{"id":1873,"date":"2016-02-11T17:51:11","date_gmt":"2016-02-11T17:51:11","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1873"},"modified":"2024-08-21T13:09:26","modified_gmt":"2024-08-21T13:09:26","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-receptores-associacao-de-geradores-com-receptores-lei-de-pouillet","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/receptores-associacao-de-geradores-com-receptores-lei-de-pouillet\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-receptores-associacao-de-geradores-com-receptores-lei-de-pouillet\/","title":{"rendered":"Receptores \u2013 Associa\u00e7\u00e3o de geradores com receptores \u2013 Lei de Pouillet – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Receptores \u2013 Associa\u00e7\u00e3o de geradores com receptores \u2013 Lei de Pouillet<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>\u00a0Quando i=2A\u00a0 —\u00a0 U=22V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 22=E\u2019 + r\u2019.2 (I)\u00a0 —\u00a0 quando i=5 A\u00a0 —\u00a0 U=25V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 25=E\u2019 + r\u2019.5 (II)\u00a0 —\u00a0 resolvendo o sistema composto por (I) e (II)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0a) C\u00e1lculo da corrente el\u00e9trica no receptor\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=i.U\u00a0 —\u00a0 1.600=i.200\u00a0 —\u00a0 i=8A\u00a0 —\u00a0 P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=r\u2019.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10.(8)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=640W<\/b><\/span><\/span><\/p>\n

b) Pot\u00eancia \u00fatil\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.600=640 + P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=960W\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=E\u2019.i\u00a0 —\u00a0 960=E\u2019.8\u00a0 —<\/b><\/span><\/span>\u00a0 E\u2019=120V<\/b><\/span><\/span><\/p>\n

c) \u03b7=E\u2019\/U=120\/200=0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u03b7=60%<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0a) Equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 50=E\u2019+1.2,5\u00a0 —<\/b><\/span><\/span>\u00a0E\u2019=47,5V<\/b><\/span><\/span><\/p>\n

b) O volt\u00edmetro indica o valor da for\u00e7a contra-eletromotriz do receptor\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=50V<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 440=E\u2019 + 25.0,4\u00a0 —\u00a0 E\u2019=440 \u2013 10\u00a0 —\u00a0 E\u2019=430V\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D\u00a0<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0O gr\u00e1fico do gerador \u00e9 o da reta descendente\u00a0 —\u00a0 icc=20A — icc=E\/r\u00a0 —\u00a0 20=E\/r\u00a0 —\u00a0 E=20r\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o do gerador\u00a0\u00a0 —\u00a0 quando i=10A \u2013 U=20V\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 20=E \u2013 r.10\u00a0 —\u00a0 20=20r \u2013 10r \u00a0—\u00a0 r=2\u2126\u00a0 —\u00a0 E=20r=20.2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E=40V<\/b><\/span><\/span><\/p>\n

O gr\u00e1fico do receptor \u00e9 o da reta ascendente\u00a0 —\u00a0 observe que E\u2019=10V\u00a0 —\u00a0 quando 1=10A \u2013 U=20V\u00a0 —\u00a0 equa\u00e7\u00e3o do receptor\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

U=E\u2019 + r\u2019.i\u00a0 —\u00a0 20=10 +r\u2019.10\u00a0 —\u00a0 r\u2019=10\/10 \u00a0—\u00a0 r\u2019=1\u2126\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Chave aberta\u00a0 —\u00a0 se voc\u00ea ligar um volt\u00edmetro ideal (resist\u00eancia interna infinita) nos terminais de um gerador ele indicar\u00e1 sua for\u00e7a eletromotriz (no caso\u00a0<\/b><\/span><\/span>E=20V<\/b><\/span><\/span>), pois ele funciona como se estivesse em circuito aberto\u00a0 —\u00a0 \u00a0se voc\u00ea ligar um volt\u00edmetro nos terminais de um receptor desligado (chave k aberta) como por exemplo de um liquidificador, ferro el\u00e9trico, etc., ele indicar\u00e1 zero\u00a0 Chave fechada\u00a0 — circular\u00e1 corrente de valor\u00a0 —\u00a0 i=(\u2211E – \u2211E\u2019)\/\u2211R=(20 \u2013 15)\/\/(3 + 2)\u00a0 —\u00a0 i=1A\u00a0 —\u00a0 \u00a0c\u00e1lculo de U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0pelo gerador\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=E \u2013 r.i=20 \u2013 3.1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=17V\u00a0<\/b><\/span><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=E\u2019 + r\u2019.i=15 + 2.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=17V\u00a0\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0a) i=(3 \u2013 1,5)\/((10 + 20)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=0.05A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

b) pelo gerador\u00a0 —\u00a0 U=E \u2013 r.i=3 \u2013 20.0,05\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=2V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>ou, pelo receptor\u00a0 —\u00a0 U=E\u2019 + r\u2019+.i=1,5 + 1=.0,05\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=2V<\/b><\/span><\/span><\/p>\n

c) Haver\u00e1 desgaste da pilha de maior fem, que \u00e9 a de\u00a0<\/b><\/span><\/span>3V<\/b><\/span><\/span>\u00a0 —\u00a0 Observa\u00e7\u00e3o\u00a0 —\u00a0 esse tipo de liga\u00e7\u00e3o \u00e9 inconveniente, pois sempre haver\u00e1 corrente entre elas com a de 3V funcionando como gerador e fornecendo corrente para a de 1,5V que funciona como receptor. Ent\u00e3o, o desgaste \u00e9 inevit\u00e1vel.<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Resolvendo R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0paralelo com R<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sub>paralelo com R<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0para chegar \u00e0 uma malha \u00fanica\u00a0 — R\u2019=2\u2126\u00a0 —\u00a0 i=(4 \u2013 2)\/(2 + 0,5 + 0,5 + 2)\u00a0 —\u00a0 i=2\/5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=0,4 A<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0 U=R\u2019.i=2.0,4\u00a0 —\u00a0<\/b><\/span><\/span>U=0,8V<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Deixando uma malha \u00fanica\u00a0 —\u00a0 20 \u2126 paralelo com (10 \u2126 + R)\u00a0 —\u00a0 R\u2019=20.(10 + R)\/20 + 10 + R\u00a0 —\u00a0 R\u2019=(200 + 20R)\/(30 + R)\u00a0 —\u00a0 i=(55 \u2013 5)\/(200 + 20R)\/(30 + R) + 10\u00a0 —\u00a0 2=50\/(500 + 30R)\/(30 + R)\u00a0 —\u00a0 1.000 + 60R=1.500 + 50R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R=50\u2126<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0Observe que o sentido da corrente \u00e9 no hor\u00e1rio\u00a0 —\u00a0 i=(+6 +6 -2)\/(4 + 2 +4)\u00a0 —\u00a0 i=10\/10\u00a0 —\u00a0 i=1A\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 2=U\/1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>U=2V<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0i=(12 \u2013 6)\/(4 + 2)\u00a0 —\u00a0 i=1A\u00a0 —\u00a0 pelo gerador (E=12V e r=2\u2126)\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=E \u2013 r.i=12 \u2013 2.1\u00a0 —\u00a0 U=10V\u00a0 —\u00a0 ou, pelo receptor (E\u2019=6V e r\u2019=4\u2126)\u00a0 —\u00a0 U=E\u2019 + r\u2019.i=6 + 4.1\u00a0 —\u00a0 U=10V\u00a0 — \u00a0observe que V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u00e9 maior que V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=10\u00a0 —\u00a0 15<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=5V<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0entre A e B\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A\u00ad\u00a0<\/b><\/span><\/span><\/sub>\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=R.i=2.2\u00a0 —\u00a0 U<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=4V\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=4\u00a0 —\u00a0 4 \u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>\u00a0 —\u00a0 gerador\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0> V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 U<\/b><\/span><\/span>CB<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.i=10 \u2013 0,5.2\u00a0 —\u00a0 U<\/b><\/span><\/span>CB<\/b><\/span><\/span><\/sub>=9V\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=9\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 0 = 9\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=9V<\/b><\/span><\/span>\u00a0 —\u00a0 receptor\u00a0 —\u00a0 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0> V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 U<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.i=3 + 1.2\u00a0 —\u00a0 U<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>=5V\u00a0 —\u00a0 U<\/b><\/span><\/span>CD<\/b><\/span><\/span><\/sub>= V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5=9 \u2013 V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=4V<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Observe que V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0(unidos por fio ideal)<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Substituindo os dois geradores de 3V que est\u00e3o em paralelo, o circuito fica com malha \u00fanica e <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

conforme a figura abaixo\u00a0 —i=((1,5+9) \u2013 (3+1,5))\/(1,75 + 1,25)= 6\/3\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>i=2 A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Como eles est\u00e3o associados em s\u00e9rie, a tens\u00e3o nos terminais das pilhas (U<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>=4,5V) \u00e9 igual \u00e0 soma das tens\u00f5es nos terminais da resist\u00eancia R (U\u2019) e do motor (U<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=4V)\u00a0 —\u00a0 4,5=U\u2019 + 4\u00a0 —\u00a0 U\u2019=0,5V\u00a0 —\u00a0 R=U\u2019\/i\u00a0 —\u00a0 1=0,5\/i\u00a0 —\u00a0 i=0,5 A\u00a0 —\u00a0 P<\/b><\/span><\/span>motor<\/b><\/span><\/span><\/sub>=i.U=0,5.4\u00a0 —\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Receptor\u00a0 —\u00a0 figura 1\u00a0 —\u00a0 reta ascendente\u00a0 —\u00a0 E\u2019=5V\u00a0 —\u00a0 i=2 A \u2013 U=9V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 9=5 + r\u2019.2\u00a0 — r\u2019=2\u2126\u00a0 —\u00a0 gerador\u00a0 —\u00a0 reta descendente\u00a0 —\u00a0 figura 2\u00a0 —\u00a0 E=10V\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 20=10\/r\u00a0 —\u00a0 r=0,5\u2126\u00a0 — i=(10 \u2013 5)\/((0,5 + 2)\u00a0 —\u00a0 i=5\/2,5\u00a0 —\u00a0 i=2 A\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I\u00a0 —\u00a0 receptor\u00a0 —\u00a0 E\u2019=20V\u00a0 —\u00a0 i=5A \u2013 U=40V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 40=20 + r\u2019.5\u00a0 —\u00a0 r\u2019=4\u2126\u00a0 —\u00a0 II\u00a0 —\u00a0 resistor\u00a0 —\u00a0 R=U\/i=40\/5\u00a0 —\u00a0 R=8\u2126\u00a0 —\u00a0 III\u00a0 —\u00a0 gerador\u00a0 —\u00a0 E=40V\u00a0 —\u00a0 i=5A\u00a0\u00a0 —\u00a0 i<\/b><\/span><\/span>cc<\/b><\/span><\/span><\/sub>=E\/r\u00a0 —\u00a0 5=40\/r\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

r=8\u2126\u00a0 —\u00a0 montando o circuito\u00a0 —\u00a0i=(40 \u2013 20)\/(8 + 8 + 4)\u00a0 —i=20\/20\u00a0 —\u00a0 i=1A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 U<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>=40 \u2013 8i (I)\u00a0 —\u00a0 equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=E\u2019 + r\u2019.i\u00a0 —\u00a0 <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

U=20 + 4.i (II)\u00a0 —\u00a0 na intersec\u00e7\u00e3o\u00a0 —\u00a0 U<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 substituindo (I) em (II)\u00a0 —\u00a0 40 \u2013 8i=20 + 4i\u00a0 —\u00a0 20=12i\u00a0 —\u00a0 i=20\/12\u00a0 —\u00a0 i=5\/3 A\u00a0 —\u00a0 U=20 + 4.5\/3\u00a0 —\u00a0 U=80\/3V\u00a0 —<\/b><\/span><\/span>\u00a0 R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Gerador\u00a0 —\u00a0 reta descendente\u00a0 —\u00a0 E=100V\u00a0 —\u00a0 i=4 A\u00a0 —\u00a0 U=20V\u00a0 —\u00a0 equa\u00e7\u00e3o do gerador\u00a0 —\u00a0 U=E \u2013 r.i\u00a0 —\u00a0 20=100 \u2013 r.4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>r=20\u2126\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 receptor\u00a0 —\u00a0 E\u2019=40V\u00a0 —\u00a0 i=2 A\u00a0 —\u00a0 U=60V\u00a0 —\u00a0 equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 60=40 + r\u2019.2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0r\u2019=10\u2126\u00a0 —\u00a0<\/b><\/span><\/span>c\u00e1lculo da corrente i pela lei de Pouillet\u00a0 —\u00a0 i=(100 \u2013 40)\/(20 + 10)\u00a0 —\u00a0 i=2 A \u00a0—\u00a0 observe que para essa corrente U \u00e9 comum ao gerador e ao receptor\u00a0 —\u00a0 rendimento do gerador\u00a0 —\u00a0 \u03b7=U\/E=60\/100\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b7<\/b><\/span><\/span>gerador<\/b><\/span><\/span><\/sub>=0,6=60%<\/b><\/span><\/span>\u00a0 —\u00a0 \u03b7<\/b><\/span><\/span>receptor<\/b><\/span><\/span><\/sub>=E\u2019\/U=40\/60\u00a0 —\u00a0<\/b><\/span><\/span>\u03b7<\/b><\/span><\/span>receptor<\/b><\/span><\/span><\/sub>=0,67=67%<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) O circuito est\u00e1 esquematizado abaixo onde est\u00e3o indicados os sentidos de percurso e da <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

corrente\u00a0 —\u00a0 + 11 + 0,1i + 0,1i \u2013\u00a0\u00a012=0 —\u00a0 i=1\/0,2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0i=5 A<\/b><\/span><\/span><\/p>\n

b) Pela parte superior\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=+ 12 \u2013 0,10.5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=11,5V<\/b><\/span><\/span>\u00a0 —\u00a0 ou pela parte inferior\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=+ 11 + 0,10.5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0– V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=11,5V<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe o esquema abaixo\u00a0 —\u00a0 figura a\u00a0 —\u00a0 U=2V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=R\/2\u00a0 —\u00a0 i\u2019=i\/2\u00a0 —\u00a0 i=2i\u2019 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u00a0 —\u00a0 R\/2=2V\/i\u00a0 —\u00a0 i=4V\/R\u00a0 —\u00a0 2i\u2019=4V\/R\u00a0 —\u00a0 i\u2019=2V\/R\u00a0 —\u00a0 figura b\u00a0 —\u00a0 U=V\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=2R\u00a0 —\u00a0 R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>=U\/i\u2019\u2019\u00a0 —\u00a0 2R=V\/i\u2019\u2019\u00a0 —\u00a0 i\u2019\u2019=V\/2R\u00a0 —\u00a0 i\u2019\/i\u2019\u2019=2V\/R x 2R\/V\u00a0 —\u00a0 I\u2019\/i\u2019\u2019=4\u00a0 —<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) O gr\u00e1fico 1 refere-se a um gerador e, portanto, representa a bateria\u00a0 —\u00a0 o \u00a0gr\u00e1fico 2 representa um receptor e, portanto, \u00e9 o ventilador\u00a0 —\u00a0 \u00a0gr\u00e1fico 3 representa um resistor e, portanto, \u00e9 o chuveiro.<\/b><\/span><\/span><\/p>\n

b) O rendimento do receptor, ou seja, do ventilador\u00a0 —\u00a0\u00a0 U = 10 + 2i\u00a0 —\u00a0 para i = 2 A\u00a0 —\u00a0 U = 14 V\u00a0 —\u00a0 rendimento\u00a0 —\u00a0 \u03b7 = 10\/14 \u00a0\u00a0—\u00a0<\/b><\/span><\/span>\u00a0\u03b7 = 71,4%\u00a0<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Equa\u00e7\u00e3o do receptor\u00a0 —\u00a0 U=E\u2019<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>+ r\u2019.i\u00a0 —\u00a0 100=E\u2019 + 2.5\u00a0 —\u00a0 E\u2019=90V\u00a0 —\u00a0 P=U.i=100.5\u00a0 —\u00a0 P=500W\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span> R- E\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01. Correta\u00a0 —\u00a0 se i=0, ou seja, se o gerador estiver em circuito aberto, tem-se:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

02. Correta\u00a0 —\u00a0 se o gerador estiver ligado a uma associa\u00e7\u00e3o de resistores de resist\u00eancia equivalente R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>\u00a0a pot\u00eancia m\u00e1xima fornecida \u00e0 essa<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

associa\u00e7\u00e3o ser\u00e1\u00a0 —\u00a0 U<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=Req.i<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E\/2=R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub>.(E\/2r)\u00a0 — \u00a0r=R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/p>\n

04. Falsa\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 o rendimento el\u00e9trico de um receptor \u00e9 definido pela raz\u00e3o entre sua pot\u00eancia \u00fatil e sua pot\u00eancia consumida\u00a0 —\u00a0 \u03b7=P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03b7=E.i\/U.i\u00a0 —\u00a0 \u03b7=E\/U
\n16- Correta\u00a0 —\u00a0 veja teoria
\n<\/b><\/span><\/span>R- (01 + 02 + 16)=19<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01. Percorrendo i circuito no sentido hor\u00e1rio e supondo a corrente tamb\u00e9m no sentido hor\u00e1rio\u00a0 —\u00a0 i \u2013 15 + 3i + 4i + 5 + i + i=0\u00a0 —\u00a0 i=1 A\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

02- V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=3.1=3V\u00a0 —\u00a0 V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=3V\u00a0 —\u00a0 V<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0> V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

04. U=E \u2013 ri=15 \u2013 1.1\u00a0 —\u00a0 U=14V\u00a0 —\u00a0 P=i.u=1.14\u00a0 —\u00a0 P=14W\u00a0 —\u00a0 Verdadeira<\/b><\/span><\/span><\/p>\n

08. P=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=4W\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

R- 4<\/b><\/span><\/span><\/p>\n

 <\/p>\n

28-<\/b><\/span><\/span><\/span> Pot\u00eancias de um receptor:<\/b><\/span><\/span><\/p>\n

\"\"\u00a0Pot\u00eancia total (Pt) \u2013 corresponde \u00e0 energia total recebida pelo receptor da fonte externa (gerador) por unidade de tempo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Pot\u00eancia dissipada (Pd) \u2013 refere-se \u00e0 pot\u00eancia consumida (perdida, dissipada, sob forma t\u00e9rmica) pela resist\u00eancia interna r\u2019 do receptor e fornecida por:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Pot\u00eancia \u00fatil (Pu) \u2013pot\u00eancia aproveitada pelo receptor para seu funcionamento normal, sob forma n\u00e3o t\u00e9rmica.Exemplo: num ventilador \u00e9 a parcela da energia total que \u00e9 transformada em energia exclusivamente mec\u00e2nica.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Rendimento (\u03b7) de um receptor\u00a0\u2013 definido como sendo a raz\u00e3o entre sua pot\u00eancia \u00fatil (Pu) e sua pot\u00eancia total Pt):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=E\u2019.i\u00a0 —\u00a0 2000=E\u2019.10\u00a0 —\u00a0 E\u2019= 200V\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=U.i=220.10=2200W\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2200=2000 + r\u2019.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 200=r\u2019.100\u00a0 —\u00a0 r\u2019=2\u2126\u00a0 —\u00a0 \u03b7=P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=2000\/2200\u00a0 —\u00a0 0,909\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span> O processo de resolu\u00e7\u00e3o est\u00e1 explicado em fisicaevestibular.com.br (eletricidade-eletrodin\u00e2mica-circuitos compostos)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(I) partindo de P e percorrendo a malha no sentido hor\u00e1rio\u00a0 —\u00a0 1.i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 14 + 1i=0\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ i = 14 (a)\u00a0 —\u00a0 (II) partindo de P e percorrendo a malha no sentido hor\u00e1rio\u00a0 —\u00a0 3i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 14 \u2013 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 3i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= -14 (b)\u00a0 —\u00a0 i=i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0(c)\u00a0 —\u00a0 resolvendo o sistema formado por (a), (b) e (c)\u00a0 —\u00a0 i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-2 A (sentido correto-anti-hor\u00e1rio)\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=8 A sentido correto-hor\u00e1rio)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

i=6 A (sentido correto-hor\u00e1rio)\u00a0 —\u00a0 P<\/b><\/span><\/span>R2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=3.4=12W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Receptores \u2013 Associa\u00e7\u00e3o de geradores com receptores \u2013 Lei de Pouillet 01-\u00a0j\u00a0Quando i=2A\u00a0 —\u00a0 U=22V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 22=E\u2019 + r\u2019.2 (I)\u00a0 —\u00a0 quando i=5 A\u00a0 —\u00a0 U=25V\u00a0 —\u00a0 U=E\u2019 + r\u2019.i\u00a0 —\u00a0 25=E\u2019 + r\u2019.5 (II)\u00a0 —\u00a0 resolvendo o sistema composto por (I) e (II)\u00a0 —\u00a0\u00a0R- C 02-\u00a0a) C\u00e1lculo da<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1839,"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-1873","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1873","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=1873"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1873\/revisions"}],"predecessor-version":[{"id":10701,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1873\/revisions\/10701"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1839"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}