Potencia dissipada
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Potencia dissipada
A figura mostra um circuito em que um gerador de fem 24 V alimenta um circuito ideal com resistores de 2,0 Ω, 2,0 Ω e 3,0 Ω, e um motor de fcem 6,0 V
![Potencia dissipada 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)
A potência dissipada pelo motor, em W, deve ser de:
Gabarito: 24W
A potência dissipada pelo motor, em W, deve ser de:
Gabarito: 24W
anbongo1- Padawan
- Mensagens : 66
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Idade : 27
Localização : Goiás
Re: Potencia dissipada
2,0 Ω // 3,00 ---> R = 1,2 Ω
Sejam i1 a corrente no gerador (de cima para baixo), i2, 13 a corrente em R e em M, para cima
Como R está em paralelo com M a ddp de R vale U = E' = 6 V --> U = R.i2 --> 6 = 1,2.i2 --> i2 = 5 A
Malha esquerda ---> 2.i1 = E - U ---> 2.i1 = 24 - 6 ---> i1 = 9 A
i3 = i1 - i2 ---> i3 = 9 - 5 ---> i3 = 4 A
P(M) = E'.i3 ---> P(M) = 6.4 ---> P(M) = 24 W
Sejam i1 a corrente no gerador (de cima para baixo), i2, 13 a corrente em R e em M, para cima
Como R está em paralelo com M a ddp de R vale U = E' = 6 V --> U = R.i2 --> 6 = 1,2.i2 --> i2 = 5 A
Malha esquerda ---> 2.i1 = E - U ---> 2.i1 = 24 - 6 ---> i1 = 9 A
i3 = i1 - i2 ---> i3 = 9 - 5 ---> i3 = 4 A
P(M) = E'.i3 ---> P(M) = 6.4 ---> P(M) = 24 W
Elcioschin- Grande Mestre
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