# Simple circuit question: nodal analysis

"simple circuit"

I'm trying to find vout as a function of v1,r,c, and v2, but I'm i'm kind of stuck on what to do.
I'm pretty sure you need to do nodal analysis, but I've never really encountered a circuit like this where I need to find the voltage when it goes to ground.

My initial attempt looks like (v1-vx)/R+(v2-vx)/Zc=0 where vx is the voltage at that node. I'm not sure if there's more to the equation or how to find bout.

Any help would be appreciated.

Well, notice that using KCL we can see that:

$$0=\text{I}_1+\text{I}_2\tag1$$

And we also know that:

$$\text{I}_1=\frac{\displaystyle\text{V}_1-\text{V}_\text{o}}{\displaystyle\text{R}}\space\wedge\space\text{I}_2=\frac{\displaystyle\text{V}_2-\text{V}_\text{o}}{\displaystyle\frac{1}{\text{sC}}}=\text{sC}\left(\text{V}_2-\text{V}_\text{o}\right)\tag2$$

Combining, gives:

$$0=\frac{\displaystyle\text{V}_1-\text{V}_\text{o}}{\displaystyle\text{R}}+\text{sC}\left(\text{V}_2-\text{V}_\text{o}\right)\space\Longleftrightarrow\space\text{V}_\text{o}=\frac{\displaystyle\text{V}_1+\text{V}_2\text{sCR}}{\displaystyle1+\text{sCR}}\tag3$$

You've got all you need, you just need to solve that equation for Vx, everytime you work with voltages in nodes you're getting the voltage between that node and ground, so all you have to do is solve for Vx and that is the answer.

The ground sign here means the reference voltage (0 Volts), which means every voltage told in the circuit are actually referenced according to that. What you call V1 is actually the potential difference between the leftmost node and the "ground", which you should consider as 0 Volts.

So your path to solution is correct. Just remember that what you call voltage should always be referenced from somewhere on the circuit.

Your answer is correct if you have a known expression of Zc. I hope you aren't assuming it to be 1/jwC as this will be valid only for sinusoid input system. A more general approach will be to write I = C dv/dt where I is (v1-vout)/R. Now solve this differential equation for vout.

For transfer function in the s-domain, replace Zc by 1/sC and solve for Vx. I think you already have an answer.