How do I simplify the circuit below?
simulate this circuit – Schematic created using CircuitLab
I know that R2 and R3 form a voltage divider, but I can't figure out the input impedance that In sees. I think it is (R1||[R2+R3]), but am not sure how to show that. Given V1 and VIn, I'm not sure how to determine VOut, or what the Thevenin equivalent circuit is.
Your schematic with some comments added:
simulate this circuit – Schematic created using CircuitLab
Assuming you leave \$V_{_\text{OUT}}\$ unloaded, then \$V_{_\text{IN}}\$ sees \$R_1\$ going to one ideal voltage source and sees \$R_2+R_3\$ going to another ideal voltage source. So the resulting impedance seen from the point of view of \$V_{_\text{IN}}\$ (\$V_{_\text{OUT}}\$ unloaded) would in fact be just as you said: \$R_1\vert\vert\left(R_2+R_3\right)\$.
You can demonstrate this fact by injecting a \$1\:\text{A}\$ current source into \$V_{_\text{IN}}\$. (There are more mathematical ways -- graph theoretic -- which provide more insight, too. But it would take me longer and I'd rather not spend the time if I can avoid it.) Suppose you don't inject a current and measure a voltage at \$V_{_\text{IN}}\$ and then inject \$1\:\text{A}\$ and measure a voltage at \$V_{_\text{IN}}\$, again? The impedance/resistance is the difference between these voltages (divided by \$1\:\text{A}\$ to keep the units correct, but it doesn't make a numerical difference.)
The KCL equation, including the optional injected current, would be \$\frac{V_{_\text{IN}}-V_1}{R_1}+\frac{V_{_\text{IN}}}{R_2+R_3}=I_{_\text{INJ}}\$. This solves out as \${V_{_\text{IN}}}_{(I_{_\text{INJ}})}=\frac{\left(V_1+I_{_\text{INJ}} \,R_1\right)\left(R_2+R_3\right)}{R_1+R_2+R_3}\$. We want to know \$\frac{{V_{_\text{IN}}}_{(1\:\text{A})}-{V_{_\text{IN}}}_{(0\:\text{A})}}{1\:\text{A}-0\:\text{A}}\$. But that's \$\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}\$. Which also happens to be what you get for \$R_1\,\vert\vert\left(R_2+R_3\right)\$.
You could go further. You could set up two KCL equations with two optionally injected currents, one at \$V_{_\text{IN}}\$ and one at \$V_{_\text{OUT}}\$, and use that to help you see what the impedance at \$V_{_\text{IN}}\$ looks like with differing loads on \$V_{_\text{OUT}}\$. If you need help with that, let me know.
Edit: I left this question hanging for a couple of hours, mid-edit, before finishing it. I see Spehro already answered it now that I finally posted this up. So I guess you have two thoughts. :)
Here is what your circuit looks like with another voltage source and assuming no load on Vout:
simulate this circuit – Schematic created using CircuitLab
One way to solve this is with superposition. Short V1 and solve for Vout (maybe you want to solve for Vx first and then find Vout). Then short Vin and solve for Vout. Add the two.
In this case, it's pretty easy to just write down the answer in the form
Vout = (V1/R1 + Vin/Rsource) * (some stuff) * (more stuff)
by looking at it, but it's better to do things methodically at the beginning.
According to the “Substitution Theorem” of network theory we can replace any passive part or branch of a network by a fixed voltage source if this source has the same value as the voltage across this branch. According to this theorem, there will be no change of all branch voltages and currents within the network.
With respect to your circuit example, this would mean that the resistor would be replaced by a corresponding voltage source BETWEEN the nodes labelled as "in" and "out".
how to show that.? \$\endgroup\$