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ep19z02c - 170 Elektrotechnika podstawowa Zad 2-6...

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Elektrotechnika podstawowa 170 Zad. 2-6. Kondensator o pojemno ś ci C 1 = 1 μ F zostal naladowany do napi ę cia U = 60 V, a nast ę p- nie odl ą czony od ź ródla i pol ą czony – jak na rysunku – z nienaladowanym kondensatorem o po- jemno ś ci C 2 = 2 μ F. Oblicz warto ść napi ę cia na tak pol ą czonych kondensatorach oraz warto ść energii wytraconej w przewodach podczas ladowania C 2 z C 1 . Uklad w stanie I Uklad w stanie II U C Q = 1 1 ' , 0 ' 2 = Q 2 1 2 1 ' ' " " Q Q Q Q = A. Rozwi ą zanie metod ą ukladu zast ę pczego (pol ą czenie równolegle) 3 2 1 12 = = C C C μ F , 60 " " " 1 2 1 12 = = = U C Q Q Q μ C , 20 " 12 12 12 = = C Q U V, 20 12 2 1 = = = U U U V. B. Rozwi ą zanie metod ą równa ń ukladu (opis „formalny”) 2 1 U U = 2 1 2 1 ' ' " " Q Q Q Q = czyli 0 2 1 = - U U U C U C U C = 1 2 2 1 1 Otrzymane równanie liczbowe = 60 0 2 1 1 1 2 1 U U - rozwi ą zuje si ę metod ą Cramera: 3 2 1 1 1 = = - W , 60 2 60 1 0 1 = = - W , 60 60 1 0 1 2 = = W ; 20 1 1 = = W W U V, 20 2 2 = = W W U V.
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