I'm currently taking Physics II and I'm tasked to solve for the voltage $$U_R(t)$$ in this circuit.

The RLC-Circuit

It's driven by a current source $$I(t) = I_0*e^{i*\omega*t}$$ and I can neglect the switch on process. I'm currently at the point of having the differential equation to solve this problem but I'm not entirely sure on how to solve it yet.

I wanted to ask if my thought process was correct so far and if I landed at the correct differential equation or am I on the completely wrong path?

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  • 2
    \$\begingroup\$ Markus, Brethlosze's solution is a "hybrid" Laplace approach and it is (though a little harder to see exactly why) equivalent to your solution. Chu simply stayed in the time domain, like you did, and used KCL. I liked it, as well. You should feel comfortable with both techniques. Your approach, slightly more written out, is also just fine. Do you know enough of Laplace to see why Brethlosze's approach is roughly equivalent? \$\endgroup\$
    – jonk
    Apr 27, 2020 at 1:04

3 Answers 3


Yes, your analysis is correct.

It would have been quicker to solve for the resistor current and then multiply through by \$\small R\$ to get the resistor voltage, \$v_R\$, as below (I've used \$\small j=\sqrt{-1}\$ to avoid confusion with current, \$i)\$.

Let \$i\$ be the resistor current, then the voltage across the current source is: $$L\frac{d}{dt}(I_0e^{j\omega t}-i)=\frac{1}{C}\int i\:dt+Ri$$ differentiate,

$$L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{1}{C}i=L\frac{d^2}{dt^2}(I_0e^{j\omega t})$$

multiply by \$\frac{R}{L}\$,

$$\frac{d^2v_R }{dt^2}+\frac{R}{L}\frac{dv_R}{dt}+\frac{1}{LC}v_R=-R\omega^2I_0e^{j\omega t}$$


As you could figure out, your source is expressed as a phasor, hence we can assume safely the solution can be expressed with phasors and impedances, in consistency with the requirement of ignoring transients.

Hence, the current through the resistor \$R\$, by simply using a current divider is: $$I_R={ Z_L\over Z_L+Z_C+Z_R}I ={j \omega L \over j\omega L +{1\over j \omega C} +R}I \\ ={-\omega^2 LC\over 1+j\omega RC-\omega^2 LC}I $$

So, the voltage will be: $$ V_R=I_RZ_R={-\omega^2 RLC\over 1+j\omega RC-\omega^2 LC}I_0e^{i \omega t} $$

Which is equivalent to your solution.

Hence the \$V_R\$ voltage depends solely on the pasive values on the circuits, in the amplitude of the current source, and in the frequency (and phase, here equal to zero) of the current source.

Note that this is not the canonical series or parallel RLC, hence the maximum is not the simplest case, and we need to take the derivative of the absolute complex value.

Taking the absolute value (\$I_0=1\$): $$ \left|V_R)\right| =\left|{-\omega^2 RLC\over 1+j\omega RC-\omega^2 LC}I_0e^{i \omega t}\right| ={\omega^2 RLC\over \sqrt{ 1-2\omega^2 LC+\omega^2 R^2C^2+L^2C^2\omega^4}} $$

Taking the derivative: $$ \frac{d}{d\omega}(V_R) =\frac{d}{d\omega}\left({\omega^2 RLC \over \sqrt{1-2\omega^2 LC+\omega^2 R^2C^2+L^2C^2\omega^4}} \right) ={LRC\omega((R^2C^2-2LC)\omega^2+2) \over \left(L^2C^2\omega^4+R^2C^2\omega^2-2LC\omega^2+1\right)^{3/2}} $$

We have:

  • A minimum value \$|V_R(\omega_{min})|=0\$ at \$\omega_{min}=0\$
  • A final asymptotic value of \$|{V_R}_{\infty}|=RI_0\$ at \$\omega \to \infty\$,

Hence, the value \$\omega=\sqrt{\frac{2}{2LC-R^2C^2}}\$ is indeed the maximum we are looking for.

And also note. From this expression, the maximum happens only when \$2L>R^2C\$. If this do not happens, you only have the asymptotic value as limit.

  • 2
    \$\begingroup\$ If you'd stayed with the s-domain, I think this would be been just slightly cleaner (closer to Laplace) and therefore easier to see why it is equivalent to the OP's answer. But it's correct, now. Thanks for fixing it up! (Change of name, noted.) \$\endgroup\$
    – jonk
    Apr 27, 2020 at 0:58
  • 1
    \$\begingroup\$ Thank you for that! We didn‘t go trough Laplace in our lectures but I get that I can use a current divider, this seems indeed much easier. Another task was to find when $U_R$ has its maximum. So I guess I have to calculate the magnitude of the complex function above and then differentiate it in respect to $\omega$ and set it to zero? \$\endgroup\$ Apr 27, 2020 at 6:36
  • 1
    \$\begingroup\$ Ok, I‘m not entirely sure, but shouldn‘t de amplitude be at its maximum when the driving frequency is equal to the eigenfrequency of the system? So when $$\omega = 1/\sqrt{L*C}$$ ? I assumed that $$\omega_0 = 1/{L*C}$$ \$\endgroup\$ Apr 27, 2020 at 11:14


If you aren't familiar with Laplace, getting familiar with exactly why Laplace helps with differential equations is worth some of your time. Find some good book that covers it in a very simple, introductory way (or a video that does the same.) It's actually quite wonderful to see in action. And once you get it using a few of the simpler examples, you can just "accept" that the more difficult cases also work and just use a table from then on. It's like integration, that way. Some integration takes a lot of intuition and imagination to work out on your own -- or, you can just look it up in a table or book, as someone else did the hard work for you. Laplace is kind of like that.

Laplace also relies a lot on partial fraction decomposition. (This is just breaking down a ratio where the denominator includes multiple factors into a sum of fractions which include simpler denominators -- if you don't mind a sloppy way of saying it.) It makes looking up the solution easier, because now you can do it in smaller pieces that are easier to look up. So you'll also want some skills there, too, if you do much with Laplace.

Laplace is more commonly applied solving complex differential equations and, occasionally, convolution. But it does really make things simpler. So it's worth the trouble to at least get a feel for it.

A reason why Laplace is so important in electronics is that capacitors and inductors are ideally represented as differential or integral equations and this makes Laplace stand out as a likely handy approach. And since those components abound in electronics, using Laplace is an obvious fit.

(Sidebar: That's not the only place where Laplace is widely used, as most physical systems involve differentials requiring solutions with exponential relationships. There is a deep relationship between exponentials and sampling, which is exactly what happens when a particle collision occurs. So it abounds in natural/physical systems, as \$d\,x=x\cdot \text{d}\ln x\$. But that's for another time.)

Calculus, as Newton envisioned it, introduces the infinitesimal variable to algebra (which only deals in finite variables.) You can stay in the time domain, by preserving the ratios of two infinitesimal variables where the one in the divisor relates to time, or you can trivially remove the time domain entirely, by simply multiplying it out. For example, one relationship between current and capacitance is expressed as \$I_C=\frac{\text{d}\,Q}{\text{d}\,t}=C\:\frac{\text{d}\, V_C}{\text{d}\, t}\$. But you can multiply that by \$\text{d}\,t\$ to remove time and find the timeless relationship of \$\text{d}\,Q=C\:\text{d}\,V\$. (Or, conversely, you can introduce time arbitrarily by simply dividing both sides by the infinitesimal of time, \$\text{d}\,t\$, to get back the earlier equation.) It's just simple algebra, with the addition of infinitesimal variables.

But look at this a different way. It's true that: \$I_C=\frac{\text{d}\,Q}{\text{d}\,t}=C\:\frac{\text{d}\, V_C}{\text{d}\, t}\$. But with Laplace, as I'll mention more about shortly, multiplying by s means "taking the time derivative."

Let's apply that "trick" to the capacitor's time-derivative equation: \$I_C=C\:\frac{\text{d}\, V_C}{\text{d}\, t}=C\:s\: V_C\$. Now re-work this around to get: \$Z_C=\frac{V_C}{I_C}=\frac1{s\,C}\$!! This is the Laplace version, but look at how simple it is to turn complicated-looking time-derivative equations into something so much simpler to work with! Apply it now to \$V_L=L\:\frac{\text{d}\, I_L}{\text{d}\, t}= L\:s\: I_L\$. So \$Z_L=\frac{V_L}{I_L}=s\, L\$! Nice.

For now, this may just look like some magic-wand-waving... and it is, kind of. But it actually works, too. There are good reasons for it and this is why I want you to really go and study Laplace. It may seem a little daunting, at first, but the "trick" is really, really powerful. So it's worth the time to gain some comfort with the ideas.

Now, in electronics usage, \$s=\sigma+j\,\omega\$. As you probably already know, multiplication in the complex domain involves both rotation and scaling. Usually, timeless polar coordinates here are represented as \$e^s\$ and you introduce time by simply multiplying the exponent by \$t\$ to get \$e^{s\,t}\$. If \$\sigma=0\$ then the factor is \$1\$ and there's no spiral scaling (less than 0 would be spiraling inward and greater than 0 would be spiraling outward, over time, as rotation over time takes place due to \$\omega\$.) So if you are only interested in the frequency response, you set \$\sigma=0\$ and move forward with \$s=j\,\omega\$ (as Brethlosze did.) For systems with \$\sigma<0\$ you can expect them to diminish over time. For systems with \$\sigma>0\$ you can expect them to increase over time (which usually in electronics isn't a good thing as eventually the system must exceed its limits.)

Since \$\text{d}\,e^{s\,t}=s\,e^{s\,t}\,\text{d}\,t\$, multiplying by s is similar to taking the derivative. Dividing by s is similar to taking the integral.

So, keeping in mind how Laplace can simplify otherwise complicated-looking differential equations into simpler-looking algebraic ones, let's now see how this may apply to help out in your case.

Your problem

In your case, the current divides along two branches; one branch is through just an inductor and the other branch is through a series pair made up of a resistor and a capacitor. The simple approach that Brethlosze chose to take is one often used by folks familiar electronics. But let me clean it up a little bit.

The impedance for the three passive components is \$Z_C=\frac1{s\,C}\$, \$Z_L=s\,L\$, and \$Z_R=R\$. Given the splitting of current into two branches, you'll find that the splitting ratio is the conductance of the branch of interest divided by the sum of the total conductance. In this case, this is \$\frac{I_R}{I}=\frac{\frac1{Z_C+Z_R}}{\frac1{Z_C+Z_R}+\frac1{Z_L}}=\frac{L\,C\,s^2}{L\,C\,s^2+R\,C\,s+1}=\frac{s^2}{s^2+\frac{R}{L}\,s+\frac1{L\,C}}\$.

So \$V_R=I_R\,R = I\,R\,\frac{s^2}{s^2+\frac{R}{L}\,s+\frac1{L\,C}}\$. But we can just multiply both sides by the denominator to get this result (remembering that multiplication by s means taking the derivative):


Now, if you decide to set \$\sigma=0\$ and only focus on frequency \$\omega\$ then \$s=j\,\omega\$:

$$\begin{align*}\ddot{V_R}+\frac{R}{L}\,\dot{V_R}+\frac1{L\,C}\,V_R&=I\,R\,s^2\\\\&=\left(j\,\omega\right)^2 I\,R\\\\&=-\omega^2\,I\,R\end{align*}$$

You can, at this point, insert your \$I=I_0\,e^{j\,\omega\,t}\$ and get the same answer as you achieved.


I just wanted to hammer home a few more thoughts about Laplace and why it is handy. You can do all of the hard work, staying in the time domain. But it's often easier to switch over to the Laplace (s) domain and treat things a little more simply before working backward into the time domain, again.

  • 1
    \$\begingroup\$ Wow, that was really beautiful, thank you for this awesome explanation! You really waked my interest in the topic, I'll definitely look into it. Seems like it would be a pretty powerful tool. At the moment we are also going over convolution in our math class so this maybe could come in handy there. \$\endgroup\$ Apr 28, 2020 at 8:27
  • \$\begingroup\$ @markus Thanks for the kind words. There is so much of interest surrounding this area of complex analysis, including Euler's gamma function (which has a surprising relationship to prime numbers), quaternions, and much more. Just enjoy. \$\endgroup\$
    – jonk
    Apr 28, 2020 at 8:47

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