v(t) = 5t volts, L = 50 H, C = 450 ?F, R = 6 k?. The downward current in the ind

Question

v(t) = 5t volts, L = 50 H, C = 450 ?F, R = 6 k?. The downward current in the inductor is 3 mA at t = 0. Find the magnitude and direction of the current flowing in all four elements as a function of time.

Explanation / Answer

given Io =3mA we know current through inductor I=I0 e^-t/tou...............>1 where I0=3mA tou = L/R =50/6000 =0.00833 1/tou=120 substitute all values in 1 I =0.003*e^-120t------------------>2 voltage across inductor Vl=L di/dt =-50 *0.003*-120e^-120t =-18e^-120t----------------------->3 we know in parallel voltage is same so Vl=Vc=Vr Vc=18e^-120t current through capacitor Ic = Cdv/dt =450*10^-6*-18*(-120)e^-120t =0.972 e^-129t----------------->4 current through resistor is I=Vr/R =18e^-120t/6000 =0.003e^-120t -------------->5 2,4,5 are answers (currents) and direction positivee means downwards negative means upwards

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