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I have the below circuit. enter image description here

I'm solving for Vx. Vx is just Vout for another voltage divider. So, the top left R is R1 the bottom is R2. Therefore shouldn't Vx be R/3R (Vs) = 1/3 Vs?

But it's actually enter image description here

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  • \$\begingroup\$ The blue text is correct. The current's path from Vx to ground (or Vout -) is through the parallel combination of R and 2R. \$\endgroup\$ – Chu Nov 2 '19 at 19:52
  • \$\begingroup\$ which R's are they though? do we not need the top left R? Isn't that in series w the voltage divider? \$\endgroup\$ – Melanie Sanders Nov 2 '19 at 20:00
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    \$\begingroup\$ imho the title is not representative of the question here... what are the two voltage dividers that are in parallel here? If anything it's two cascaded voltage dividers... \$\endgroup\$ – vicatcu Nov 2 '19 at 20:06
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    \$\begingroup\$ Yes. The top left resistor is, say, \$\small R_1\$, the combination of the other three resistors is, say, \$\small R_2=\frac{R\times 2R}{R+2R}=\frac{2}{3}R\$, hence \$\small V _x =\frac{R_2}{R_1+R_2}V_{s}\$ \$\endgroup\$ – Chu Nov 2 '19 at 20:13
  • \$\begingroup\$ All of the current flowing into \$V_\text{X}\$ from \$V_\text{S}\$ comes through a single \$R\$. All of the current flowing out of \$V_\text{X}\$ to ground flows through two paths; an \$R\$ path and a \$2\,R\$ path. So:$$\frac{V_\text{S}-V_\text{X}}{R}=\frac{V_\text{X}}{R}+\frac{V_\text{X}}{2\,R}$$Must be so, yes? Multiply through by \$2\,R\$ and get:$$2\left(V_\text{S}-V_\text{X}\right)=2\,V_\text{X}+V_\text{X}$$And,$$2\,V_\text{S}=2\,V_\text{X}+V_\text{X}+2\,V_\text{X}=5\,V_\text{X}\implies 2\,V_\text{S}=5\,V_\text{X}$$ \$\endgroup\$ – jonk Nov 3 '19 at 5:02
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schematic

simulate this circuit – Schematic created using CircuitLab

This circuit proves intuitively that Vx = V1.(2/5)

Alternatively Millman's theorem can be used or just combining resistors in parallel and series in potential divider fashion also proves this.

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