\u00a0massa de m = 4 kg.<\/b><\/span><\/span><\/p>\n

Da equa\u00e7\u00e3o fundamental da calorimetria\u00a0\"\"\u00a0Q = m.c.\u2206\u03b8 = 4.1.(80 \u2013 20) = 4.60\u00a0\"\"\u00a0Q = W = 240 kcal.<\/b><\/span><\/span><\/p>\n

Com a chave em A\u00a0\"\"\u00a0P<\/b><\/span><\/span>A<\/b><\/span>\u00a0=\u00a0\"\"=\u00a0\"\"<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0P<\/b><\/span><\/span>A<\/b><\/span>\u00a0= 400 W.<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>A<\/b><\/span>\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>2.40 =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a0R =<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>R =<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>18 \u2126<\/b><\/span><\/span><\/p>\n

27- <\/b><\/span><\/span><\/span>Sendo o\u00a0<\/b><\/span><\/span>volt\u00edmetro ideal\u00a0<\/b><\/span><\/span>ele possui\u00a0<\/b><\/span><\/span>resist\u00eancia interna infinita <\/b><\/span><\/span>(isolante), por ele n\u00e3o circula corrente el\u00e9trica, ent\u00e3o voc\u00ea pode\u00a0<\/b><\/span><\/span>retir\u00e1-lo do circuito,\u00a0<\/b><\/span><\/span>que fica conforme a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span>sequ\u00eancia a seguir\u00a0<\/b><\/span><\/span>est\u00e1 sendo calculada a resist\u00eancia el\u00e9trica do <\/b><\/span><\/span>resistor equivalente<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Veja na\u00a0<\/b><\/span><\/span>figura<\/b><\/span><\/span>\u00a0abaixo que a\u00a0<\/b><\/span><\/span>corrente el\u00e9trica\u00a0<\/b><\/span><\/span>que circula pelo<\/b><\/span><\/span>amper\u00edmetro\u00a0<\/b><\/span><\/span>(<\/b><\/span><\/span>indica\u00e7\u00e3o do mesmo<\/b><\/span><\/span>) \u00e9 de\u00a0<\/b><\/span><\/span>2,5A\u00a0<\/b><\/span><\/span>e veja tamb\u00e9m as\u00a0<\/b><\/span><\/span>correntes <\/b><\/span><\/span>que circulam por\u00a0<\/b><\/span><\/span>cada resistor:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o, voltagem)\u00a0<\/b><\/span><\/span>entre\u00a0<\/b><\/span><\/span>A e B<\/b><\/span><\/span>\u00a0\u00e9 fornecida por<\/b><\/span><\/span>UAB=R.i<\/b><\/span><\/span>=0,3.1,25=<\/b><\/span><\/span>0,375V.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o, voltagem)\u00a0<\/b><\/span><\/span>entre\u00a0<\/b><\/span><\/span>A e C<\/b><\/span><\/span>\u00a0\u00e9 fornecida por <\/b><\/span><\/span>U<\/b><\/span><\/span>AC <\/b><\/span><\/sub><\/span>= R.i <\/b><\/span><\/span>= 0,6.1,25 = <\/b><\/span><\/span>0,75V.<\/b><\/span><\/span><\/p>\n

U<\/b><\/span><\/span>AB<\/b><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span> —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span>\u00a0= 0,375V — <\/b><\/span><\/span>\u00a0U<\/b><\/span><\/span>AC<\/b><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0\u2013V<\/b><\/span><\/span>C <\/b><\/span><\/sub><\/span>—<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>VA\u00a0\u2013 VC\u00a0= 0,75V.<\/b><\/span><\/span><\/p>\n

UAC\u00a0\u2013 UAB<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0,75 \u2013 0,375\u00a0<\/b><\/span><\/span>= 0,375V — \u00a0<\/b><\/span><\/span>(V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0\u2013 VC) \u2013 (V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u00a0\u2013 VB) = 0,375 =\u00a0<\/b><\/span><\/span>VA<\/b><\/span><\/span>\u00a0\u2013 VC<\/b><\/span><\/span>\u2013 VA<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>+\u00a0VB —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span> <\/b>0,375V.<\/b><\/span><\/span><\/p>\n

Observe na figura que o\u00a0<\/b><\/span><\/span>volt\u00edmetro<\/b><\/span><\/span>\u00a0est\u00e1 inserido entre os pontos\u00a0<\/b><\/span><\/span>B e C <\/b><\/span><\/span>ent\u00e3o ele\u00a0<\/b><\/span><\/span>indicar\u00e1 a ddp<\/b><\/span><\/span>\u00a0entre esses dois pontos, ou seja, indicar\u00e1 <\/b><\/span><\/span>U<\/b><\/span><\/span>BC <\/b><\/span><\/sub><\/span>= 0,375V.<\/b><\/span><\/span><\/p>\n

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

28- <\/b><\/span><\/span><\/span>Observe que o dispositivo\u00a0<\/b><\/span><\/span>de 8V (maior valor) \u00e9 o gerador\u00a0<\/b><\/span><\/span>(a corrente sai de seu polo positivo) e o de\u00a0<\/b><\/span><\/span>6 V \u00e9 o receptor\u00a0<\/b><\/span><\/span>(a corrente sai de seu polo negativo)\u00a0\"\"\u00a0<\/b><\/span><\/span>U\u00a0<\/b><\/span><\/span>= 8 \u2013 6\u00a0<\/b><\/span><\/span>= 2V.<\/b><\/span><\/span><\/p>\n

Os\u00a0<\/b><\/span><\/span>resistores est\u00e3o associados em s\u00e9rie\u00a0\"\"\u00a0Req<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 3 + 3 + 4\u00a0<\/b><\/span><\/span>= 10 \u2126.<\/b><\/span><\/span><\/p>\n

Req\u00a0=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>10 =<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>i = 0,2 A (corrente em cada resistor)\u00a0\"\"\u00a0P4\u2126\u00a0= R.i2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 4.(0,2)2\u00a0= 4.0,04\u00a0\"\"<\/b><\/span><\/span><\/p>\n

P4\u2126\u00a0= 0,16 W.<\/b><\/span><\/span><\/p>\n

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

29- <\/b><\/span><\/span><\/span>Vamos come\u00e7ar o exerc\u00edcio\u00a0<\/b><\/span><\/span>analisando o valor de corrente dado no enunciado<\/b><\/span><\/span>, atrav\u00e9s da f\u00f3rmula da resist\u00eancia:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aonde: R \u00e9 a resist\u00eancia el\u00e9trica = 10k\u2126.<\/b><\/span><\/span><\/p>\n

U \u00e9 a tens\u00e3o da corrente (ddp).<\/b><\/span><\/span><\/p>\n

i \u00e9 a corrente el\u00e9trica = 1 mA.<\/b><\/span><\/span><\/p>\n

Isolando o U:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

U = 10 V<\/b><\/span><\/span><\/p>\n

Esse resultado \u00e9 importante para nos mostrar que\u00a0<\/b><\/span><\/span>os outros dois resistores n\u00e3o podem estar em s\u00e9rie<\/b><\/span><\/span>, visto que se estivessem\u00a0<\/b><\/span><\/span>cada um teria uma tens\u00e3o de 10 V, somados teriam uma tens\u00e3o geral de 30 V, maior que a do enunciado<\/b><\/span><\/span>. Como os resistores n\u00e3o podem estar em s\u00e9rie<\/b><\/span><\/span>eles s\u00f3 podem estar em paralelo<\/b><\/span><\/span>. Formando um circuito exemplificado na imagem abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/b><\/span><\/span>\u00a0resist\u00eancia total vai ser a soma do resistor R1 com a soma da resist\u00eancia em s\u00e9rie dos outros dois resistores<\/b><\/span><\/span>, vamos calcular essa resist\u00eancia em s\u00e9rie pela f\u00f3rmula abaixo:<\/b><\/span><\/span><\/p>\n

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

Substituindo os valores:<\/b><\/span><\/span><\/p>\n

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

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

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

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

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

Elevando ambos os lados por -1:<\/b><\/span><\/span><\/p>\n

R =\u00a0\"\"\u00a0\u2126<\/b><\/span><\/span><\/p>\n

Somando com o outro resistor para achar a resist\u00eancia total:<\/b><\/span><\/span><\/p>\n

R =\u00a0\"\"\u00a0+ 10.10\u00b3<\/b><\/span><\/span><\/p>\n

R =\u00a0\"\"\u00a0\u2126<\/b><\/span><\/span><\/p>\n

R =\u00a0\"\"\u2126<\/b><\/span><\/span><\/p>\n

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

30- <\/b><\/span><\/span><\/span>01)<\/b><\/span><\/span>\u00a0Os resistores equivalentes nos dois ramos do circuito possuem, respectivamente, resist\u00eancia el\u00e9trica de 30 \u2126 e de 5 \u2126.<\/b><\/span><\/span><\/p>\n

01. Correta<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>ramo A\u00a0\"\"\u00a0s\u00e9rie\u00a0\"\"\u00a0Req\u00a0= 10 + 10 + 10 = 30 \u2126\u00a0\"\"\u00a0ramo B\u00a0\"\"paralelo\u00a0\"\"\u00a0Req\u00a0= 10.10\/(10 + 10) = 100\/20\u00a0\"\"\u00a0Req\u00a0= 5 \u2126.<\/b><\/span><\/span><\/p>\n

02)<\/b><\/span><\/span>\u00a0A corrente el\u00e9trica que flui no ramo\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/i><\/span><\/span>\u00e9 de 2 A.<\/b><\/span><\/span><\/p>\n

02. Falsa\u00a0\"\"\u00a0Req\u00a0= U\/i\u00a0\"\"\u00a0<\/b><\/span><\/span>30\/7 = 10\/i\u00a0\"\"<\/b><\/span><\/span>\u00a0i = 2,33 A\u00a0\"\"\u00a0iA<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 10\/30<\/b><\/span><\/span>\u00a0= 0,33 A\u00a0\"\"\u00a0iB<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 10\/5<\/b><\/span><\/span>\u00a0= 2 A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Veja acima que<\/b><\/span><\/span>\u00a0iA\u00a0= 0,33 A.<\/b><\/span><\/span><\/p>\n

04)<\/b><\/span><\/span>\u00a0A pot\u00eancia dissipada no ramo\u00a0<\/b><\/span><\/span>B\u00a0<\/b><\/i><\/span><\/span>\u00e9 de 20 W.<\/b><\/span><\/span><\/p>\n

04. Correta\u00a0\"\"\u00a0P = R.iB2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 5.22<\/b><\/span><\/span>\u00a0= 20 W.<\/b><\/span><\/span><\/p>\n

08)<\/b><\/span><\/span>\u00a0A pot\u00eancia dissipada e a corrente el\u00e9trica no ramo\u00a0<\/b><\/span><\/span>B\u00a0<\/b><\/i><\/span><\/span>s\u00e3o maiores que as observadas no ramo\u00a0<\/b><\/span><\/span>A<\/b><\/i><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

08. Correta<\/b><\/span><\/span>\u00a0\"\"\u00a0de 04,\u00a0<\/b><\/span><\/span>PB\u00a0= 20W<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>PA<\/b><\/span><\/span>\u00a0= R.iA2\u00a0= 30.0,33\u00a0<\/b><\/span><\/span>= 9,9 W.<\/b><\/span><\/span><\/p>\n

16)<\/b><\/span><\/span>\u00a0A diferen\u00e7a de potencial entre os terminais do segundo resistor do ramo\u00a0<\/b><\/span><\/span>A\u00a0<\/b><\/i><\/span><\/span>\u00e9 de aproximadamente 3,3<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span><\/span><\/p>\n

16. Correta\u00a0\"\"\u00a0R = U\/iA\u00a0\"\"\u00a0<\/b><\/span><\/span>10 = U\/0,33<\/b><\/span><\/span>\u00a0\"\"\u00a0U = 3,3 V.<\/b><\/span><\/span><\/p>\n

R- (01 + 04 + 08 + 16) = 29<\/b><\/span><\/span><\/span><\/p>\n

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

Veja um resumo da teoria a seguir:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"\u00a0<\/span>Se\u00a0i=0<\/b><\/span><\/span>,\u00a0ou seja, se o gerador estiver em\u00a0<\/b><\/span><\/span>circuito aberto<\/b><\/span><\/span>, tem-se:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Se\u00a0<\/b><\/span><\/span>U=0<\/b><\/span><\/span>, ou seja, se o gerador estiver em\u00a0<\/b><\/span><\/span>curto circuito<\/b><\/span><\/span>, tem-se:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Os\u00a0<\/b><\/span><\/span>pares de valores U=E e icc=E\/r<\/b><\/span><\/span>\u00a0determinam\u00a0<\/b><\/span><\/span>dois pontos no gr\u00e1fico Uxi<\/b><\/span><\/span>, que unidos<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

por um segmento de reta (fun\u00e7\u00e3o do primeiro grau) fornece a<\/b><\/span><\/span>curva caracter\u00edstica de um gerador.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

32- <\/span><\/span><\/span>Na\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o 1<\/b><\/span><\/span>, como o\u00a0<\/b><\/span><\/span>volt\u00edmetro \u00e9 ideal, possui resist\u00eancia interna infinita<\/b><\/span><\/span>, isolante), voc\u00ea pode\u00a0<\/b><\/span><\/span>retir\u00e1-lo\u00a0<\/b><\/span><\/span>e calcular a\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente conforme a\u00a0<\/b><\/span><\/span>sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o 2<\/b><\/span><\/span>\u00a0o amper\u00edmetro est\u00e1 ligado em\u00a0<\/b><\/span><\/span>s\u00e9rie<\/b><\/span><\/span>\u00a0e, como ele \u00e9 ideal<\/b><\/span><\/span>(resist\u00eancia interna nula)<\/b><\/span><\/span>\u00a0voc\u00ea pode\u00a0<\/b><\/span><\/span>curto-circuit\u00e1-lo<\/b><\/span><\/span>\u00a0e calcular a <\/b><\/span><\/span>resist\u00eancia equivalente conforme a sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

Req=U \/ i\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

Sendo os\u00a0<\/b><\/span><\/span>aparelhos de medida ideais<\/b><\/span><\/span>\u00a0voc\u00ea pode\u00a0<\/b><\/span><\/span>\u201ccurto-circuitar\u201d<\/b><\/span><\/span>\u00a0o <\/b><\/span><\/span>amper\u00edmetro,retirar o volt\u00edmetro<\/b><\/span><\/span>\u00a0e calcular a\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span>\u00a0na seq\u00fc\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

34- <\/b><\/span><\/span><\/span>Observe que\u00a0<\/b><\/span><\/span>as duas l\u00e2mpadas acesas de mesma pot\u00eancia P = 60 W\u00a0<\/b><\/span><\/span>e sob\u00a0<\/b><\/span><\/span>tens\u00e3o de U = 12 V<\/b><\/span><\/span>\u00a0est\u00e3o\u00a0<\/b><\/span><\/span>associadas em paralelo\u00a0<\/b><\/span><\/span>e, ambas tamb\u00e9m em\u00a0<\/b><\/span><\/span>paralelo com a bateria de U = 12 V<\/b><\/span><\/span>\u00a0o que \u00e9 \u00f3bvio, pois a\u00a0<\/b><\/span><\/span>tens\u00e3o U \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

Assim, usando os\u00a0<\/b><\/span><\/span>valores nominais fornecido<\/b><\/span><\/span>s a\u00a0<\/b><\/span><\/span>corrente el\u00e9trica i em cada l\u00e2mpada ser\u00e1 a mesma e fornecida por P = i.U\u00a0\"\"\u00a0<\/b><\/span><\/span>60 = i.12\u00a0\"\"<\/b><\/span><\/span>\u00a0i = 5 A.<\/b><\/span><\/span><\/p>\n

Veja na\u00a0<\/b><\/span><\/span>figura abaixo\u00a0<\/b><\/span><\/span>que a\u00a0<\/b><\/span><\/span>corrente que passa pelo fus\u00edvel\u00a0<\/b><\/span><\/span>deve valer\u00a0<\/b><\/span><\/span>if<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 2.i =\u00a0<\/b><\/span><\/span>2.5\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>if\u00a0= 10 A.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Das\u00a0<\/b><\/span><\/span>alternativas,<\/b><\/span><\/span>\u00a0aquela cujo\u00a0<\/b><\/span><\/span>fus\u00edvel apresenta menor amperagem<\/b><\/span><\/span>(intensidade de corrente el\u00e9trica)<\/b><\/span><\/span><\/p>\n

capaz de\u00a0<\/b><\/span><\/span>proteger esse circuito \u00e9 o de 12 A.<\/b><\/span><\/span><\/p>\n

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

35- <\/b><\/span><\/span><\/span>
\nNa\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o 1<\/b><\/span><\/span>, como o\u00a0<\/b><\/span><\/span>volt\u00edmetro \u00e9 ideal, possui resist\u00eancia interna infinita<\/b><\/span><\/span>, isolante), voc\u00ea pode\u00a0<\/b><\/span><\/span>retir\u00e1-lo\u00a0<\/b><\/span><\/span>e calcular a\u00a0<\/b><\/span><\/span>resist\u00eancia equivalente conforme a\u00a0<\/b><\/span><\/span>sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o 2<\/b><\/span><\/span>\u00a0o amper\u00edmetro est\u00e1 ligado em\u00a0<\/b><\/span><\/span>s\u00e9rie<\/b><\/span><\/span>\u00a0e, como ele \u00e9 ideal<\/b><\/span><\/span>(resist\u00eancia interna nula)<\/b><\/span><\/span>\u00a0voc\u00ea pode\u00a0<\/b><\/span><\/span>curto-circuit\u00e1-lo<\/b><\/span><\/span>\u00a0e calcular a<\/b><\/span><\/span>resist\u00eancia equivalente conforme a sequ\u00eancia abaixo:<\/b><\/span><\/span><\/p>\n

Req=U \/ i\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

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

Resolu\u00e7\u00e3o Comentada Eletrodin\u00e2mica – 2017 \u2013 2016 \u2013 2015. 01- 02- \u00a0Assim como no c\u00e1lculo de\u00a0\u0394S em um gr\u00e1fico V x \u0394t, nossa carga tamb\u00e9m pode ser calculada pela \u00e1rea do gr\u00e1fico, pois como\u00a0a varia\u00e7\u00e3o da dist\u00e2ncia \u00e9 a multiplica\u00e7\u00e3o entre tempo e velocidade, a carga \u00e9 a multiplica\u00e7\u00e3o entre intensidade da corrente e o tempo, logo: N\u00e3o podemos esquecer<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3302,"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-3304","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3304","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=3304"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3304\/revisions"}],"predecessor-version":[{"id":3306,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3304\/revisions\/3306"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3302"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3304"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}