How is open-circuit voltage calculated?
It's not hard, but you need to have developed some very modest appreciation for voltage, current, and resistance by working through a few different examples for this to become more obvious. One of the very specific things you need to really understand better, perhaps, is the fact that when a current flows through a resistor, a voltage drop develops. Now, even my use of the word "develops" there might throw you off. It also could be just as well said that it "a difference in voltage potential must be present across it." It doesn't matter whether you think of it as the current creating the voltage, or the voltage creating the current, so long as you know that Ohm's Law tells you that there will be such a voltage if there is such a current, and visa versa.

Okay. Granting that. How can you compute the currents? Well, take your left side schematic. It's a series loop of three resistors. So the current through this loop will be:

$$I_{TOTAL} = \frac{U_{TOTAL}}{R_{TOTAL}}$$

But you know that:

$$R_{TOTAL}=R_1+R_3+R_4$$

So the current is:

$$I_{TOTAL} = \frac{U_{TOTAL}}{R_1+R_3+R_4}$$

That is where that denominator comes from, by the way.

Now, the next problem is the voltage at \$U_i\$. One side is referenced to the bottom node and the other side is in between two resistors. You could come at this from either side, approaching a computation that either adds up the voltages on the left side or else adds them up on the right. Either way, you'll get the right answer. Let's just go straight from the right side where there are two resistors there.

To compute the voltage across them, you need to know the current through them. But you know that, as we just computed it above. So let's do the sums:

$$\begin{align*}
U_i &amp;= I_{TOTAL}\cdot R_4 + I_{TOTAL}\cdot R_3 \\
 \\
 &amp;= \frac{U_{TOTAL}}{R_1+R_3+R_4}\cdot R_4+\frac{U_{TOTAL}}{R_1+R_3+R_4}\cdot R_3 \\
 \\
 &amp;= U_{TOTAL}\cdot\left(\frac{R_4}{R_1+R_3+R_4} +\frac{R_3}{R_1+R_3+R_4} \right) \\
 \\
 &amp;= U_{TOTAL}\cdot\frac{R_3+R_4}{R_1+R_3+R_4}
\end{align*}$$

Do you see that now?



Regarding \$R_i\$, I'll take a diagram approach. It wastes tons of HTML space, but a diagram is worth a thousand words they say.







simulate this circuit &ndash; Schematic created using CircuitLab

You can see the transitions there. There's only one more to go:







simulate this circuit

That's it.

So, just extracting from that schematic, the two equations are:

$$\begin{align*}
U_{TH} &amp;= \frac{U}{R_1}\cdot\frac{R_1\cdot\left(R_3+R_4\right)}{R_1+R_3+R_4} = U\cdot\frac{R_3+R_4}{R_1+R_3+R_4} \\
 \\
R_{TH} &amp;= \frac{R_1\cdot\left(R_3+R_4\right)}{R_1+R_3+R_4} 
\end{align*}$$

You'll notice that we re-aquired our voltage calculation from this, too. I hadn't realized that this is the approach you needed from the start, I guess. Hope that helps.