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In the picture below the switch \$S_1\$ is open for a long time, and I need to find the expression of \$i(t)\$ for \$t>0\$ (when \$S_1\$ closes):

enter image description here

Since this is an RLC circuit, I need to solve a second order ODE. I found \$i(0)=4\mathrm{ A}\$ and $$\frac{di(0)}{dt}=28\mathrm{ A/s}$$ as initial conditions to be able to solve that ODE (if someone could double check those values, that would be great).

After \$S_1\$ is closed, the \$30\Omega\$ resistor is short-circuited and the \$10\Omega\$ and \$20\Omega\$ resistors are in parallel, so we get the following circuit:

enter image description here

I have tried using both Kirchhoff's laws but couldn't find that ODE. Any ideas how can I proceed here?

Attempt using KCL:

enter image description here

NODE A:

$$6=i_c+i_2$$

$$6=0.15\frac{dv_c}{dt}+\frac{v_c-0}{6.667} \tag{1}$$

NODE 0:

$$i_2=i+i_3$$

$$\frac{v_c-0}{6.667}\ = i + \frac{0-v_L}{15}\tag{2}$$

Replacing (2) in (1):

$$6=0.15\frac{dv_c}{dt}+i - \frac{v_L}{15}$$

$$6=0.15\frac{dv_c}{dt}+i -\frac{1}{15}\frac{di_3}{dt}$$

$$6=0.15\frac{dv_c}{dt}+i -\frac{1}{15}\frac{d(i_2-i)}{dt}$$

$$6=0.15\frac{dv_c}{dt}+i +\frac{1}{15}\frac{di}{dt}-\frac{1}{15}\frac{di_2}{dt}$$

$$6=0.15\frac{dv_c}{dt}+i +\frac{1}{15}\frac{di}{dt}-\frac{1}{15}\frac{d(6-i_c)}{dt}$$

$$6=0.15\frac{dv_c}{dt}+i +\frac{1}{15}\frac{di}{dt}+\frac{1}{15}\frac{di_c}{dt}$$

This is as far as I could go trying to get that second order ODE for \$i(t)\$.

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  • \$\begingroup\$ @Antonio51 I first tried finding the ODE using KCL, but didn't succeed. Then i tried KVL, but could't make it either. \$\endgroup\$
    – Murilo
    Sep 7 at 11:50
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    \$\begingroup\$ Check this. When switch is closed, R-L alone at the left, R-C-I alone at the right ... Two First Order equations. \$\endgroup\$
    – Antonio51
    Sep 7 at 12:01
  • \$\begingroup\$ Show us your work. There is no point in asking someone to redo everything that you have done. \$\endgroup\$ Sep 7 at 12:35
  • \$\begingroup\$ @ElliotAlderson The work is there. I have found all the initial and final conditions of my desired variable i(t). I have found the equivalent circuit after the switch have been closed. But i still can't find a second order ODE for i(t), like i have read in books we should find to solve a RLC circuit, or two first order ODE like Antonio sugested (never heard we could do that, but i tried anyway). I have added a new picture showing my attempt using KCL in the top nodes. \$\endgroup\$
    – Murilo
    Sep 7 at 13:09
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    \$\begingroup\$ Have you the initial condition before closing the switch? When you close the switch, it is obvious that the first loop (L-R) is alone with a "initial" inductor current. So, after closing the switch, current in the self is an exponentially decreasing function starting at 2 A -> you have the first part of your i(t) current ... \$\endgroup\$
    – Antonio51
    Sep 7 at 14:06

2 Answers 2

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Check first this, switch open. Dynamic DC analysis. Microcap v12.
This will give you the "initial" condition of the inductor current and voltage capacitor ... for the two "independent" loops ... after closing the switch.

After closing the switch, "i(t)" is composed of two parts: current1 from the left, current2 from the right, from independent loops.

enter image description here

Since this is an RLC circuit, I need to solve a second order ODE.

Are you sure?

Perhaps two first-order independent equations, then one add the solutions?

This simulation could help you.
Check the variables you need.

enter image description here

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  • \$\begingroup\$ I finally solved it! The two first order ODE helped a lot, until yesterday, i didn't knew we could do that to solve a RLC circuit. One more question, wich software did you use to simulate that circuit? It's a free software? or commercial one? \$\endgroup\$
    – Murilo
    Sep 8 at 12:31
  • \$\begingroup\$ FULL FREE software with many examples. Check the link in answer. \$\endgroup\$
    – Antonio51
    Sep 8 at 12:34
  • \$\begingroup\$ I tried click the link, but nothing happened. I will Google it, Sepctrum Software Microcap? \$\endgroup\$
    – Murilo
    Sep 8 at 12:46
  • \$\begingroup\$ Spectrum Software - Micro-Cap 12. Analog simulation, mixed ...spectrum-soft.com go to Download ... microcap v12 full CD \$\endgroup\$
    – Antonio51
    Sep 8 at 12:56
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    \$\begingroup\$ Link corrected in answer ... Search also newsletters \$\endgroup\$
    – Antonio51
    Sep 8 at 13:01
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Intuitive analysis.

enter image description here

Fig a)
switch open, 6A source , 10 Ohm load = 60V source. Vsw=30V, IL= 2A enter image description here

Fig b)
switch closed. 6A source, 10||20 = 6.67 ohm load, Vsource= I*R = 40V
Parallel L/R discharge loop on left , Parallel RC discharge loop on right of switch from 60 to 40 V. You figure out the details of exponential slopes. ;)

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