Find the charge on the capacitor in an LRC-series circuit at t = 0.02 s when L =

Question

Find the charge on the capacitor in an LRC-series circuit at t = 0.02 s when L = 0.05 h, R = 1 ?, C = 0.04 f, E(t) = 0 V, q(0) = 7 C, and i(0) = 0 A. (Round your answer to four decimal places.) Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.)

Explanation / Answer

Lq" + Rq' + q/C = E(t) L = 0.05 = 1/20 (1/20)q" + 2q' + 100q = 0 => q" + 40q' +2000q = 0 m^2 + 40m + 2000 = 0 //The auxiliary equation m = -20 +/- 40i //After applying quadratic formula q(t) = A(e^-20t)cos(40t) + B(e^-20t)sin(40t) //A and B are constants q(0) = 5 = A => A = 5 q'(t) = -20A(e^-20t)cos(40t) - 40A(e^-20t)sin(40t) _ 40B(e^-20t)cos(40t) - 20B(e^-20t)sin(40t) i(0) = q'(0) = 0 = -20A + 40B => B = (5/2) So, q(t) = 5(e^-20t)cos(40t) + (5/2)(e^-20t)sin(40t) q(0.015) = 4.568C //Which is correct in the book. The charge on the capacitor at t = 0.015 //Now to find the first time at which the charge equals zero: q(t) = 0 = 5(e^-20t)cos(40t) + (5/2)(e^-20t)sin(40t) //and solve for t. 0 = 5cos40t + (5/2)sin(40t) => -5cos(40t) = (5/2)sin(40t) => -5 = 5sin(40t)/2cos(40t) => -2 = sin40t/cos40t => -2 = tan(40t) => -2 + arctan(40) = t => t = -0.544s ... I don't know what you are doing here 40*t = arctan(-2) Although arctan is defined within -90° to +90° - we have to seek answer in the domain of 90° to 180°. So 40*t = -1.10715 + pi = 2.03444 t = 0.050861098 However, I am not getting the answer in the back of the book Now, the time can't be negative, and the answer in the back of the book says 0.544s.

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