8.4 The Natural Response of a Series/Parallel RLC Circuit

8.3 Finding Initial Values• Example 2 (cont.)d + d +From ( B) vo(0 ) =−2 vR(0 ) LL( C)dt dtFrom ( A) and ( C)d + d +vR(0 ) =− 2 vR(0 ) + 2dtdtd 2 v+∴ (0 ) VR=dt 3 SC.T. Pan 238.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20Vi Lt →∞3A2Ω4Ω+v R ( )-+v c ( )-20Vi L2∴iL( ∞ ) = 3A× = 1A2+4v ( ∞ ) =−20Vcv ( ∞ ) = 3 A× (2ΩP4 Ω ) = 4VRC.T. Pan 24