3 - Resistor Voltage Divider

$$I_0+I_1+I_2=0\qquad,\qquad\text{! Notice the directions of the currents !}$$
$${\large \cases{I_0=\frac{V_0-V_X}{R_0}\\ {\ }\\ I_1=\frac{V_1-V_X}{R_1}\\ {\ }\\ I_2=\frac{V_2-V_X}{R_2}}}$$
$$\frac{V_0-V_X}{R_0}+\frac{V_1-V_X}{R_1}+\frac{V_2-V_X}{R_2}=0$$
$$\frac{V_0R_1R_2-V_XR_1R_2+V_1R_0R_2-V_XR_0R_2+V_2R_0R_1-V_XR_0R_1}{R_0R_1R_2}=0\qquad|\qquad×R_0R_1R_2$$
$$\frac{V_0R_1R_2+V_1R_0R_2+V_2R_0R_1}{R_1R_2+R_0R_2+R_0R_1}=V_X$$
Note: the $V_X$ is the unmarked voltage point on the schematic !

It is easy to see the calculation scheme easily adjusts for N voltages all connecting to the single $V_X$
$$V_{X_N}=\frac{\displaystyle{\sum_{i=0}^{N-1}V_i\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}{\displaystyle{\sum_{i=0}^{N-1}\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}$$
thus BWD chek ::
for N=2 :
$$V_X=\frac{V_1R_0+V_0R_1}{R_0+R_1}$$
incase of the $V_0=0$ it reduces to a trivial 2 resistor divider formula :
$$V_X=V_1\frac{R_0}{R_0+R_1}$$

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Simulation example/chk. for N=4 :

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