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I am reading this page that contains the transient solutions for RLC series circuits: over-damped, critically damped and under-dumped.

This site presents these graphics for the 3 cases:

enter image description here enter image description here enter image description here

Then I have decided to create a simulation using multisim, at https://www.multisim.com/content/LzLcDKNVvCzQeRUc6B9GTU/rlc/open/

My simulations are far from giving me these curves.

If I understood the theory correctly, these are my values:

  1. Over damped = R^2 > 4L/C... I choose R = 10Ω, C = 1C, L = 2.25H
  2. critically dumped = R^2 = 4L/C... I choose R = 3Ω, C = 1C, L = 2.25H
  3. under damped = R^2 < 4L/C... I choose R = 1, C = 1C, L = 2.25H

These are the curves I get:

OVER

enter image description here

UNDER

enter image description here

CRITICALLY

enter image description here

My curves are completely off.

Solid curves are voltage across capacitor. Dashed curve = current. V1 is a pulse, 10 VDC amplitude, 1ms.

What am I missing?

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    \$\begingroup\$ What stimulus signal was used in the reference diagrams, and what stimulus signal did you use in your simulations? \$\endgroup\$
    – The Photon
    Commented Jul 18, 2018 at 17:17
  • \$\begingroup\$ What circuit did you use and what are the graph nodes on that circuit? For instance PR2: V(3) is meaningless to me. Also what does "C = 1C" mean? Do you mean 1 farad? \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2018 at 17:30
  • \$\begingroup\$ just click on the link multisim.com/content/LzLcDKNVvCzQeRUc6B9GTU/rlc/open and you will be able to see my simulation and its parameters. It is public. \$\endgroup\$
    – Duck
    Commented Jul 18, 2018 at 17:39
  • \$\begingroup\$ solid curve = voltage across capacitor. Dashed curve = current. V1 is a pulse, 10 V amplitude, 1ms. \$\endgroup\$
    – Duck
    Commented Jul 18, 2018 at 17:42
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    \$\begingroup\$ The author of the linked article considers the homogeneous case which means the input stimulus is zero, and only initial conditions serve to create the transient. However, the initial conditions are not stated, hence the exact graphs cannot be derived and the graphs presented in the article are somewhat arbitrary. I suggest you read a different article \$\endgroup\$
    – Chu
    Commented Jul 18, 2018 at 18:05

2 Answers 2

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As far as I can tell your simulations match what I get from this online tool. I only did the under-damped case and the tool only provides Vout across the capacitor but it looks very similar to your graph: -

enter image description here

Looking at the transient voltage peak I get an overshoot of 1.329 after about 5 seconds and this pretty much ties with your graph except yours peaks at 13.3 volts due to the transient step being 10 volts.

What am I missing?

I don't know - they look good to me.

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The webpage made a slight error in the reduction of the general equation for critically damped.

enter image description here

\$i(t)=Ae^{m_1​t}+Be^{m_2​t}\$

for m1=m2=-R/2L

\$i(t)=(A+Bt)e^{-\dfrac{Rt}{2L}}\$ given on site is incorrect , it shud be.... \$i(t)=(A+B)e^{-\dfrac{Rt}{2L}}\$

Intuitively, you know series inductor current cannot start instantly.it will rise asymptotically towards Imax=V/R but then V decays from the capacitor charging up.

A simpler way to analyze it is the 1st order time constants.

The current rise time starts as T=L/R and after the peak the decay time T~RC so projecting the trianglular peak should take you towards the V/R current maximum when critically damped or slightly underdamped. enter image description here

To err is human, although your Sim is correct.

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