RLC hw problems

[EE students only]

Discussion of RLC circuit equation

• homogeneous linear DE case
  underdamped RLC circuit parameters: R = 16 ohms, C = 1/40 = .025 = 25 millifarads, L = 8 henries, E₀= 17 volts: evaluate τ₀, ω₀, T₀, ω₀ τ₀,  τ, ω, T, ω τ, A₀ = E₀/(ω L), and I(t) from the handout formulas (either resolve the IVP with Maple or plug into the formulas for the solution);
  Maple:
  [Make a single plot showing the current and its amplitude envelope for 4 decay times t = 0.. 4τ for this new problem.]
  First you have to solve this DEQ with the initial conditions given on the sheet handout. You may use the formulas on the sheet by hand or use maple dsolve to get the current i(t), which will be of the form: i(t) = exp(-kt)(c1 cos(ω t) + c2 sin(ω t).
  plot [i(t), exp(-kt), -exp(-kt)] , the latter two curves providing the envelope curves, in gray say, using the color option:
  color=[red,gray,gray]
  what is the characteristic time for this new set of numbers?
  
   Maple solution and plots [pdf solution (second page)]
   
• nonhomogeneous (driven) linear DE case
  maple plot: for the driven RLC problem [underdamped RLC circuit parameters: R = 16 ohms, C = 1/40 farads = .025 farads (= 25 millifarads), L = 8 henries, with a voltage source E(t) = 4 sin(2 t) and initial conditions I(0) = 0 = I '(0)],
  plot I, I_p (the steady state part of the solution) and E/2L on the same plot and see how long it takes for the transient to essentially be zero to the pixel accuracy
  again you have to solve this first either by hand or with dsolve, but when you plot E(t) on the same axes with your solution I(t), and the purely sin and cosine part I_p (t), the amplitude 4 of E(t) will squash the other two curves, so just plot E(t)/2L so you can compare the driving function to the response function (in phase). the exponential term damps out of the full solution so that your solution curve merges into the steady state function after how long? use the color option to distinguish the 3 curves, say color=[red,blue,green]. so the solution red, starts out at the origin but then approaches the steady state blue curve, which lags behind (oops, slightly leads) the green curve in phase.
  
  [alternate wording for this problem:
  repeat the previous undriven RLC problem now with a voltage source E(t) = 4 sin(2 t) and initial conditions I(0) = 0 = I '(0), solving it ignoring the solution formulas derived on the handout; then check your result with these formulas.
  *RLC Maple plot template: for this RLC problem, plot I, I_particular (the steady state part of the solution) and E/(2L) on the same plot and see how long it takes for the transient (difference between these two currents) to essentially be zero to the pix
  [USA rms voltage 120 Volts means peak voltage 120 sqrt(2) = 169.71V]
  ]
   
  Maple solution and plots [pdf solution]