(5.1.45)
This is a series circuit problem.
Find the charge on the capacitor in an LRC series circuit at t = 0.01s when L = 0.05h, R = 2 ohms, C = 0.01f, E(t) = 0V, q(0) = 5C, and i(0) = 0A. Determine the first time at which the charge equals zero on the capacitor.
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
Now, the time can't be negative, and the answer in the back of the book says 0.544s.
So, where did I go wrong?
Also, I'm assuming that i(t) = q'(t) which works for the first part of the problem - Is this correct?
Best Regards,
--Hank Stalica
This is a series circuit problem.
Find the charge on the capacitor in an LRC series circuit at t = 0.01s when L = 0.05h, R = 2 ohms, C = 0.01f, E(t) = 0V, q(0) = 5C, and i(0) = 0A. Determine the first time at which the charge equals zero on the capacitor.
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
Now, the time can't be negative, and the answer in the back of the book says 0.544s.
So, where did I go wrong?
Also, I'm assuming that i(t) = q'(t) which works for the first part of the problem - Is this correct?
Best Regards,
--Hank Stalica