{"id":1835,"date":"2016-01-25T00:43:26","date_gmt":"2016-01-25T00:43:26","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1835"},"modified":"2024-08-21T13:06:58","modified_gmt":"2024-08-21T13:06:58","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-circuitos-com-geradores","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/resolucao-comentada-de-circuitos-com-geradores\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-circuitos-com-geradores\/","title":{"rendered":"Circuitos com geradores – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

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

01<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0Os dois resistores est\u00e3o em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=(R + 20)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 (R + 20)=6\/0,1\u00a0 —\u00a0 R + 20=60\u00a0 —\u00a0 R=40\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0a) Com a chave C fechada, L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e L<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o em paralelo\u00a0 —\u00a0 R\u2019=30\/2=15\u2126\u00a0 —\u00a0 R\u2019fica em s\u00e9rie com a resist\u00eancia de R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=30\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=15 + 30\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=45\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 45=18\/i\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i=0,4A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>chave fechada\u00a0 —\u00a0 corrente que passa por i<\/b><\/span><\/span>L1<\/b><\/span><\/span><\/sub>==0,4A e por L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>L2<\/b><\/span><\/span><\/sub>=0,4\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i<\/b><\/span><\/span>L2<\/b><\/span><\/span><\/sub>=0,2A<\/b><\/span><\/span><\/p>\n

b) Chave aberta\u00a0 —\u00a0 s\u00f3 passa corrente por L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e por L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, que est\u00e3o em s\u00e9rie\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=30 + 30\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=60\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 60=18\/i<\/b><\/span><\/span>L1<\/b><\/span><\/span><\/sub>=\u00a0 —\u00a0 i<\/b><\/span><\/span>L1<\/b><\/span><\/span><\/sub>=18\/60\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>i<\/b><\/span><\/span>L1<\/b><\/span><\/span><\/sub>=0,3A\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 chave aberta\u00a0 —\u00a0 por L<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>\u00a0passa 0,3A\u00a0\u00a0 —\u00a0 chave fechada\u00a0 —\u00a0 por L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0passa 0,4A —\u00a0 abrindo-se a chave, por<\/b><\/span><\/span>L<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>passa menor corrente e ela brilha menos<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0a) Chave C aberta\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 2R =12\/i\u00a0 —\u00a0 i=6\/R\u00a0\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R.i\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=(6\/R).R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=6V<\/b><\/span><\/span><\/p>\n

b) Chave C fechada\u00a0 —\u00a0 observe a sequ\u00eancia abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 3R\/2=12\/i\u2019\u00a0 —\u00a0 i\u2019=24\/3R\u00a0 —\u00a0 i\u2019=8\/R\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R.i\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R\/2.8\/R\u00a0 —<\/b><\/span><\/span>\u00a0U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=4V<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Quando R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0os pontos A e B est\u00e3o em curto circuito e<\/b><\/span><\/span>\u00a0U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span><\/p>\n

Quando R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=6\u2126\u00a0 —\u00a0 observe as seq\u00fc\u00eancias abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 4=1,2\/i\u00a0 —\u00a0 i=0,3A\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R.i=2.0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=0,6V<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0O \u201cfio terra\u201d indica que o potencial no ponto B \u00e9 nulo (V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0\u00ad)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1 + 2 + 4\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=7\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 7=14\/i\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=4.2\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 0=8\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0\u00a0V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=8V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>.i\u00a0 —\u00a0 0 \u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=2.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>= – 4V\u00a0 —\u00a0 R- D\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Obs: Se a polaridade do gerador fosse invertida, o sentido da corrente tamb\u00e9m seria invertido e os valores seriam V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= – 8V e V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0A liga\u00e7\u00e3o terra torna o potencial do ponto A nulo\u00a0 —\u00a0 (V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0)\u00a0 —\u00a0 c\u00e1lculo de i\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 12=12\/i\u00a0 —\u00a0 i=1A\u00a0 —\u00a0 chamando o terminal negativo do gerador de C e o positivo de B\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 U<\/b><\/span><\/span>AC<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=R.i\u00a0 —\u00a0 0 \u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=9.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>= – 9V\u00a0 — \u00a0R- C\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Obs.\u00a0 —\u00a0 observe que\u00a0 —\u00a0 U<\/b><\/span><\/span>BA<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R.i\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 0=3.1\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=3V\u00a0 —\u00a0 observe ainda que V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=3 \u2013 (-9)=12V<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0a) No fio neutro o potencial \u00e9 nulo\u00a0 —\u00a0\u00a0\u00a0 diferen\u00e7a de potencial (tens\u00e3o) entre os pontos A e P (quatro l\u00e2mpadas de cima) \u00e9 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=110 \u2013 0=110V, e nas duas l\u00e2mpadas de baixo (entre P e B) \u00e9 V<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0 \u2013 (-110)=110V\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- 110V<\/b><\/span><\/span><\/p>\n

b) Se o fio quebrar em P\u00a0 —\u00a0 curto circuito\u00a0 —\u00a0 refazendo o circuito (figuras abaixo)\u00a0 —\u00a0 as 4 l\u00e2mpadas de cima est\u00e3o em paralelo (r\/4) e as duas de baixo tamb\u00e9m est\u00e3o em paralelo (r\/2)\u00a0 –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=110 \u2013 (- 110)\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=220V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3r\/4\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 3r\/4=220\/i\u00a0 —\u00a0 i=4.220\/3r\u00a0 —\u00a0 i=880\/3r\u00a0 —\u00a0 tens\u00e3o (U<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>) na parte de baixo (duas l\u00e2mpadas em paralelo R<\/b><\/span><\/span>rq<\/b><\/span><\/span><\/sub>=r\/2) vale \u00a0—\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 r\/2=U<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>\/880\/3r\u00a0 —\u00a0 U<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>=r\/2×880\/3r\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>=147V<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=0,1 + 1,9 + (18\/3) + 2\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=10\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 10=12\/i\u00a0 —\u00a0 i=1,2 A\u00a0 —\u00a0 veja nas seq\u00fc\u00eancias abaixo que a corrente que flui pelo resistor de 9\u2126 \u00e9 i\u2019\u2019=0,8 A\u00a0 —P=i.U\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=0,8.7,2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=5,76W<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0a) chave fechada\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=r + 3,5\u00a0 —\u00a0 i=3 A\u00a0 —\u00a0 U=\u00a0\u03b5=12V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 r + 3,5=12\/3\u00a0 —<\/b><\/span><\/span>\u00a0r=0,5\u2126<\/b><\/span><\/span><\/p>\n

b) equa\u00e7\u00e3o do gerador (bateria)\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 U=12 \u2013 0,5.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=10,5V<\/b><\/span><\/span><\/p>\n

c) P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>=U.i=10,5.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>\u00fatil<\/b><\/span><\/span><\/sub>=31,5W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b7=U\/\u00a0\u03b5=10,5\/12\u00a0 —\u00a0 \u03b7=0,875\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b7=87,5%<\/b><\/span><\/span><\/p>\n

d) pot\u00eancia liberada (dissipada) pelo resistor\u00a0 —\u00a0 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3,5.(3)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=31,5W\u00a0 —\u00a0 energia dissipada na \u00e1gua em \u0394t=10min=10.60\u00a0 — \u0394 t=600s\u00a0 —\u00a0 W=P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>.\u0394t=31,5.600\u00a0 —<\/b><\/span><\/span>\u00a0W=18.900J<\/b><\/span><\/span><\/p>\n

e) como 1J=0,24cal\u00a0 —\u00a0 W=18.900.x0,24\u00a0 —\u00a0 W=Q=4.356cal\u00a0 —\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0 4.356=240.1. \u0394\u03b8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394\u03b8=18,15<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 4\/3=\u00a0\u03b5\/9\u00a0 —\u00a0\u00a0\u03b5=12V<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da resist\u00eancia de cada l\u00e2mpada\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 0,5=(1)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=2\u2126\u00a0 —\u00a0 Chave aberta\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 6 + 2r=(1,5 + 1,5)\/0,3\u00a0 —\u00a0 2r=4\u00a0 —\u00a0 r=2\u2126\u00a0 —\u00a0 chave fechada\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 6 + 2=3\/i\u00a0 —\u00a0 i=3\/8=0,375\u00a0 —\u00a0 i=375mA\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A ddp U que o gerador fornece ao circuito esterno (resist\u00eancia R=12\u2126) vale\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 12=U\/2\u00a0 —\u00a0 U=24V\u00a0 —\u00a0 \u03b7=80%=0,8\u00a0 —\u00a0 \u03b7=U\/E\u00a0 —\u00a0 0,8=24\/E\u00a0 —\u00a0 E=24\/0.8\u00a0 —\u00a0 E=30V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 24=30 \u2013 r.2\u00a0 —\u00a0 r=3,0\u2126\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R= E<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Chave aberta\u00a0 —\u00a0 como as l\u00e2mpadas s\u00e3o id\u00eanticas, a tens\u00e3o V em cada l\u00e2mpada \u00e9 a mesma\u00a0 —\u00a0 chave fechada\u00a0 —\u00a0 como a tens\u00e3o em L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0continua a mesma, a tens\u00e3o em L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>tamb\u00e9m ser\u00e1 a mesma e as duas l\u00e2mpadas continuar\u00e3o com o mesmo brilho\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>C\u00e1lculo da resist\u00eancia da l\u00e2mpada\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 0,6=9\/R\u00a0 —\u00a0 R=15\u2126\u00a0 —\u00a0 corrente na l\u00e2mpada\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 15=2,5\/i\u00a0 —\u00a0 i=0,1666A\u00a0 —\u00a0 equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0\u2013 r<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>.i\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

2,5=<\/b><\/span><\/span>(1,5 + 1,5) \u2013 2r.0,1666\u00a0 —\u00a0 r=0,5\/0,333\u00a0 —\u00a0 r=1,5\u2126\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>As tr\u00eas pilhas em s\u00e9rie fornecem uma fem total de E=4,5V\u00a0 —\u00a0 como todos os elementos est\u00e3o em s\u00e9rie a tens\u00e3o total fornecida (E=4,5V0 \u00e9 igual\u00a0 \u00e0 soma das tens\u00f5es parciais\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

E= U<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0+U<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4,5=4 + U<\/b><\/span><\/span>R\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 0,5=R.i\u00a0 —\u00a0 0,5=1.i\u00a0 —\u00a0 i=0,5 A\u00a0 —\u00a0 pot\u00eancia do motor\u00a0 —\u00a0 P=U.i=4.0,5\u00a0 —\u00a0 P=2,0W\u00a0 —<\/b><\/span><\/span>–\u00a0 R- D<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0A resist\u00eancia equivalente \u00e9 a de 6 salsichas em paralelo\u00a0 —\u00a0 Req=440.000\/6\u2126\u00a0 —\u00a0 Pot\u00eancia empregada pelo sistema\u00a0 —\u00a0 P=U2\/Req(110)2\/(440.000\/6)\u00a0 —\u00a0 P=0,165W\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

energia em 20s\u00a0 —\u00a0 W=P.\u0394t=0,165.20\u00a0 —\u00a0 W=3,3J\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0a) A pot\u00eancia P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u00e9 dada por\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>\uf020<\/b><\/span><\/span>\u00a0—\u00a0 \u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= (120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 1200 W<\/b><\/span><\/span><\/p>\n

b) De acordo com o enunciado P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 2 . P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 2R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=12\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4\u2126 e R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=8\u2126<\/b><\/span><\/span><\/p>\n

c) A pot\u00eancia P \u00e9 dada por\u00a0 —\u00a0 P=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\/P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=(U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\/U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P\/P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=(1\/4 + 1\/8)\/1\/12\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P\/P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4,5<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0a) Como a tens\u00e3o entre X e Y \u00e9 U<\/b><\/span><\/span>XY<\/b><\/span><\/span><\/sub>=16V, a corrente no resistor de 8\u2126 ser\u00e1\u00a0 —\u00a0 R=U<\/b><\/span><\/span>XY<\/b><\/span><\/span><\/sub>\/i\u00a0\u00a0 —\u00a0 8=16\/i\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0 a tens\u00e3o entre os pontos M e N, onde i=2 A vale U<\/b><\/span><\/span>MN<\/b><\/span><\/span><\/sub>=R\u2019.i=(8 + 7).2<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 U<\/b><\/span><\/span>MN<\/b><\/span><\/span><\/sub>=30V\u00a0 —\u00a0 no resistor de 30\u2126 a tens\u00e3o tamb\u00e9m \u00e9 U<\/b><\/span><\/span>MN<\/b><\/span><\/span><\/sub>=30V\u00a0 —\u00a0 R=U<\/b><\/span><\/span>MN<\/b><\/span><\/span><\/sub>\/i\u2019\u00a0 —\u00a0 30=30\/i\u2019\u00a0 —\u00a0 i\u2019=1 A\u00a0 —\u00a0 i<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=2 + 1\u00a0 —\u00a0 i<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=3 A\u00a0 —\u00a0\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=15 paralelo com 30(10\u2126) + 30\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=40\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 40=E\/3\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E=120V<\/b><\/span><\/span><\/p>\n

b) E=240V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=40\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 40=240\/i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=6 A\u00a0 —\u00a0 como uma resist\u00eancia \u00e9 o dobro da outra, as correntes nos ramos de 30\u2126 e (7\u2126 e 8\u2126) ser\u00e3o o dobro uma da outra de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0maneira que sua soma seja 6 A\u00a0 —\u00a0 i=4 A e i\u2019= 2 A\u00a0 —\u00a0 U<\/b><\/span><\/span>XY<\/b><\/span><\/span><\/sub>=R.i=8.4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0U<\/b><\/span><\/span>XY<\/b><\/span><\/span><\/sub>=32V<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>C\u00e1lculo da tens\u00e3o U em L<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0onde i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=500mA=0,5 A\u00a0 —\u00a0 P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.U\u00a0 —\u00a0 2=0,5.U\u00a0 —\u00a0 U=4V\u00a0 —\u00a0 esta tens\u00e3o (U=4V) \u00e9 a mesma para as tr\u00eas l\u00e2mpadas (est\u00e3o em paralelo)\u00a0 —\u00a0 L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.U\u00a0 —\u00a0 2=i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.4\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,5 A\u00a0 —\u00a0 L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.U\u00a0 —\u00a0 1=i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.4\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0,25 A\u00a0 —\u00a0 i=i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=0,5 + 0,5 + 0,25\u00a0 —\u00a0 i=1,25 A\u00a0 —\u00a0 equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 4=E \u2013 0,2.1,25\u00a0 —\u00a0 E=4,25V\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Do enunciado\u00a0 —\u00a0 U=E\/3\u00a0 —\u00a0 E=3U\u00a0 —\u00a0 U=R.i\u00a0 —\u00a0 U=6i\u00a0 —\u00a0 E=3.U\u00a0 —\u00a0 E=3.6i=18i\u00a0 —\u00a0 equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0\u00a0 6i= 18i \u2013 r.i\u00a0 —\u00a0 r=12\u2126\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=3R paralelo com 3R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=1,5R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 1,5R=\u00a0\u03b5\/i\u00a0 —\u00a0 i=\u00a0\u03b5\/1,5R\u00a0 —\u00a0 em cada ramo de resist\u00eancias iguais (3R) a corrente i=\u00a0\u03b5\/1,5R se divide pela metade\u00a0 —\u00a0\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u00a0\u03b5\/3R e<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03b5\/3R\u00a0 — V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R.\u00a0\u03b5\/3R\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u00a0\u03b5\/3 ( I )\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2R.\u00a0\u03b5\/3R\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 2\u03b5\/3 ( II )\u00a0 —\u00a0 subtraindo (II) de (I)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 (V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)= 2\u03b5\/3 \u2013\u00a0\u03b5\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u03b5\/3<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Falsa\u00a0 —\u00a0 L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00e9 percorrida por corrente maior que a de L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e de L<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sub>e brilha coma maior intensidade\u00a0 (i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>+ i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

b) Correta\u00a0 —\u00a0 sendo as l\u00e2mpadas id\u00eanticas\u00a0 —\u00a0\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=i\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>+ i<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i + i\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2i e i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 s\u00e3o iguais<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 apenas L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e L<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0est\u00e3o submetidas ao mesmo potencial<\/b><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 a pot\u00eancia dissipada por L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00e9 maior que a dissipada por L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, pois L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00e9 percorrida por maior corrente<\/b><\/span><\/span><\/p>\n

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

23-<\/b><\/span><\/span><\/span>\u00a001. Falsa\u00a0 —\u00a0 l\u00e2mpadas id\u00eanticas em s\u00e9rie\u00a0 —\u00a0 percorridas pela mesma corrente\u00a0 —\u00a0 brilham por igual<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 veja 01<\/b><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 brilham por igual\u00a0 —\u00a0 mesma ddp e mesma corrente<\/b><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 \u00e9 percorrida por maior corrente, pois tem menor resist\u00eancia<\/b><\/span><\/span><\/p>\n

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

Corretas\u00a0 —\u00a0 02 e 08<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Colocando os valores e construindo o gr\u00e1fico<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

26-<\/b><\/span><\/span><\/span> Observe que, como a corrente total \u00e9 1,0A e se, por R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0passa 0,5 A pelo resistor de 0,5\u2126 passar\u00e1 tamb\u00e9m 0,5 A\u00a0 —\u00a0 por R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

passar\u00e1 1,0 A e por R<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=0,75\u00aa\u00a0 —\u00a0 como R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e 8\u2126 est\u00e3o em paralelo eles est\u00e3o submetidos \u00e0 mesma diferen\u00e7a de potencial U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, tal que\u00a0 —\u00a0 R=U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/i\u00a0 —\u00a0 8=U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/0,5\u00a0 —\u00a0 U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4V\u00a0 —\u00a0 em B\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 12=1.U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12V\u00a0 —\u00a0 U<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ U<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= U<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=24V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

4 + 12 + U<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=24\u00a0 —\u00a0 U<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=8V\u00a0 —\u00a0 a pot\u00eancia el\u00e9trica dissipada no resistor R<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0vale\u00a0 —\u00a0 P<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=0,75.8\u00a0 —\u00a0 P<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=6W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span> Modo A\u00a0 —\u00a0 chave C\u00a0 ligada no ponto A\u00a0 —\u00a0 como as l\u00e2mpadas est\u00e3o ligadas em s\u00e9rie, cada uma delas ficar\u00e1 sob ddp de 60V , dissipando uma pot\u00eancia de 60W —\u00a0 c\u00e1lculo da resist\u00eancia el\u00e9trica R de cada uma\u00a0 —\u00a0 P=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 60=60<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=60\u2126\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=60 + 60=120\u2126\u00a0 —\u00a0 pot\u00eancia total dissipada\u00a0 —\u00a0 P<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=120<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/120\u00a0 —\u00a0 P<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=120W\u00a0 —\u00a0 energia el\u00e9trica consumida em 4h\u00a0 —\u00a0 P<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=W\/\u2206t\u00a0 —\u00a0 120=W\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=480Wh=0,48kWh<\/b><\/span><\/span><\/p>\n

Modo B\u00a0 —\u00a0 chave C ligada ao ponto B\u00a0 —\u00a0 L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0fica apagada (n\u00e3o passa corrente por ela)\u00a0 —\u00a0 para que a l\u00e2mpada L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

resist\u00eancia R<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=60\u2126 dissipe uma pot\u00eancia de 15W, a corrente atrav\u00e9s dela deve ser\u00a0 —\u00a0 P<\/b><\/span><\/span>L2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 15=60.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 i=\u221a(0,25)\u00a0 —<\/b><\/span><\/span><\/p>\n

i=0,5 A (\u00e9 a mesma corrente que passa por R, R e 60\u2126 est\u00e3o em s\u00e9rie)\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=(R + 60)\u00a0 —\u00a0 U=120V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R + 60 = 120\/0,5\u00a0 —\u00a0 R + 60=240\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=180\u2126<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\u00a0\u00a0a) Aquecedor\u00a0 —\u00a0 P<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=990W\u00a0 —\u00a0 \u2206t=3h\u00a0 —\u00a0 P<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\/\u2206t<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 990=W<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 W<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=2.970 Wh\u00a0 —\u00a0 ferro\u00a0 —\u00a0 P<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>=980W\u00a0 — \u00a0\u2206t=3h\u00a0 —\u00a0 P<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>\/\u2206t<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 980=W<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 W<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>=2.940 Wh\u00a0 —\u00a0 2 l\u00e2mpadas\u00a0 —\u00a0 P<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=2.60=120W\u00a0 — \u2206t=3h\u00a0 —\u00a0 P<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/\u2206t<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 120=W<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 W<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=360 Wh\u00a0 —\u00a0 chuveiro\u00a0 —\u00a0 P<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>=4.400W\u00a0 —\u00a0 \u2206t=12min=12\/60=0,2h\u00a0 —\u00a0 P<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>\/\u2206t<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4.400=W<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>\/0,2\u00a0 —\u00a0 W<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>=880 Wh\u00a0\u00a0 —\u00a0 energia total\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=2.970 + 2.940 + 360 + 880\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=7.150Wh\u00a0 —\u00a0 W<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=7,15kWh.<\/b><\/span><\/span><\/p>\n

b) C\u00e1lculo da intensidade de corrente el\u00e9trica em cada aparelho\u00a0 —\u00a0 aquecedor\u00a0 —\u00a0 P<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 990=i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>.110\u00a0 —\u00a0 i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>=9 A\u00a0 —\u00a0 ferro\u00a0 —\u00a0 P<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 980=i<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>.110\u00a0 —\u00a0 i<\/b><\/span><\/span>fe<\/b><\/span><\/span><\/sub>=98\/11 A\u00a0 —\u00a0 cada l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 60=i<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.110\u00a0 —\u00a0 i<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=6\/11 A\u00a0 —\u00a0 chuveiro\u00a0 —\u00a0 P<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>.U<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4.400=i<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>.220\u00a0 —\u00a0 i<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>=20,0 A\u00a0 —\u00a0 observe atentamente o esquema da figura e o circuito abaixo onde foi tirado o fio terra, j\u00e1 que por<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

ele n\u00e3o passa corrente el\u00e9trica\u00a0 — a corrente i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0no fio fase 1 se divide nas correntes que passam pelo aquecedor e pelo chuveiro, ou seja, i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\u00a0+ i<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub>= 9 + 20=29,0 A\u00a0 — a corrente i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0no fio fase 2 recebe as correntes que passam pelo ferro,\u00a0 pelas duas l\u00e2mpadas e pelo chuveiro, ou seja, i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= 6\/11 + 6\/11 + 98\/11 + 20\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=29,98\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u224830 A.<\/b><\/span><\/span><\/p>\n

c) Observe nas figuras abaixo que no ponto P chegam as correntes do fio neutro i<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0e do aquecedor i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\u00a0e e sai a corrente i\u2019 que corresponde<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e0s somas das correntes que passam pelo ferro e pelas l\u00e2mpadas\u00a0 —\u00a0 i<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>+i<\/b><\/span><\/span>aq<\/b><\/span><\/span><\/sub>\u00a0= i\u2019\u00a0 —\u00a0 i\u2019=98\/11 + 6\/11 + 6\/11=10 A\u00a0 —\u00a0 i<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0+ 9=10\u00a0 — \u00a0i<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>=1 A.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Circuitos com geradores \u00a0 01–\u00a0Os dois resistores est\u00e3o em s\u00e9rie\u00a0 —\u00a0 Req=(R + 20)\u00a0 —\u00a0 Req=U\/i\u00a0 —\u00a0 (R + 20)=6\/0,1\u00a0 —\u00a0 R + 20=60\u00a0 —\u00a0 R=40\u2126\u00a0 —\u00a0\u00a0R- C\u00a0\u00a0 02-\u00a0a) Com a chave C fechada, L2\u00a0e L3\u00a0est\u00e3o em paralelo\u00a0 —\u00a0 R\u2019=30\/2=15\u2126\u00a0 —\u00a0 R\u2019fica em s\u00e9rie com a resist\u00eancia de R1=30\u2126\u00a0 —\u00a0 Req=15 + 30\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1833,"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-1835","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1835","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=1835"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1835\/revisions"}],"predecessor-version":[{"id":10699,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1835\/revisions\/10699"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1833"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}