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Analisis Rangkaian Elektrik - Ee-cafe.org

Analisis Rangkaian Elektrik - Ee-cafe.org

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I 1 (s)Penyelesaian :Misalkan : V2(s)= 1→VC( s)= V2(s)= 1 →1IC( s)= = sC1/ sC→ IL(s)= IC( s)= sC →2VL(s)= sL × sC = LCs22LCs + 1→VR(s)= VL(s)+ VC( s)= LCs + 1 → IR(s)=R22*LCs + 1 LCs + RCs + 1⇒ I1( s)= IR(s)+ I L(s)= + sC =RR1 R⇒ Ks= =*2I1( s)LCs + RCs + 1R⇒V2(s)= KsI1(s)=I1(s)2LCs + RCs + 14.8.2. Metoda SuperposisiCOTOH-4.5: Dengan menggunakan metoda superposisi, carilahteganganinduktor v o (t)pada rangkaianberikut ini.Penyelesaian :Au(t)I R (s)I L (s)<strong>Rangkaian</strong> kita transformasikan ke kawasan s menjadi+−RRsLI C (s)L+v o−1/sCR+V 2 (s)−BsinβtAsR++Bβ− sL VRo2 2s + β−Jika sumber arus dimatikan, maka rangkaian menjadi :94 Sudaryatno Sudirham, <strong>Analisis</strong> <strong>Rangkaian</strong> Listrik (2)

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