The following parallel RLC circuit is example 8.8 in Fundamentals of Electric Circuits by Charles Alexander.

enter image description here

In the example in this text, the current through the inductor i(t) is found to be:

$$i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978})$$

This solution is found by directly solving the second order differential equation. I want to solve this same circuit using Laplace transforms. Following the methods in the textbook, I have performed a Laplace transform on this circuit:


simulate this circuit – Schematic created using CircuitLab

Please note that even though 4 A current supply is always I, I have converted it to from 4 A to 4/s; I hope that is correct. Also, the initial current through the inductor is 4 A, leading to another 4/s current supply in the opposite direction.

In this circuit, the current through the inductor is:

$$ I(s) = \frac{0.75 s}{s(s^2 + 12.5 s + 6.25)}$$

The inverse Laplace transform of I(s) is:

$$i(t) = 0.0655(e^{-0.5218t} - e^{-11.978})$$

In this second solution, the "4" is missing. I have a two questions:

Should Laplace only give the transient solution?

Even though the current supply on the left does not change at t=0, why must it change to 4/s in the Laplace domain?

  • \$\begingroup\$ What is 30.u(-t) ? \$\endgroup\$
    – Antonio51
    Apr 14, 2023 at 13:33
  • \$\begingroup\$ @Antonio51 Sorry, I should add that to my question. u(t) is the step function, so u(-t) indicates that the voltage supply shuts off at t=0. \$\endgroup\$
    – EElmo
    Apr 15, 2023 at 1:11

2 Answers 2


I wish you would have disclosed your process. It would have been appreciated.

Grounding the bottom node, find this KCL for the switched node:

$$\begin{align*} \frac{v_t}{R_1}+\frac{v_t}{R_2}+C_1\frac{\text{d}\,v_t}{\text{d}t}+\frac1{L_1}\int v_t\:\text{d}t&= 0\:\text{A} \\\\ \frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t&= 0\:\text{A} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\mathscr{L}\left\{\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t\right\}+\mathscr{L}\left\{\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\mathscr{L}\left\{\frac{\text{d}}{\text{d}t}v_t\right\}+\frac1{L_1\,C_1}\mathscr{L}\left\{\vphantom{\frac{\text{d}}{\text{d}t}}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \left\{\vphantom{\frac{\text{d}}{\text{d}t}}s^2V_s-sv_{_0}-v_{_0}^{'}\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\left\{\vphantom{\frac{\text{d}}{\text{d}t}}sV_s-v_{_0}\right\}+\frac1{L_1\,C_1}V_s&= \mathscr{L}\left\{0\:\text{A}\right\} \end{align*}$$


$$\begin{align*} \left[s^2+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)s+\frac1{L_1\,C_1}\right]V_s&= \left[s+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\right]v_{_0}+v_{_0}^{'} \end{align*}$$

Since \$v_{_0}=15\:\text{V}\$ and the current supplied by \$C_1\$ at \$t=0\$ must be \$-1.5\:\text{A}\$, it's clear that \$\frac{\text{d}}{\text{d}t}v_{t=0}=\frac{-1.5\:\text{A}}{8\:\text{mF}}=-187.5\:\frac{\text{V}}{\text{s}}\$. The above then becomes:

$$\begin{align*} \left[s^2+12.5s+6.25\right]V_s&= 15s \end{align*}$$

Now for your mistake. I suspect that you elected to believe that \$V_s=s L_1 I_s\$ and found that:

$$\begin{align*} s\left[s^2+12.5s+6.25\right]I_s&= 0.75s \end{align*}$$

However, that's not correct.

Instead \$V_s=\mathscr{L}\left\{v_t\right\}=L_1 \mathscr{L}\left\{i_t^{'}\right\}=L_1\left[sI_s-i_{_0}\right]\$.

So, as \$L_1=20\:\text{H}\$:

$$\begin{align*}\require{cancel} \left[s^2+12.5s+6.25\right]L_1\left[sI_s-i_{_0}\right]&= 15s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s-i_{_0}\right]&= 0.75s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s\right]&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ s\left[s^2+12.5s+6.25\right]I_s&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ I_s&=\frac{0.75\cancel{s}}{\cancel{s}\left[s^2+12.5s+6.25\right]}+\frac{\cancel{\left[s^2+12.5s+6.25 \right]}4}{s\cancel{\left[s^2+12.5s+6.25\right]}} \\\\ &=\frac{0.75}{s^2+12.5s+6.25}+\frac4{s} \end{align*}$$

So, your Laplace equation should be \$I_s=\frac4{s}+\frac{0.75}{s^2+12.5\, s+6.25}\$.

If you had solved this in the time domain, you'd have been using annihilation and finding a need for hyperbolic sine (hyperbolic because the system is over-damped.)

$$i_t=4.0 + 0.130930734141595\cdot\exp\left(-6.25\,t\right)\cdot\sinh\left(5.7282196186948\,t\right)$$

The way you wrote it is also correct as the hyperbolic sine has a built-in \$\frac12\$ factor and expanding it would change 0.130930734141595 to 0.0654653670707975 (and provide your difference pair of exponentials.)

  • 1
    \$\begingroup\$ This is it exactly, this was my mistake. \$\endgroup\$
    – EElmo
    Apr 13, 2023 at 18:48
  • 1
    \$\begingroup\$ @EElmo Thanks for letting me know. And I thought as much. The reason I bothered writing (for such a short question that really didn't expose all of your thinking) is that I could almost see exactly where you made a mistake. You gave me enough for that much. But just barely enough. And it didn't dawn on me until a few hours after, doing other housework here. A little more writing wouldn't have hurt and it might have sped my ability to spot it. Just some kind advice for next time. Also, I'm impressed. You've pretty much mastered this kind of work! I don't believe many here can say as much. \$\endgroup\$ Apr 13, 2023 at 21:15
  • 1
    \$\begingroup\$ @EElmo Just FYI. I really like the 9th edition of Fundamentals of Differential Equations by Nagle, Saff & Snider. The pacing is excellent and very well-managed, in my opinion, for self-learners as well as for those taking classes. (You don't need it, though.) Beyond that, almost anything Gilbert Strang writes is pretty darned decent (if a little bit unexpectedly folksy -- which is something I actually like about him but some others don't, so much.) He provides insights and how to gather them up that many here at this site should acquire (directed graph math patterns and Kalman, e.g.) \$\endgroup\$ Apr 13, 2023 at 21:23
  • 1
    \$\begingroup\$ I will buy this book. Are older editions ok? \$\endgroup\$
    – EElmo
    Apr 15, 2023 at 11:33
  • \$\begingroup\$ @EElmo I only have the 9th edition. I can't speak to earlier versions. In the 9th edition, the section that covers the above is "7.5 Solving Initial Value Problems" in chapter 7 which is on Laplace Transforms. \$\endgroup\$ Apr 15, 2023 at 20:09

It seems that it should be this ...

enter image description here

Final current into L1 = 4 + 30/20 = 5.5 A ?

Or this ?

enter image description here

  • \$\begingroup\$ Antonio51 - Hi, (a) Your comment to the downvoter was sarcastic & unfriendly so it broke the Code of Conduct & has therefore been deleted. Please read that important rule. Don't break any part of it again. (b) The FAQ article about being downvoted explains what type of comment response is allowed, if you are sure you want to comment after a downvote. (c) If I had to guess a reason for the downvote, it's because the answer did not address the specific question about using Laplace transforms. An answer should answer that question. TY \$\endgroup\$
    – SamGibson
    Apr 14, 2023 at 13:53

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