What is the best method for solving this circuit analysis problem in the S-domain?

Question

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.

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.

Answer 1

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\$:

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

Or,

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

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:

Here's a simulation result:

The data points match exactly.

Answer 2

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.

Answer 3

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