I'm just curious what the best method for solving for voltage across the capacitor might be.

My initial idea was to use nodal analysis but I ran into a roadblock in that I am not entirely sure how the dependent source interacts with the nodal equations.

I was thinking to just give it a placeholder variable of I and hope that it could be worked out but I encountered a dead-end from that approach.

Probably I'm just not applying nodal analysis correctly.

This image shows the problem along with my initial work.

enter image description here

What I was getting from Node V1 was:

V1(s)/100k + (V1 + 0.0012)/0.06(s) + 15I(S) = 0

Node V2:

-15I(S) + V2/200 + V2/80 = 0

Then I would just plug in I = -V/80 but plugging that in didn't help seem to help.

  • \$\begingroup\$ Why not solve it using a simulator? \$\endgroup\$
    – Andy aka
    Commented Mar 16, 2023 at 11:18
  • \$\begingroup\$ Thxforhelping: You've not responded to anyone yet. I can help and show you how to get the time domain equation that will exactly predict what you'll see in a simulator, too. But without responding, I don't feel much need to write. Say something? \$\endgroup\$ Commented Mar 17, 2023 at 3:49
  • \$\begingroup\$ @periblepsis Hello I'm just asking how to set up the nodal equations, I'm not asking for anyone to solve it for me entirely. If you want to help, feel free to do so. \$\endgroup\$ Commented Mar 17, 2023 at 6:35
  • \$\begingroup\$ @Thxforhelping Okay. There's really only one nodal to worry about because the other node can be defined with respect to the first. \$\endgroup\$ Commented Mar 17, 2023 at 7:06
  • \$\begingroup\$ @Thxforhelping You needed a relationship between the two nodes. \$\endgroup\$ Commented Mar 18, 2023 at 18:58

3 Answers 3


The schematic with initial conditions and just following the moment after \$t=0\$ looks like this:


simulate this circuit – Schematic created using CircuitLab

Easy to work out that \$V_{_\text{A}}=\frac{V_{_\text{O}}}{1-\frac{15\:\Omega}{R_2}}\$. (Because \$V_{_\text{A}}+15\:\Omega\cdot I=V_{_\text{O}}\$ and \$I=-\frac{V_{_\text{A}}}{R_2}\$.)

So the KCL (nodal) is just:

$$C_1\frac{\text{d}}{\text{d}\,t}V_{_\text{O}}+\frac1{L_1}\int V_{_\text{O}}\:\text{d}\,t+\frac{V_{_\text{O}}}{\left(1-\frac{15\:\Omega}{R_2}\right)R_1}+\frac{V_{_\text{O}}}{\left(1-\frac{15\:\Omega}{R_2}\right)R_2}=0\:\text{A}$$

Removing the integral (take derivative), collecting terms, then in s-space:

$$\left[s^2+\frac{1}{\left(1-\frac{15\:\Omega}{R_2}\right)\cdot\left(R_1\mid\mid R_2\right)\cdot C_1}s+\frac{1}{L_1\,C_1}\right]V_{_\text{O}}=0\:\text{A}$$

Conveniently homogeneous.

Any homogeneous s-space eq. with a response of the form \$\left[s^2+a\,s+b\right]\$ solves by setting \$\alpha=-\frac12 a\$ and \$\beta=\sqrt{b-\frac14 a^2}\$, to get equivalent \$\left[\frac{\text{d}}{\text{d}\,t}-\alpha\right]^2+\beta^2\$, and finding that \$V_{_\text{O}}=\exp\left(\alpha\,t\right)\cdot\left(A_1\,\cos\left(\beta\,t\right)+A_2\,\sin\left(\beta\,t\right)\right)=A_1\,\exp\left(\alpha\,t\right)\cdot\left(\cos\left(\beta\,t\right)+\frac{A_2}{A_1}\,\sin\left(\beta\,t\right)\right)\$, since in this case \$b\ge \frac14 a^2\$ and the system is under-damped. (Otherwise, looking at hyperbolics for the over-damped case.)

\$A_1=100\:\text{V}\$ (just substitute in \$t=0\$.) So that leaves only \$A_2\$. At \$t=0\$, \$\frac1{L_1}\int V_{_\text{O}}\:\text{d}\,t=0\:\text{A}\$. So, from the first equation at \$t=0\$:

$$\left[\frac{\text{d}}{\text{d}\,t}V_{_{\text{O}}}\right]_{t=0}+\frac{V_{_{\text{O}\:t=0}}}{\left(1-\frac{15\:\Omega}{R_2}\right)\cdot\left(R_1\mid\mid R_2\right)\cdot C_1}=0\:\text{A}\\\\ \alpha\,A_1+\beta\,A_2+\frac{V_{_{\text{O}\:t=0}}}{\left(1-\frac{15\:\Omega}{R_2}\right)\cdot\left(R_1\mid\mid R_2\right)\cdot C_1}=0\:\text{A}$$


$$A_2=-\frac{V_{_{\text{O}\:t=0}}}{\beta}\cdot\left(\frac{1}{\left(1-\frac{15\:\Omega}{R_2}\right)\cdot\left(R_1\mid\mid R_2\right)\cdot C_1}+\alpha\right)$$

Plugging in \$A_1=100\:\text{V}\$, \$A_2\approx -151.26\:\text{V}\$, \$\alpha\approx -\frac1{928.57\:\mu\text{s}}\$, and \$\beta\approx 711.971455\:\frac{\text{rad}}{\text{s}}\$ find:

$${V_{_\text{O}}}_{\left(t\right)}=100\:\text{V}\cdot\exp\left(\frac{-t}{928.57\:\mu\text{s}}\right)\cdot\left[\cos\left(711.971455\:\frac{\text{rad}}{\text{s}}\cdot t\right)-1.5126\cdot\sin\left(711.971455\:\frac{\text{rad}}{\text{s}}\cdot t\right)\right]$$

Just falls out.

Here's a plot from Desmos of the above equation:

enter image description here

Here's a simulation result:

enter image description here

The data points match exactly.

  • \$\begingroup\$ Wow, this is some wizardry. Thanks for showing me how to solve it entirely. Really I'm only confused about how you figured out Va = Vo / (1+15/R2) Also this might not be clear but switch operation allows for capacitor and inductor to both be connected in parallel with the source, therefore the inductor forms a short which means 100V/5kohms of current through the short and no voltage across the open capacitor. \$\endgroup\$ Commented Mar 18, 2023 at 23:00
  • \$\begingroup\$ @Thxforhelping I added a short node about how to resolve your confusion on Va. As far as the rest of what you wrote, I'm not sure I'm reading you well. When the switch flips at t=0, as I gathered your schematic, the inductor and capacitor are indeed in parallel. The schematic I drew up shows what it looks like (I think) after the switch has moved from the earlier state to the new state. The switch removes the earlier voltage source and resistor and connects the now-charged capacitor to the rest of your circuit. That's what I think. But did I screw up? No wizardry. Just brain-dead math. \$\endgroup\$ Commented Mar 19, 2023 at 0:40
  • \$\begingroup\$ That's what I mean, that for t < 0 the capacitor will not charge because of the short formed by the current through the inductor. Thanks for adding the short node and for clarifying your process. It was very helpful! \$\endgroup\$ Commented Mar 21, 2023 at 1:32
  • \$\begingroup\$ @Thxforhelping Hmm. I personally don't think it makes any sense that the inductor is connected prior to t=0. What exactly is the switching doing if you think that is true? \$\endgroup\$ Commented Mar 21, 2023 at 4:28

You can deal with a dependent voltage source in the nodal equations the same way you'd deal with any other voltage source:

  1. Add an equation relating the voltages of the nodes on the two sides of the source. For example if the node on the left is A and the one on the right is B, you'd write \$v_A-v_B=15i\$.

  2. Consider those two nodes as a supernode and write an equation for the total current entering (or leaving) the two nodes taken together, rather than individual equations for each of the nodes.

  • \$\begingroup\$ Thanks for your help, that's a great point. If I went with the first method, would my method of two node voltages work and just use that relationship between Va - Vb = 15i? Method 2 is interesting as well, would that look like iC + iL = i, then another that looks like i + iR1 = I(S)? Then combine them and get iC + iS = iL + iR1? \$\endgroup\$ Commented Mar 17, 2023 at 6:47
  • \$\begingroup\$ Those aren't two different methods of dealing with the voltage source. They're two steps that are both required to deal with a voltage source in nodal analysis. \$\endgroup\$
    – The Photon
    Commented Mar 17, 2023 at 17:49
  • \$\begingroup\$ So for this particular dependent voltage source, I would need both KCL and the Nodal analysis equations in order to solve it. Thanks for the help! \$\endgroup\$ Commented Mar 18, 2023 at 23:08
  • \$\begingroup\$ The equations used in nodal analysis are KCL equations. (Plus supernode equations when voltage sources are present) \$\endgroup\$
    – The Photon
    Commented Mar 19, 2023 at 0:40

A hint: that 15i should be treated as a voltage. You have thought it's a current.

Learn also how to write into the equations

  • initial voltages of the capacitors and
  • initial currents of inductors

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