{"id":3310,"date":"2017-10-01T13:42:56","date_gmt":"2017-10-01T13:42:56","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3310"},"modified":"2024-09-04T13:59:24","modified_gmt":"2024-09-04T13:59:24","slug":"resolucoes-eletrodinamica-2014-2013","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/eletricidade\/eletrodinamica-2014-2013\/resolucoes-eletrodinamica-2014-2013\/","title":{"rendered":"Resolu\u00e7\u00f5es – Eletrodin\u00e2mica \u2013 2014 – 2013"},"content":{"rendered":"

Resolu\u00e7\u00f5es Vestibulares Recentes<\/b><\/span><\/span><\/span><\/p>\n

Eletrodin\u00e2mica \u2013 2014 – 2013<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da resist\u00eancia el\u00e9trica\u00a0R\u00a0da l\u00e2mpada pelas\u00a0especifica\u00e7\u00f5es\u00a0—\u00a0P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0— 24=12<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R —<\/b><\/span><\/span><\/span><\/p>\n

R=144\/24 —\u00a0R=6\u2126.<\/b><\/span><\/span><\/span><\/p>\n

Sendo essa\u00a0R=6 \u2126 constante, quando a l\u00e2mpada estiver\u00a0sob ddp (tens\u00e3o) de 6V\u00a0a\u00a0corrente i\u00a0que circula por ela ser\u00e1 —\u00a0R=U\/i\u00a0— 6=6\/i —\u00a0i=1 A.<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia el\u00e9trica dissipada\u00a0pela l\u00e2mpada quando sob\u00a0ddp de U=6V\u00a0—\u00a0P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 6.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0P=6W.<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na figura fornecida que a\u00a0resist\u00eancia equivalente\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>\u00a0est\u00e1 entre os pontos\u00a0A e B:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=48\u2126<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Como os ramos superior e inferior est\u00e3o associados em\u00a0paralelo a tens\u00e3o U=120V \u00e9 a mesma em cada um\u00a0— sendo a resist\u00eancia R=20\u2126 de cada barra (resistor) a mesma, a\u00a0tens\u00e3o em cada resistor\u00a0do ramo\u00a0superior\u00a0vale\u00a0U<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=120\/4=30V\u00a0e em cada\u00a0resistor do ramo inferior\u00a0vale\u00a0U<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=120\/6=20V<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0pot\u00eancia el\u00e9trica\u00a0dissipada em cada resistor (barra) do ramo\u00a0superior\u00a0vale\u00a0P<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=30<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/20=45W\u00a0e em\u00a0cada resistor\u00a0do ramo\u00a0inferior\u00a0vale\u00a0P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/20=20W<\/b><\/span><\/span><\/span><\/p>\n

Cores laranja;\u00a0uma\u00a0no ramo superior (45W) e\u00a0outra\u00a0no inferior (20W) — P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=45 + 20 —\u00a0P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=65W<\/b><\/span><\/span><\/span><\/p>\n

Cores azul;\u00a0duas\u00a0no ramo superior (2x45W=90W) —\u00a0P<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=90W<\/b><\/span><\/span><\/span><\/p>\n

Cores vermelha;\u00a0uma\u00a0no ramo superior (45W) e\u00a0outra\u00a0no inferior (20W) — P<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=45 + 20 —\u00a0P<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=65W<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando a chave na\u00a0posi\u00e7\u00e3o B\u00a0voc\u00ea ter\u00e1 o\u00a0circuito da figura abaixo\u00a0cuja sequ\u00eancia nos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornece o\u00a0resistor equivalente R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=20 + R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U(E)\/i\u00a0— 20 + R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=100\/2 — 20 + R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=50 —\u00a0R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=30\u2126.<\/b><\/span><\/span><\/span><\/p>\n

Veja que o\u00a0enunciado afirma\u00a0que, com a chave em B onde\u00a0i=2A, a pot\u00eancia el\u00e9trica dissipada pela l\u00e2mpada \u00e9 de P<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=60W,\u00a0o que nos\u00a0forneceria uma resist\u00eancia\u00a0de —\u00a0P=R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 60=R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=15\u2126.<\/b><\/span><\/span><\/span><\/p>\n

Como o\u00a0exerc\u00edcio afirma que a\u00a0resist\u00eancia da l\u00e2mpada \u00e9 constante,\u00a0essa pot\u00eancia fornecida de\u00a060W \u00e9 incompat\u00edvel\u00a0e vamos\u00a0ignor\u00e1-la na continua\u00e7\u00e3o do exerc\u00edcio\u00a0e considerar a resist\u00eancia da l\u00e2mpada como seno\u00a0R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=30\u2126.<\/b><\/span><\/span><\/span><\/p>\n

Colocando agora a\u00a0chave em A, a nova resist\u00eancia equivalente ser\u00e1\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=45 + 20 + 30=95\u2126\u00a0(figura).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=E\/i\u2019\u00a0— 95=100\/i\u2019 —\u00a0i\u2019=100\/95\u22481A\u00a0— pot\u00eancia dissipada pelo circuito —\u00a0P=R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=95.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=95W.<\/b><\/span><\/span><\/span><\/p>\n

Energia el\u00e9trica dissipada pelo circuito durante\u00a0\u0394t=2min=2.60=120s\u00a0—\u00a0W=P.\u0394t=95.120 —<\/b><\/span><\/span><\/span><\/p>\n

W=11400J=1,1.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A) Falsa\u00a0— desligando-se L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>,\u00a0L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0manter\u00e1 seu brilho\u00a0j\u00e1 que ser\u00e1 percorrida pela mesma corrente que antes pois, como est\u00e3o em paralelo a tens\u00e3o U \u00e9 a mesma antes e depois de desligar L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e a\u00a0potencia de L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0tamb\u00e9m ser\u00e1 a mesma j\u00e1 que P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R e R tamb\u00e9m \u00e9 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

B) Falsa\u00a0—\u00a0pela segunda lei de Ohm\u00a0R=\u03c1\u2113\/S\u00a0e, sendo o material (\u03c1) o mesmo e o comprimento (\u2113) do filamento o mesmo, a\u00a0l\u00e2mpada 2 que \u00e9 mais grossa (maior S), possuir\u00e1 menor resist\u00eancia el\u00e9trica R e, consequentemente ser\u00e1 pecorrida por maior corrente el\u00e9trica brilhando mais<\/b><\/span><\/span><\/span><\/p>\n

C) Falsa\u00a0—\u00a0os filamentos possuem\u00a0resist\u00eancias el\u00e9tricas diferentes possuindo diferentes brilhos.<\/b><\/span><\/span><\/span><\/p>\n

D- Correta\u00a0— veja (B)<\/b><\/span><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Usando os valores nominais\u00a0de cada l\u00e2mpada id\u00eanticas voc\u00ea pode calcular o valor da\u00a0resist\u00eancia el\u00e9trica de cada uma pela express\u00e3o\u00a0P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0— 100=100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R —\u00a0R=100\u2126.<\/b><\/span><\/span><\/span><\/p>\n

Observe na\u00a0figura abaixo\u00a0que a l\u00e2mpada\u00a0L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0de resist\u00eancia R=100 \u2126, \u00e9 percorrida por i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, e pelo enunciado\u00a0sob pot\u00eancia de 100W, tal que —\u00a0P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 100=100. i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1 A\u00a0(corrente por L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0quando sob pot\u00eancia de 100W).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na\u00a0figura acima\u00a0que a l\u00e2mpada\u00a0L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0de resist\u00eancia R=100 \u2126, \u00e9 percorrida por i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, e pelo enunciado\u00a0sob pot\u00eancia de 64W, tal que —\u00a0P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 64=100. i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,8A. (corrente por L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0quando sob pot\u00eancia de 64W)<\/b><\/span><\/span><\/span><\/p>\n

No\u00a0trecho s\u00e9rie R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(R=100\u00a0\u2126),\u00a0percorridos por i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1 A e sob ddp de U=200V\u00a0(paralelos \u00e0 fonte) voc\u00ea ter\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=(R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 100)\u00a0—\u00a0U=200V e i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1 A\u00a0—\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0— R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 100 = 200\/1 —\u00a0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=100 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

No\u00a0trecho s\u00e9rie R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0(R=100\u00a0\u2126),\u00a0percorridos por i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,8 A e sob ddp de U=200V\u00a0(paralelos \u00e0 fonte) voc\u00ea ter\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 100)\u00a0—\u00a0U=200V e i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,8 A\u00a0—\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0— R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 100 = 200\/0,8 — R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 100 = 250 —<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=150 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Se R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 100 \u2126,\u00a0L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0dissipar\u00e1 uma pot\u00eancia de\u00a0100W\u00a0e\u00a0se\u00a0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 150 \u2126,\u00a0L<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0dissipar\u00e1 uma pot\u00eancia de\u00a064W,<\/b><\/span><\/span><\/span><\/p>\n

conforme pedido no enunciado.<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Voc\u00ea deve se lembrar de que a pot\u00eancia el\u00e9trica m\u00e1xima que um gerador (no caso associa\u00e7\u00e3o de geradores) fornece ao circuito externo (no caso, reostato R) ocorre quando a resist\u00eancia externa R \u00e9 igual \u00e0 resist\u00eancia interna r da associa\u00e7\u00e3o de geradores, ou seja, quando R=r<\/b><\/span><\/span><\/span>equivalente<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0na associa\u00e7\u00e3o s\u00e9rie\u00a0— r<\/b><\/span><\/span><\/span>equ<\/b><\/span><\/span><\/sub><\/span>=r + r=2r —\u00a0R=2r\u00a0—\u00a0na associa\u00e7\u00e3o paralelo\u00a0— r<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=(r.r)\/(r + r)=r\/2 — r<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=r\/2 —\u00a0R=r\/2\u00a0—\u00a0veja na figura abaixo que a resposta corresponde \u00e0 alternativa C:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o s\u00e9rie<\/b><\/span><\/span><\/span><\/p>\n

\"\"Todos os resistores s\u00e3o ligados um em seguida ao outro.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A resist\u00eancia equivalente (R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>)\u00a0\u00e9 igual \u00e0\u00a0soma das resist\u00eancias parciais\u00a0\u00a0—\u00a0\u00a0no circuito 1 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=R; no 2, R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=2R\u00a0e no 3 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=3R — a resist\u00eancia equivalente da associa\u00e7\u00e3o\u00a0aumenta.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A tens\u00e3o (diferen\u00e7a de potencial) U, constante, nos terminais da associa\u00e7\u00e3o (entre A e B) \u00e9 igual \u00e0 soma das tens\u00f5es em cada resistor, ou seja,\u00a0no circuito 1 U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U\/1; no 2 U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=U\/2 e no 3 U<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=U\/3 — a\u00a0diferen\u00e7a de potencial nos extremos de um mesmo resistor\u00a0diminui.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Pela lei de Ohm, como R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i, a\u00a0corrente el\u00e9trica i\u00a0\u00e9\u00a0inversamente\u00a0proporcional \u00e0 resist\u00eancia equivalente\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>, 0u seja, a medida que voc\u00ea acrescenta resistores,\u00a0R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>\u00a0aumenta\u00a0e,\u00a0consequentemente\u00a0i diminui.<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>— Observa\u00e7\u00e3o: Voc\u00ea poderia chegar \u00e0 mesma conclus\u00e3o observando a tabela fornecida e lembrando que, no caso, o volt\u00edmetro marca a ddp em cada resistor e o amper\u00edmetro a corrente total no circuito.<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da resist\u00eancia equivalente\u00a0R<\/b><\/span><\/span><\/span>eqi<\/b><\/span><\/span><\/sub><\/span>\u00a0antes\u00a0de estabelecer o curto circuito:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eqi<\/b><\/span><\/span><\/sub><\/span>=7R\/3\u00a0— P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eqi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eqi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(7R\/3) —\u00a0P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=3E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/7R (I)<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da resist\u00eancia equivalente\u00a0R<\/b><\/span><\/span><\/span>eqf<\/b><\/span><\/span><\/sub><\/span>\u00a0depois\u00a0de estabelecer o curto circuito:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eqf<\/b><\/span><\/span><\/sub><\/span>=2R\u00a0— P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eqf<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2R (II)<\/b><\/span><\/span><\/span><\/p>\n

(II)\/(I) —\u00a0P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=(E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2R)x(7R\/3E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)=7\/6<\/b><\/span><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span>\u00a0a) <\/b><\/span><\/span><\/span>Observe no circuito que\u00a0o resistor R e a l\u00e2mpada tipo led est\u00e3o associados em s\u00e9rie\u00a0e uma das\u00a0caracter\u00edsticas\u00a0da associa\u00e7\u00e3o s\u00e9rie \u00e9 que a\u00a0corrente el\u00e9trica\u00a0que circula em cada aparelho\u00a0\u00e9 a mesma, ou seja,\u00a0i<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span><\/sub><\/span>\u00a0= i<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0— o\u00a0enunciado fornece que a tens\u00e3o (ddp) na l\u00e2mpada led \u00e9 U<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span><\/sub><\/span>=2,5V\u00a0e, pelo gr\u00e1fico, voc\u00ea<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

observa que i<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span><\/sub><\/span>=0,04 A — assim, <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0i<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0,04 A<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Outra caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie \u00e9 que a tens\u00e3o total (da bateria U=4,5V) \u00e9 a soma das tens\u00f5es<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de cada elemento do circuito s\u00e9rie — assim, U = U<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span><\/sub><\/span>\u00a0+ U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0— 4,5 = 2,5 + U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0— U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2,0V — R=U<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

R=2\/0,04 —\u00a0<\/b><\/span><\/span><\/span>R=50\u2126<\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>Corrente el\u00e9trica total i no circuito, agora com\u00a0U=3V e R=50\u2126\u00a0pode ser determinada por\u00a0P=i.U, onde P \u00e9 fornecida\u00a0P=60mW=60.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>W\u00a0— 60.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0= i.3 — i=60.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/3 —\u00a0i=20.10<\/b><\/span><\/span><\/span>-3\u00a0<\/b><\/span><\/span><\/sup><\/span>A\u00a0— essa corrente \u00e9 a mesma que flui pelo resistor R=50\u2126 (caracter\u00edstica da associa\u00e7\u00e3o s\u00e9rie) — em R — P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=50.(20.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a050.400.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0— P=20000.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0=20.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>P=20mW<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Segunda lei de Ohm: Atrav\u00e9s de pesquisa, Ohm descobriu que a resist\u00eancia el\u00e9trica de um condutor, mantida a temperatura constante,\u00a0\u00e9 diretamente proporcional ao comprimento L\u00a0do fio,\u00a0\u00e9 inversamente proporcional \u00e0 \u00e1rea de se\u00e7\u00e3o reta transversal S\u00a0e\u00a0depende do material\u00a0que constitui o corpo (essa depend\u00eancia do valor da resist\u00eancia em rela\u00e7\u00e3o ao material que constitui o resistor, chama-se\u00a0resistividade\u00a0do material e representa-se pela letra grega \u03c1 \u201cr\u00f4\u201d).<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Da curva S do gr\u00e1fico voc\u00ea pode calcular a resist\u00eancia equivalente da associa\u00e7\u00e3o s\u00e9rie R<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>\u00a0e da curva P do gr\u00e1fico voc\u00ea pode calcular a resist\u00eancia equivalente da associa\u00e7\u00e3o s\u00e9rie R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

b) Inicialmente voc\u00ea deve determinar, utilizando o gr\u00e1fico1, a intensidade do campo magn\u00e9tico H,\u00a0sendo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecido f=8MHz=8.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>Hz e R\u22484\u2126\u00a0— pelo gr\u00e1fico —\u00a0H=35Oe\u00a0— em seguida, com esse valor<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>04-\u00a0C\u00e1lculo das\u00a0resist\u00eancias\u00a0—\u00a0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8\/2=2\u2126\u00a0—\u00a0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4\/2=2\u2126\u00a0—\u00a0R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=4\/2=2\u2126\u00a0— R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo das\u00a0pot\u00eancias\u00a0—\u00a0P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4.8=32W\u00a0—\u00a0P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2.4=8W\u00a0—\u00a0P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=2.4=8W\u00a0—\u00a0P<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub><\/span>> P<\/b><\/span><\/span><\/span>23<\/b><\/span><\/span><\/sub><\/span><\/p>\n

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

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja no circuito abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Rela\u00e7\u00e3o entre P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

Como\u00a0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=R\u00a0e, como eles est\u00e3o em\u00a0paralelo\u00a0est\u00e3o submetidos\u00a0\u00e0 mesma ddp (tens\u00e3o) U\u00a0eles s\u00e3o percorridos pela\u00a0mesma corrente el\u00e9trica i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=i\u00a0—\u00a0P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— portanto\u00a0P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia em R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado\u00a0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/2=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/2=R\/2\u00a0—\u00a0i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=i + i=2i\u00a0— P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(R\/2).(2i)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(R\/2)x4i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

2Ri<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— mas, Ri<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 2P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 2P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span><\/p>\n

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

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

\"\"<\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O volt\u00edmetro ideal (resist\u00eancia interna infinita), associado corretamente (em paralelo \u00e0s pilhas e \u00e0 l\u00e2mpada) indica a ddp\u00a0Ufornecida pelas duas pilhas \u00e0 l\u00e2mpada que \u00e9 fornecida —\u00a0U=3,00V\u00a0— o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

amper\u00edmetro ideal (resist\u00eancia interna nula) indica a corrente el\u00e9trica i que circula pela l\u00e2mpada, tamb\u00e9m fornecida —\u00a0i=0,50 A<\/b><\/span><\/span><\/span><\/p>\n

— c\u00e1lculo da pot\u00eancia\u00a0P\u00a0da l\u00e2mpada, que\u00a0n\u00e3o \u00e9 o valor nominal,\u00a0nas condi\u00e7\u00f5es do exerc\u00edcio — P=i.U=0,50×3,00 —\u00a0P=1,50W\u00a0— o enunciado afirma, sabe-se que a\u00a0pot\u00eancia dissipada\u00a0por efeito Joule no filamento da l\u00e2mpada corresponde a\u00a01\/4 do valor nominal\u00a0P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>, indicado pelo fabricante, ou seja, P=P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>\/4 —<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=4P=4×1,50 —\u00a0P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=6,0W\u00a0— pela express\u00e3o\u00a0P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0voc\u00ea observa que, sendo a resist\u00eancia R da l\u00e2mpada a mesma, a pot\u00eancia nominal\u00a0P<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 diretamente proporcional ao quadrado da tens\u00e3o (ddp) nominal\u00a0U<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>, ou seja, como a pot\u00eancia foi quadruplicada, a tens\u00e3o deve ser dobrada, (2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4 para manter a igualdade — Un=2.3,00 —\u00a0U<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=6,0V<\/b><\/span><\/span><\/span><\/p>\n

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

Corrente el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Intervalo de tempo durante o qual a bateria est\u00e1 descarregando, at\u00e9 acabar, ou\u00a0 seja descarregar sua carga m\u00e1xima que \u00e9 de q=75 A.h\u00a0 —\u00a0 i=q\/\u2206t\u00a0 —\u00a0 75A.h = 50A\/\u2206t\u00a0 —\u00a0 \u2206t=75A.h\/50A\u00a0 —\u00a0 \u2206t=1,5h\u00a0 —\u00a0 supondo a velocidade de V=60km\/h constante\u00a0 —\u00a0 V=\u2206S\/\u2206t\u00a0 —\u00a0 60=\u2206S\/1,5\u00a0 —\u00a0 \u2206S=90,0km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Deve funcionar assim, se um deles mant\u00e9m a l\u00e2mpada acesa, o outro, quando acionado, deve apag\u00e1-la e vice versa, mantendo sempre o circuito fechado. Analise atentamente cada alternativa:<\/b><\/span><\/span><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 tem apenas um fio ligado \u00e0 rede el\u00e9trica externa o que impedir\u00e1 a l\u00e2mpada de acender, n\u00e3o fechando o circuito, qualquer que seja a posi\u00e7\u00e3o das chaves.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 a l\u00e2mpada est\u00e1 apagada\u00a0 —\u00a0 se voc\u00ea mudar a chave da esquerda ela continuar\u00e1 apagada\u00a0 —\u00a0 se voc\u00ea mudar a chave da direita ela continuar\u00e1 apagada\u00a0 —\u00a0\u00a0 se voc\u00ea mudar as duas chaves, provocar\u00e1 um curto-circuito.<\/b><\/span><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 a l\u00e2mpada est\u00e1 apagada\u00a0 —\u00a0 se voc\u00ea mudar a chaves da direita ou da esquerda ela continuar\u00e1 apagada.<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 a l\u00e2mpada s\u00f3 acender\u00e1 com as duas chaves na posi\u00e7\u00e3o superior\u00a0 —\u00a0 se voc\u00ea mudar uma delas, a outra n\u00e3o acender\u00e1 a l\u00e2mpada.<\/b><\/span><\/span><\/span><\/p>\n

e) Correta\u00a0 —\u00a0 esse \u00e9 o \u00fanico caso em que voc\u00ea consegue acender ou apagar a l\u00e2mpada invertendo apenas uma das<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

chaves, n\u00e3o importando a posi\u00e7\u00e3o da outra.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Pot\u00eancia el\u00e9trica \u2013 Energia el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

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

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em \u2206t=2min=120s, a \u00e1gua recebeu Q=24000cal de energia t\u00e9rmica, que, em joules vale\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1cal \u2013 4J\u00a0 —\u00a0 24000cal \u2013 W (J)\u00a0 —\u00a0 W=24000×4\u00a0 —\u00a0 W=96000J\u00a0 —\u00a0 Pot\u00eancia el\u00e9trica do aquecedor\u00a0 —\u00a0 P=W\/\u2206t=96000\/120\u00a0 —\u00a0 P=800W\u00a0 —\u00a0 P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 800=120<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 R=14400\/800\u00a0 —\u00a0 R=18\u2126\u00a0 —\u00a0 pelo enunciado, em \u2206t=2min=120s, a temperatura da \u00e1gua variou de 20\u00b0C a 80\u00b0C\u00a0 —\u00a0 \u2206\u03b8=(80 \u2013 20)=60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da calorimetria\u00a0 —\u00a0 Q=m.c.\u2206\u03b8\u00a0 —\u00a0 24000=m.1.60\u00a0 —\u00a0 m=400g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

52-<\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Painel\u00a0 —\u00a0 Pp=Wp\/24h\u00a0 —\u00a0 810=Wp\/24h\u00a0 —\u00a0 Wp=19440Wh (energia fornecida pelo painel em 1 dia)\u00a0 —\u00a0 energia dissipada pela l\u00e2mpada em 1 dia\u00a0 —\u00a0 Pl=Wl\/24.\u2206t\u00a0 —\u00a0 9=Wl\/24.\u2206t\u00a0 —\u00a0 Wl=216.\u2206t (W)\u00a0 —\u00a0 Wp\u00a0= Wl\u00a0 —\u00a0 19440 = 216\u2206t\u00a0 —\u00a0 \u2206t=90h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0a) <\/b><\/span><\/span><\/span>Pelo enunciado observe que, para cada 100g de bauxita voc\u00ea tem 55g de \u00f3xido de alum\u00ednio (Al<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>O<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>) e 45g de res\u00edduos\u00a0 —\u00a0 massa molar do \u00f3xido de alum\u00ednio (Al<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>O<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 2x(27) + 3x(16)=102g.mol<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 — massa de alum\u00ednio que comp\u00f5e o \u00f3xido de alum\u00ednio (alumina)\u00a0 —\u00a0 Al<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2x(27)= 54g.mol<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 102g de Al<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>O<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 54g de Al\u00a0 —\u00a0 55g de Al<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>O<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 x\u00a0 —\u00a0 x=55×54\/102\u00a0 —\u00a0 x=29,12g de alum\u00ednio em 100g de bauxita\u00a0 —\u00a0 c\u00e1lculo da massa de res\u00edduos\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 29,12g de Al \u2013 45g de res\u00edduos\u00a0 —\u00a0 32.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>t de Al \u2013 y\u00a0 —\u00a0 y=45×32.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\/29,12\u00a0 —\u00a0 y=49,45t.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u2248<\/b><\/span><\/span><\/span><\/p>\n

4,9.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>t de res\u00edduos.<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 tempo de dura\u00e7\u00e3o do banho\u00a0 —\u00a0 \u2206t=10mim=10.60=600s\u00a0 —\u00a0 pot\u00eancia el\u00e9trica do chuveiro\u00a0 —\u00a0 P=3000W\u00a0 —\u00a0 P=energia(W)\/ \u2206t\u00a0 —\u00a0 3000=W\/600\u00a0 —\u00a0 W=18.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J (energia dissipada num banho de 10min)\u00a0 —\u00a0 pelos dados, 1 latinha 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 latinha \u2013 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 n latinhas \u2013 18.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 n=3,6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/18.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=0,2.10\u00a0 —\u00a0\u00a0n=2 banhos.<\/b><\/span><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo enunciado, a efici\u00eancia \u03b5 \u00e9 fornecida pela raz\u00e3o entre a quantidade de luz produzida Q e a energia el\u00e9trica W consumida\u00a0 —\u00a0 \u03b5 =Q\/W (I)\u00a0 —\u00a0 pot\u00eancia=energia\/tempo de utiliza\u00e7\u00e3o\u00a0 —\u00a0 P=W\/\u2206t\u00a0 —\u00a0 W=P.\u2206t (II)\u00a0 —\u00a0 substituindo (II) em (I)\u00a0 —\u00a0 \u03b5 =Q\/P.\u2206t\u00a0 —\u00a0 para a l\u00e2mpada incandescente de 40W\u00a0 —\u00a0 \u03b5\u2019=600lm\/40\u2206t\u00a0 —\u00a0 o intervalo de tempo \u2206t \u00e9 o mesmo para cada alternativa.<\/b><\/span><\/span><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 l\u00e2mpada fluorescente de 8W\u00a0 —\u00a0 se uma l\u00e2mpada fluorescente de 40W emite Q=3000lm, uma de 8W emitir\u00e1 5 vezes menos, ou seja, Q=3000\/5=600lm\u00a0 —\u00a0 \u03b5=600lm\/8\u2206t\u00a0 —\u00a0 observe que \u03b5 > \u03b5\u2019.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 l\u00e2mpada fluorescente de 40W\u00a0 —\u00a0 \u03b5=3000lm\/40\u2206t\u00a0 —\u00a0 observe que\u00a0 \u03b5 > \u03b5\u2019.<\/b><\/span><\/span><\/span><\/p>\n

c) Verdadeira\u00a0 —\u00a0 l\u00e2mpada fluorescente de 8W\u00a0 —\u00a0 \u03b5=600lm\/8\u2206t\u00a0 —\u00a0 \u03b5\u2019 < \u03b5 e mesmo Q=600lm.<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 veja (b).<\/b><\/span><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 veja (B).<\/b><\/span><\/span><\/span><\/p>\n

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

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

Resistores \u2013 Primeira lei de Ohm \u2013 Pot\u00eancia el\u00e9trica em um resistor<\/b><\/span><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Resist\u00eancia el\u00e9trica\u00a0de um resistor \u2013 representa a medida da dificuldade imposta \u00e0 movimenta\u00e7\u00e3o das cargas el\u00e9trica que o atravessam (corrente el\u00e9trica), e \u00e9 definida como a raz\u00e3o entre a diferen\u00e7a de potencial (U) nos terminais do resistor e a intensidade da corrente el\u00e9trica ( i ) que o atravessa:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A express\u00e3o acima \u00e9 conhecida como\u00a0Lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Um ohm (1\u2126) \u00e9 o valor da resist\u00eancia el\u00e9trica entre dois pontos P e Q de um condutor quando, sob diferen\u00e7a de potencial de 1 volt<\/b><\/span><\/span><\/span><\/p>\n

(1V) \u00e9 percorrido por uma corrente el\u00e9trica constante de 1 amp\u00e8re (1A).<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0Representa\u00e7\u00e3o simb\u00f3lica:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Primeira lei de Ohm\u00a0\u2013 Se a resist\u00eancia el\u00e9trica de um resistor for constante, a diferen\u00e7a de potencial aplicada nos seus extremos \u00e9 diretamente proporcional \u00e0 intensidade da corrente el\u00e9trica, ou seja,U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= . . . =U\/i=constante. Nesse caso eles s\u00e3o chamados de\u00a0resistores (ou condutores) \u00f4hmicos\u00a0e, como a rela\u00e7\u00e3o R=U\/i \u00e9 uma fun\u00e7\u00e3o do primeiro grau o gr\u00e1fico U x i tem o aspecto do gr\u00e1fico da esquerda da figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Qualquer gr\u00e1fico U x i diferente do gr\u00e1fico da esquerda da figura acima n\u00e3o representa um resistor \u00f4hmico, como por exemplo, o gr\u00e1fico da direita.<\/b><\/span><\/span><\/span><\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Corrente que provoca fibrila\u00e7\u00e3o ventricular passando pelo t\u00f3rax\u00a0 —\u00a0 iv=500mA=500.10-3=0,5 A\u00a0 —\u00a0 corrente provocada pelo poraqu\u00ea de U=1500Vquando o pescador o segura com uma m\u00e3o na cabe\u00e7a e outra na cauda e que passa pelo seu t\u00f3rax de R=1000\u2126\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 1000=1500\/i\u00a0 —\u00a0 i=1500\/1000\u00a0 —\u00a0 i=1,5 A\u00a0 —\u00a0 rela\u00e7\u00e3o entre essas correntes\u00a0 —\u00a0 i\/iv=1,5\/0,5\u00a0 —\u00a0 i=3iv\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0A) <\/b><\/span><\/span><\/span>Na posi\u00e7\u00e3o INVERNO o chuveiro deve dissipar a maior pot\u00eancia (4800W)\u00a0 —\u00a0 como a ddp (tens\u00e3o) U \u00e9 constante, pela express\u00e3o P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R voc\u00ea observa que a pot\u00eancia P \u00e9 inversamente proporcional \u00e0 resist\u00eancia el\u00e9trica R e,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

assim para uma maior pot\u00eancia corresponde uma menor resist\u00eancia el\u00e9trica, o que ocorre quando a chave seletora estiver <\/b><\/span><\/span><\/span>ligada na\u00a0posi\u00e7\u00e3o Y.<\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>Na posi\u00e7\u00e3o INVERNO a pot\u00eancia deve ser P=4800W\u00a0 —\u00a0 \u2206t=10min=10\/60=1\/6h\u00a0 —\u00a0 P=W\/\u2206t\u00a0 —\u00a0 4800=W\/(1\/6)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W=0,8kWh.<\/b><\/span><\/span><\/span><\/p>\n

C) <\/b><\/span><\/span><\/span>Uma pessoa num dia\u00a0 —\u00a0 P=0,8kWh\u00a0 —\u00a0 4 pessoas num dia\u00a0 —\u00a0 P=3,2kWh\u00a0 —\u00a0 4 pessoas num m\u00eas\u00a0 —\u00a0 P=3,2×30 = 96kWh\u00a0 —\u00a0 como a resid\u00eancia consome num m\u00eas 200kWh, sem os banhos a economia ser\u00e1 de <\/b><\/span><\/span><\/span>W\u2019<\/b><\/span><\/span><\/span>=200 \u2013 96=<\/b><\/span><\/span><\/span>104kWh.<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Num chuveiro el\u00e9trico, normalmente de tens\u00e3o 220V (mantida constante), as posi\u00e7\u00f5es inverno (quente), ver\u00e3o (morna)\u00a0correspondem \u00e0 diferentes pot\u00eancias. Quando voc\u00ea muda o cursor do chuveiro, por exemplo, do\u00a0ver\u00e3o\u00a0para<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

o\u00a0inverno,\u00a0voc\u00ea\u00a0est\u00e1\u00a0diminuindo a resist\u00eancia el\u00e9trica no interior do chuveiro e, consequentemente aumentando a pot\u00eancia, fazendo-o aquecer mais, claro que mantendo a vaz\u00e3o constante. O oposto ocorre quando voc\u00ea muda o cursor do chuveiro do inverno para o ver\u00e3o. Isso acontece porque, para a mesma tens\u00e3o, a pot\u00eancia \u00e9 inversamente proporcional \u00e0 resist\u00eancia, pois, P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R, com U constante.<\/b><\/span><\/span><\/span><\/p>\n

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

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O condutor possui resist\u00eancia \u00f4hmica, pois o gr\u00e1fico \u00e9 uma reta inclinada a partir da origem e a ddp U \u00e9 diretamente proporcional\u00a0 a i de modo que R seja constante\u00a0 —\u00a0 R= U\/i= 20\/0,5 = 40\/1 = 60\/1,5 = 80\/2=40\u2126\u00a0 — P=R.i2=40.(1,5)2\u00a0P=40.2,25\u00a0 —\u00a0 P=90W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O funcionamento do\u00a0fus\u00edvel\u00a0baseia-se no princ\u00edpio segundo o qual uma corrente que passa por um condutor gera calor proporcional ao quadrado de sua intensidade\u00a0 —\u00a0 quando a corrente atinge a intensidade m\u00e1xima toler\u00e1vel, o calor gerado n\u00e3o se dissipa com rapidez suficiente, derretendo um componente e interrompendo o circuito.<\/b><\/span><\/span><\/span><\/p>\n

Disjuntores\u00a0\u00a0—\u00a0\u00a0modernamente, nos circuitos el\u00e9tricos de resid\u00eancias, edif\u00edcios e ind\u00fastrias, em vez de fus\u00edveis, utilizam-se dispositivos baseados no efeito magn\u00e9tico da corrente denominados disjuntores\u00a0 —\u00a0 em ess\u00eancia, o disjuntor \u00e9 uma chave magn\u00e9tica que se desliga automaticamente quando a intensidade da corrente no condutor supera certo valor\u00a0 —\u00a0 tem sobre o fus\u00edvel a vantagem de n\u00e3o precisar ser trocado\u00a0 —\u00a0 uma vez resolvido o problema que provocou o desligamento, basta relig\u00e1-lo para que a circula\u00e7\u00e3o da corrente se restabele\u00e7a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

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

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

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>U=270V\u00a0 —\u00a0 l=1,5m\u00a0 —\u00a0 \u03c1=1,8\u2126m\u00a0 —\u00a0 S=30cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=30.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>=3.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=\u03c1.l\/S=1,8.1,5\/3.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=900\u2126\u00a0 — R=U\/i\u00a0 —\u00a0 900=270\/i\u00a0 —\u00a0 i=270\/900=0,3 A=300mA\u00a0 —\u00a0 a corrente el\u00e9trica fatal \u00e9 de 200mA\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Segunda lei de Ohm\u00a0 —\u00a0 R=\u03c1\u2113\/S=1,8.10-9.3.105\/2.10-4=1,8.1,5.10o\u00a0 —\u00a0 R=2,7\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

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

Resistor equivalente nas associa\u00e7\u00f5es s\u00e9rie, paralelo e mista \u2013 curto circuito<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>S\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=2R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 2R=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U\/2R\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=R\/2\u00a0 —\u00a0 R\/2=U\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2U\/R\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 9\u00a0 —\u00a0 2U\/R \u2013 U\/2R=9\u00a0 —\u00a0 3U\/2R=9\u00a0 —\u00a0 3.60\/2R=9\u00a0 —\u00a0 R=10\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo enunciado A\u2019A=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e B\u2019B=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 denominando AP\u2019 de m e P\u2019B de n e observando que os <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tri\u00e2ngulos nas figuras s\u00e3o semelhantes\u00a0 —\u00a0 P\u2019P\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m\/(m + n) (I)\u00a0 —\u00a0 os dois tri\u00e2ngulos hachurados \"\"<\/b><\/span><\/span><\/span><\/p>\n

nas figuras a seguir tamb\u00e9m hachurados s\u00e3o semelhantes\u00a0 —\u00a0 P\u2019P\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=n\/(m + n) (II)\u00a0 —\u00a0 (I) + (II)\u00a0 —\u00a0 P\u2019P\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P\u2019P\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= m\/(m + n) + n\/(m + n)=(m + n)\/(m + n)=1\u00a0 —\u00a0 P\u2019P\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P\u2019P\/R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 P\u2019P.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P\u2019P.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P\u2019P=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/(R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

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

Caracter\u00edsticas de associa\u00e7\u00e3o s\u00e9rie e associa\u00e7\u00e3o paralela<\/b><\/span><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>S\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=2R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 2R=U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=U\/2R\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=R\/2\u00a0 —\u00a0 R\/2=U\/i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2U\/R\u00a0 —\u00a0 i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 9\u00a0 —\u00a0 2U\/R \u2013 U\/2R=9\u00a0 —\u00a0 3U\/2R=9\u00a0 —\u00a0 3.60\/2R=9\u00a0 —\u00a0 R=10\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as l\u00e2mpadas est\u00e3o associadas em paralelo, a tens\u00e3o U \u00e9 a mesma para cada l\u00e2mpada e vale U=120V\u00a0 —\u00a0 pot\u00eancia de cada l\u00e2mpada P=60W\u00a0 —\u00a0 corrente i em cada l\u00e2mpada\u00a0 —\u00a0 P=i.U\u00a0 —\u00a0 60=i.120\u00a0 —\u00a0 i=0,5 A\u00a0 —\u00a0 a corrente limite do fus\u00edvel (corrente total de entrada) deve ser de it=10 A\u00a0 —\u00a0 essa corrente \u00e9 dividida em n l\u00e2mpadas cada uma com 0,5 A\u00a0 —\u00a0 it=n.i\u00a0 —\u00a0 10=n.0,5\u00a0 —\u00a0 n=20 l\u00e2mpadas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

62-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>O mesmo aquecimento corresponde \u00e0 mesma quantidade de energia, assim, Wparalelo=Ws\u00e9rie\u00a0 —\u00a0 P=W\/\u2206t\u00a0 —\u00a0 W=P.\u2206t\u00a0 —\u00a0 Wp\u00a0= Ws\u00a0 —\u00a0 Pp.\u2206tp\u00a0= Ps.\u2206ts\u00a0 —\u00a0 P=U2\/R (U constante, dado do exerc\u00edcio)\u00a0 —\u00a0 (U2\/Rp).\u2206tp\u00a0= (U2\/Rs),\u2206ts\u00a0 —\u00a0\u00a0 \u2206tp\/Rp=\u2206ts\/Rs\u00a0 —\u00a0 associa\u00e7\u00e3o paralelo Req=Rp=R\/2\u00a0 —\u00a0 associa\u00e7\u00e3o s\u00e9rie Req=Rs=2R\u00a0 —\u00a0 \u2206tp\/(R\/2) = \u2206ts\/2R\u00a0 —\u00a0 \u2206tp\u00a0= \u2206ts\/4\u00a0 —\u00a0 \u2206ts=1min=60s\u00a0 —\u00a0 \u2206tp=60\/4=15s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E.<\/b><\/span><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo enunciado, se o acerto \u00e9 zero, somente a chave C \u00e9 fechada e o circuito fica conforme a <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura\u00a0 —\u00a0 o amper\u00edmetro indica a corrente no circuito de valor\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 20=120\/i\u00a0 —\u00a0 i=6 A —\u00a0 se o participante errar apenas P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, n\u00e3o passar\u00e1 corrente por ela e a corrente circular\u00e1 apenas nas l\u00e2mpadas P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e P<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, com a chave C aberta\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0\u00a0 veja a sequ\u00eancia indicada na figura acima\u00a0 onde a resist\u00eancia equivalente vale R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=40\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 40=120\/i\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>i=3 A.<\/b><\/span><\/span><\/span><\/p>\n

64-<\/b><\/span><\/span><\/span> <\/b>C\u00e1lculo da resist\u00eancia equivalente para as 25 l\u00e2mpadas, cada uma com resist\u00eancia R associadas em s\u00e9rie (i \u00e9 a mesma e a resist\u00eancia equivalente \u00e9 a soma da resist\u00eancia de todas as l\u00e2mpadas em s\u00e9rie)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eqs<\/b><\/span><\/span><\/sub><\/span>=25R\u00a0 —\u00a0 na associa\u00e7\u00e3o em paralelo que fornece a resist\u00eancia equivalente total (R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>), voc\u00ea tem 4 resist\u00eancia de 25R associadas em paralelo\u00a0 —\u00a0 1\/R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>= 1\/25R + 1\/25R +\u00a0 1\/25R +\u00a0 1\/25R=4\/25R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>=25R\/4 (I)\u00a0 — c\u00e1lculo da resist\u00eancia equivalente total pelos valores nominais fornecidos (pot\u00eancia el\u00e9trica e tens\u00e3o totais)\u00a0 —\u00a0 P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 22 = 220<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>=48400\/22\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eqtotal<\/b><\/span><\/span><\/sub><\/span>=2200\u2126 (II)\u00a0 —\u00a0 igualando (I) com (II)\u00a0 —\u00a0 25R\/4 = 2200\u00a0 —\u00a0 R=8800\/25=352\u2126\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

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

65-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Estando esses aparelhos funcionando normalmente, todos eles est\u00e3o submetidos \u00e0 mesma ddp (tens\u00e3o) que \u00e9 de U-110V e, para que isso ocorra, todos esses aparelhos devem estar associados em paralelo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como todos os aparelhos est\u00e3o funcionando dentro das especifica\u00e7\u00f5es comerciais, a pot\u00eancia total dissipada \u00e9 a soma da pot\u00eancia de cada um desses aparelhos\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=40 + 40 + 20 + 120=220W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>total.<\/b><\/span><\/span><\/sub><\/span>.U\u00a0 —\u00a0 220= i<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>.110\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

i<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=2 A.<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Tempo de funcionamento\u00a0 —\u00a0 \u2206t=30diasx4h=120h\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206t\u00a0 —\u00a0 220W=W\/120h\u00a0 —\u00a0 W=26400W=26,4kWh\u00a0 —\u00a0 custo\u00a0 —\u00a0 C$=26,4×0,30=7,92\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>C$=7,92 reais.<\/b><\/span><\/span><\/span><\/p>\n

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

C\u00e1lculo de correntes, tens\u00f5es e pot\u00eancias em associa\u00e7\u00f5es mistas<\/b><\/span><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Um volt\u00edmetro ideal (resist\u00eancia interna infinita) inserido na entrada indicaria\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>entrada\u00a0<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>.i=(R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).i\u00a0 —\u00a0 um<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

volt\u00edmetro ideal (resist\u00eancia interna infinita) inserido na sa\u00edda indicaria\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>sa\u00edda\u00a0<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>.i=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i\u00a0 —\u00a0 pelo enunciado V<\/b><\/span><\/span><\/span>sa\u00edda<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>entrada<\/b><\/span><\/span><\/sub><\/span>\/10\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i =\u00a0 (R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>).i\/10\u00a0 —\u00a0 10R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub><\/span>+ R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 9R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=9.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que entre os pontos P (de potencial +5V) e A voc\u00ea tem o resistor de 4k\u2126 em circuito aberto e que, nesse trecho n\u00e3o passa corrente\u00a0 —\u00a0 assim, o potencial no ponto A \u00e9 o mesmo que no ponto P e vale +5V\u00a0 —\u00a0 observe tamb\u00e9m que os dois resistores de 2k\u2126 est\u00e3o em paralelo e devem ser substitu\u00eddos por um \u00fanico de R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=produto\/soma=(2×2)\/(2 + 2)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=1k\u2126\u00a0 —\u00a0 haver\u00e1 passagem de corrente el\u00e9trica apenas nos trechos PBQ (veja<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0figura) e nele os resistores de 4k\u2126 e 1k\u2126 est\u00e3o em s\u00e9rie, sobrando apenas um resistor de R=4 + 1=5k\u2126 submetido a uma ddp de U=(5 \u2013 0)=5V\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 5\/i\u00a0 —\u00a0 i=5\/5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 i=1.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0A ou i=1m A\u00a0 —\u00a0 entre A e B\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i\u00a0 —\u00a0 4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=4V\u00a0\u00a0—\u00a0 entre B e Q\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>BQ<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>BQ<\/b><\/span><\/span><\/sub><\/span>\/i\u00a0 —\u00a0 1.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=U<\/b><\/span><\/span><\/span>BQ<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>BQ<\/b><\/span><\/span><\/sub><\/span>=1.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>BQ<\/b><\/span><\/span><\/sub><\/span>=1V \u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

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

Geradores\u00a0 – associa\u00e7\u00e3o de geradores<\/b><\/span><\/span><\/span><\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na sequ\u00eancia abaixo o c\u00e1lculo do resistor equivalente de resist\u00eancia R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=10\u2126 e da <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

corrente total i\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=\u03b5 \/i\u00a0 —\u00a0 10=12\/i\u00a0 —\u00a0 i=1,2 A\u00a0 —\u00a0 c\u00e1lculo da diferen\u00e7a de potencial U nos terminais do gerador de resist\u00eancia interna r=1\u2126\u00a0 —\u00a0 Usando a equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U= \u03b5 \u2013 r.i=12 \u2013 1.1,2\u00a0 —\u00a0 U=10,8V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Corrente i no motor de equa\u00e7\u00e3o U=E\u2019 + r\u2019.i\u00a0 —\u00a0 1,2=1,0 + 0,2i\u00a0 —\u00a0 i=1 A\u00a0 —\u00a0 tempo que ele demora <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

para dissipar a carga Q=1200mAh=1,2Ah fornecida pelo motor\u00a0 —\u00a0 i=Q\/t\u00a0 —\u00a0 1A = 1,2Ah\/\u2206t\u00a0 — \u2206t=1,2h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>A liga\u00e7\u00e3o em paralelo \u00a0obtida quando se liga entre si todos os p\u00f3los positivos dos geradores (o\u00a0 terminal da uni\u00e3o \u00e9 o de mais alta tens\u00e3o do circuito) e todos os p\u00f3los negativos (o terminal de uni\u00e3o \u00e9 o de mais baixa tens\u00e3o do circuito). Nesse tipo de associa\u00e7\u00e3o conv\u00e9m que todos os geradores tenham a mesma for\u00e7a eletromotriz E, para que n\u00e3o dissipem parte da energia que fornecem ao circuito externo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Caracter\u00edsticas desse tipo de associa\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A intensidade de corrente el\u00e9trica se subdivide entre os geradores<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A for\u00e7a eletromotriz da associa\u00e7\u00e3o \u00e9 igual \u00e0quela de cada um dos geradores\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span><\/p>\n

Observe que, quando voc\u00ea liga as duas baterias do exerc\u00edcio em paralelo (figura I) a tens\u00e3o do sistema \u00e9 a mesma que a da bateria carregada\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Utilizando a primeira lei de Ohm (R=V\/i) voc\u00ea pode completar a tabela\u00a0 —\u00a0 quando V=1,10V e R=4,40\u2126\u00a0 —\u00a0 4,40 = 1,10\/i\u00a0 —\u00a0 i=0,25 A\u00a0 —\u00a0\u00a0 quando V=0,96V e R=1,60\u2126\u00a0 —\u00a0 1,60 =0,96\/i\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>i=0,60 A.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) <\/b><\/span><\/span><\/span>Construindo o gr\u00e1fico a partir dos valores da tabela:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) <\/b><\/span><\/span><\/span>Voc\u00ea pode obter a for\u00e7a eletromotriz do gerador, prolongando a reta representativa at\u00e9 tocar o eixo das tens\u00f5es\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

este ponto fornece a for\u00e7a eletromotriz \u03b5 que no caso \u00e9 aproximadamente de \u03b5\u22481,20V\u00a0 —\u00a0 a resist\u00eancia interna r pode ser determinada pela equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 V = E \u2013 r.i\u00a0 —\u00a0 quando, por exemplo V=0,96V, i=0,6 A\u00a0 —\u00a0 0,96 = 1,2 \u2013 r.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>r\u22480,4\u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

Circuitos\u00a0 com geradores <\/b><\/span><\/span><\/span><\/p>\n

\u00a029-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Observe na seq\u00fc\u00eancia abaixo o c\u00e1lculo da resist\u00eancia do resistor equivalente (R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 \u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 4,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=12\/i\u00a0\u00a0\u00a0 —\u00a0 i=12\/4,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>=2,7.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>A<\/b><\/span><\/span><\/span>=2,7mA.<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Observe na figura abaixo os valores das correntes devidamente distribu\u00eddas\u00a0 —\u00a0 entre os pontos B <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

e C, R=1k\u2126 e i=1,35.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0A\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 1.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=U\/1,35.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 U=1,35.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>U=1,35V.<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na sequ\u00eancia abaixo o c\u00e1lculo da resist\u00eancia do resistor equivalente\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

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

1,5=3\/i\u00a0 —\u00a0 i=2A\u00a0 —entre os pontos A e B flui uma corrente i=2A na resist\u00eancia de R=1\u2126\u00a0 —\u00a0 R=U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=1.2=2V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da corrente el\u00e9trica i que percorre o resistor de 4\u2126 quando dissipa uma pot\u00eancia de 36W\u00a0 —\u00a0 P=R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 36=4.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 i=3A\u00a0 —\u00a0 observe que a resist\u00eancia equivalente (R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>) \u00e9 percorrida pela corrente i=3A quando submetida \u00e0 ddp (tens\u00e3o) de U=24V\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=U\/i=24\/3=8\u2126\u00a0 —\u00a0 observe na \"\"<\/b><\/span><\/span><\/span><\/p>\n

sequ\u00eancia abaixo que, como R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=8\u2126, a resist\u00eancia R\u2019 (que corresponde a R associado em paralelo com 12\u2126) deve valer R\u2019=8 – (1 + 4)\u00a0 =3\u2126\u00a0 —\u00a0 3=produto\/soma=12R\/(12 + R)\u00a0 —\u00a0 3=12R\/(12 + R)\u00a0 —\u00a0 36 + 3R=12R\u00a0 —\u00a0 9R=36\u00a0 —\u00a0 R=4\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

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

Receptores e associa\u00e7\u00e3o de geradores e de receptores <\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U=E\u2019 + r.i\u00a0 —\u00a0 pelo gr\u00e1fico\u00a0 —\u00a0 quando U=45V, i=8 A\u00a0 —\u00a0 45=E\u2019 + r.8 (I)\u00a0 —\u00a0\u00a0 quando U=40V, i=3 A\u00a0 —\u00a0 40=E\u2019 + r.3 (II)\u00a0 —\u00a0 multiplicando (II) por (-1) e somando com (I)\u00a0 —\u00a0 5=5r\u00a0 —\u00a0 r=1\u2126\u00a0 —\u00a0 subst. Em (I)\u00a0 —\u00a0 45=E\u2019 + 1.8\u00a0 —\u00a0 E\u2019=37V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A) <\/b><\/span><\/span><\/span>A tens\u00e3o U que a bateria de fem E e resist\u00eancia interna r fornece ao circuito externo\u00e9 dada pela equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 do gr\u00e1fico\u00a0 —\u00a0 quando U=10V, i=4 A\u00a0 —\u00a0 10=E \u2013 r.4 (I)\u00a0 —\u00a0 quando U=8V, i=8 A\u00a0 —\u00a0 8=E \u2013 r.8 (II)\u00a0 —\u00a0 resolvendo (I) com (II)\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>E=12V e r=0,5\u2126.<\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>Calculado em (A)\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>r=0,5\u2126.<\/b><\/span><\/span><\/span><\/p>\n

C) <\/b><\/span><\/span><\/span>Para essa particular l\u00e2mpada a corrente I=6,0A\u00a0 —\u00a0 U=12 \u2013 0,5.6=12 \u2013 3=9V\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R=9\/6=1,5\u2126.<\/b><\/span><\/span><\/span><\/p>\n

D) <\/b><\/span><\/span><\/span>Pot\u00eancia dessa particular l\u00e2mpada de R=1,5\u2126 e I=6 A\u00a0 —\u00a0 P=R.I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=1,5.36\u00a0 —\u00a0 P=54W\u00a0 —\u00a0 o rendimento do gerador (bateria) \u00e9 fornecido por \u03b7=P<\/b><\/span><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=Ui\/Ei=9\/12=0,0,75\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>\u03b7=75%<\/b><\/span><\/span><\/span><\/p>\n

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

Aparelhos de medi\u00e7\u00e3o el\u00e9trica (amper\u00edmetros, volt\u00edmetros, ponte de Wheatstone)<\/b><\/span><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, leia as informa\u00e7\u00f5es a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Ponte de Wheatstone\u00a0\u2013 trata-se de uma associa\u00e7\u00e3o de tr\u00eas resistores fixos e um vari\u00e1vel (reostato), como representado na figura.<\/b><\/span><\/span><\/span><\/p>\n

Serve, dentre outras coisas, para determinar a resist\u00eancia de um resistor. Conhecendo, por exemplo, R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e R<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, pode-se determinar R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>, da seguinte maneira:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Liga-se entre B e D um amper\u00edmetro muito sens\u00edvel (galvan\u00f4metro)<\/b><\/span><\/span><\/span><\/p>\n

\"\"Varia-se o reostato (resist\u00eancia vari\u00e1vel) R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 que a intensidade i<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0de corrente no amper\u00edmetro se torne nula (i<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0). Nessa situa\u00e7\u00e3o, diz-se que\u00a0a ponte est\u00e1 em equil\u00edbrio. Sendo\u00a0nula a corrente no amper\u00edmetro, n\u00e3o existe diferen\u00e7a de potencial entre os pontos B e D.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>AD<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0(I)\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span>=U<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0 Dividindo membro a membro (I) por (II)\u00a0 —\u00a0 (R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)=(R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>)\/(R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i\u2019)\/(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i\u2019)=(R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.i\u2019\u2019)\/(R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>.i\u2019\u2019)\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span>\u00ad\u00a0\u00a0— \u00a0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\"\"<\/p>\n

Observe no circuito fornecido que, como as l\u00e2mpadas s\u00e3o id\u00eantica, possuem a mesma resist\u00eancia R\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>IV<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 assim, trata-se de uma ponte de Wheatstone em equil\u00edbrio e pela l\u00e2mpada de resist\u00eancia R<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>\u00a0n\u00e3o passa corrente el\u00e9trica\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os aparelhos est\u00e3o ligados corretamente\u00a0 —\u00a0 o amper\u00edmetro em s\u00e9rie e o volt\u00edmetro em paralelo\u00a0 —\u00a0 considerando os aparelhos ideais, se o amper\u00edmetro estivesse ligado em paralelo voc\u00ea teria um curto circuito e, se o volt\u00edmetro estivesse ligado em s\u00e9rie n\u00e3o passaria corrente pelo condutor\u00a0 —\u00a0 o resistor n\u00e3o \u00e9 \u00f4hmico, pois R=U\/i n\u00e3o \u00e9 constante para todas as medidas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

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

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea quer medir a intensidade da corrente, por exemplo, em uma l\u00e2mpada, voc\u00ea deve inserir o amper\u00edmetro no trecho onde ela est\u00e1, pois ele \u201cl\u00ea\u201d a corrente que passa atrav\u00e9s dele\u00a0 —\u00a0 assim o amper\u00edmetro deve ser associado em s\u00e9rie no trecho onde voc\u00ea deseja medir a corrente\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

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

Capacitores – condensadores<\/b><\/span><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>S\u00e9rie\u00a0 —\u00a0 1\/Ceq=1\/C + 1\/C + 1\/C\u00a0 —\u00a0 2p=C\/3\u00a0 —\u00a0 C=6p\u00a0 —\u00a0 paralelo\u00a0 —\u00a0 Ceq=6p + 6p=12pF\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- C.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na seq\u00fc\u00eancia abaixo, em (I) foi calculada a capacit\u00e2ncia equivalente de cada linha (associadas em s\u00e9rie) e calculadas por 1\/C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=1\/C + 1\/C + 1\/C + 1\/C, cujos valores est\u00e3o fornecidos em (II)\u00a0 —\u00a0 em (II) eles est\u00e3o associados em paralelo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=30 + 20 + 10 + 5=65 \u03bcF)\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>= 65 \u03bcF\u00a0 —\u00a0 a energia el\u00e9trica acumulada entre os terminais \u00e9 fornecida por\u00a0 —\u00a0 W=C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>.U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=65.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.500<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=8125000.10-6J\u00a0 —\u00a0 W=8,125J\u00a0 —\u00a0\u00a0R- B.<\/b><\/span><\/span><\/span><\/p>\n

54-\u00a0Observe na seq\u00fc\u00eancia a seguir o c\u00e1lculo do capacitor equivalente (C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0est\u00e3o em <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

s\u00e9rie\u00a0 —\u00a0C<\/b><\/span><\/span><\/span>eq1<\/b><\/span><\/span><\/sub><\/span>=20.30\/(20 + 30)\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>eq1<\/b><\/span><\/span><\/sub><\/span>=600\/50=12\u03bcF\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>, C<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>\u00a0e C<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sub><\/span>\u00a0est\u00e3o em s\u00e9rie\u00a0 —\u00a0 1\/C<\/b><\/span><\/span><\/span>eq2<\/b><\/span><\/span><\/sub><\/span>=1\/40 + 1\/50 + 1\/60=(15 + 12 + 10)\/600\u00a0 —\u00a0\u00a0 C<\/b><\/span><\/span><\/span>eq2<\/b><\/span><\/span><\/sub><\/span>=600\/37=16,2\u03bcF\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>eq1<\/b><\/span><\/span><\/sub><\/span>, C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e C<\/b><\/span><\/span><\/span>eq2<\/b><\/span><\/span><\/sub><\/span>\u00a0 est\u00e3o em paralelo\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=12 + 16,2 + 10= 38,2\u03bcF\u00a0 —\u00a0 C<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>=Q\/U\u00a0 —\u00a0 38,2.10<\/b><\/span><\/span><\/span>-6\u00a0<\/b><\/span><\/span><\/sup><\/span>=Q\/110\u00a0 —\u00a0 Q=4202\u03bcC\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

Resolu\u00e7\u00f5es Vestibulares Recentes Eletrodin\u00e2mica \u2013 2014 – 2013 01-\u00a0C\u00e1lculo da resist\u00eancia el\u00e9trica\u00a0R\u00a0da l\u00e2mpada pelas\u00a0especifica\u00e7\u00f5es\u00a0—\u00a0P=U2\/R\u00a0— 24=122\/R — R=144\/24 —\u00a0R=6\u2126. Sendo essa\u00a0R=6 \u2126 constante, quando a l\u00e2mpada estiver\u00a0sob ddp (tens\u00e3o) de 6V\u00a0a\u00a0corrente i\u00a0que circula por ela ser\u00e1 —\u00a0R=U\/i\u00a0— 6=6\/i —\u00a0i=1 A. Pot\u00eancia el\u00e9trica dissipada\u00a0pela l\u00e2mpada quando sob\u00a0ddp de U=6V\u00a0—\u00a0P=R.i2\u00a0= 6.12\u00a0—\u00a0P=6W. R- A 02-\u00a0a) Observe na figura fornecida que a\u00a0resist\u00eancia equivalente\u00a0Req\u00a0est\u00e1 entre os<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3308,"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-3310","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3310","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=3310"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3310\/revisions"}],"predecessor-version":[{"id":3313,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3310\/revisions\/3313"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3308"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}