Question

A series circuit has a capacitor O. x 6 a res stor of 5 X 3 Ω and an inductor of The initial cha e o the capacitor is zero a V battery is connected closed at t 0, determine the charge on the capacitor at t = 0.001 seconds, att = O seconds, and at any time t Also determine the limiting charge as t → Enter the exact answer with a< b The charge at any time is given by the formula Q(t) (Ae" + Be" +C) x 10-6 coulombs, where the arcuit and the arcuits x106 coulombs as t 00)- x10-6 coulombs 00.001) x10-6 coulombs Q(0.01) =

Answer 1

G.T C = 0.25 times 10^-6 f = > 1/C = 4 times 10^6 R = 5 times 10^3 L = H Q(0) = 0 Q prime(0) = f(t) = 24 The eq of RLC circuit is L d^2 Q/dt^2 + R dQ/dt + 1/C Q = f(t) d^2Q/dt^2 + 5 times 10^3 dQ/dt + 4 times 10^6 Q = 24 [D^2 + (5 times 10^3) + (4 times 10^6)] Q = 24 (1) where D = d/dt The auxiliary eq of (1) is m^2 + (5 times 10^3) m + 4 times 10^6 = 0 (m + (1 times 10^3)) (m + (4 times 10^3) = 0 m = -10^3 = -1000 m = -4 (10^3) = -4000 Q_c (t) = C_1 e^-1000 t + c_2 e^-4000 t \r\n Q_P = 1/D^2 + (5 (10)^3 D + 4(10^6) 24 e^0.t (D = 0) = 1/0 + 0 + 4 910^6) 24 e^0.6 Q_p = 6 times 10^-6