I need to find the current going through the inductor \$(I_l)\$ for \$t>0\$ using Laplace circuit analysis, to then find \$(I_l)\$ in the time domaine.


simulate this circuit – Schematic created using CircuitLab

I came up with the following way of finding \$(I_l)\$ using current divider.

\begin{align} I_L (s) &= \frac{Z_{total}}{Z_L} \cdot I_{total}(s)\\[0.7em] &= \frac{Z_C + Z_R + Z_L}{Z_L}\cdot I_{total}(s) \\[0.7em] &= \frac{s^2 + 2s + 1}{s^3} \end{align}

and then to find \$ I_L(t) \$ I would do

\begin{align} I_L (t) &= \mathcal{L}^{-1}\{I_L(s)\} \\[0.7em] &= \mathcal{L}^{-1}\left\{ \frac{s^2 + 2s + 1}{s^3} \right\}\\[0.7em] &= \frac{t^2}{2} + 2t + 1 \end{align}

Could someone please tell me if I did it the right way ?


Your current divider equation is wrong. The equation should be:

$$I_L = \dfrac {Z_{total}}{Z_L} * I_{total} $$

Where \$Z_{total}\$ is the equivalent parallel impedance of all 3 branches and \$Z_L\$ is the impedance of the L branch.

This should give you the right result.

  • \$\begingroup\$ I fixed my math, but now my answer really doesnt look right \$\endgroup\$
    – Liam F-A
    Apr 3 '17 at 15:21
  • \$\begingroup\$ Are you sure that thats the current divider equation : en.wikipedia.org/wiki/Current_divider ? \$\endgroup\$
    – Liam F-A
    Apr 3 '17 at 15:36
  • \$\begingroup\$ @LiamF-A Yep. The equation in the Wiki article is also correct. \$R_T\$ there refers to the equivalent impedance of all the branches except the branch in question. \$\endgroup\$ Apr 4 '17 at 1:12
  • \$\begingroup\$ Ok but when I fixed my math, the answer doesnt look right. I edited it in my question \$\endgroup\$
    – Liam F-A
    Apr 4 '17 at 1:27
  • \$\begingroup\$ @LiamF-A You're supposed to find the equivalent impedance, as in parallel impedance, not just add them up. Edited my answer. \$\endgroup\$ Apr 4 '17 at 2:18

I can only help you answer your question very directly as asked, eg. Yes /No. Since you drew the thing in CircuitLab, it's only a small step to run it as a full simulation. It should allow you to model the current anywhere with respect to time, and compare it to your solution.

Alternatively if you can't get CircuitLab running for the amount of dollars you find acceptable, LTSpice is free, reliable and has a huge support base. It's great for circuit education.

Fiddling around with:-


I get:-

simulation run

The reason it's hard is because of the constant current source. That's impossible. You can't start a simulation from nothing, arrive at a steady state without passing through intermediate states. That's why LTSpice has to be tweaked with the 'startup' option. It forces the analysis to begin with a voltage of zero, otherwise how do you get from nothingness to current = 1A and whatever voltage is necessary to drive it? The tweak is in the Edit Simulation Cmd /Transient tab.

Any use?

  • \$\begingroup\$ Thanks ! I didnt know about LTSPICE. It seems like a powerfull tool. But how would I know if my equation is correct ? \$\endgroup\$
    – Liam F-A
    Apr 2 '17 at 21:30
  • \$\begingroup\$ @LiamF-A Yeh, that's what I meant by my first sentence. Who's Laplace? \$\endgroup\$
    – Paul Uszak
    Apr 2 '17 at 21:44
  • \$\begingroup\$ Pierre-Simon Laplace was the mathematician who came up with this s-domain analysis. \$\endgroup\$
    – Liam F-A
    Apr 2 '17 at 21:46
  • \$\begingroup\$ @LiamF-A I'm sorry - I was kidding... \$\endgroup\$
    – Paul Uszak
    Apr 2 '17 at 21:48
  • \$\begingroup\$ You had me worried for a minute \$\endgroup\$
    – Liam F-A
    Apr 2 '17 at 21:51

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