Recall that the differential equation for the instantaneous charge q (t) on the capacitor in an lrc-series circuit is l d 2q dt 2 + r dq dt + 1 c q = e (t). see this excerpt about lrc-series circuits. use the laplace transform to find q (t) when l = 1 h, r = 20 ω, c = 0.005 f, e (t) = 160 v, t > 0, q (0) = 0, and i (0) = 0. q (t) =

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Guapo

Physics

20 March, 07:10

Recall that the differential equation for the instantaneous charge q (t) on the capacitor in an lrc-series circuit is l d 2q dt 2 + r dq dt + 1 c q = e (t). see this excerpt about lrc-series circuits. use the laplace transform to find q (t) when l = 1 h, r = 20 ω, c = 0.005 f, e (t) = 160 v, t > 0, q (0) = 0, and i (0) = 0. q (t) =

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Rodney Dickerson · Answer 1

DE which is the differential equation represents the LRC series circuit where

L d²q/dt² + Rdq/dt + I/Cq = E (t) = 150V.

Initial condition is q (t) = 0 and i (0) = 0.

To find the charge q (t) by using Laplace transformation by

Substituting known values for DE

L*d²q/dt² + 20 * dq/dt + 1/0.005 * q = 150

d²q/dt² + 20dq/dt + 200q = 150