# Circuit with two voltage dividers in parallel

I have the below circuit.

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

• The blue text is correct. The current's path from Vx to ground (or Vout -) is through the parallel combination of R and 2R. – Chu Nov 2 '19 at 19:52
• which R's are they though? do we not need the top left R? Isn't that in series w the voltage divider? – Melanie Sanders Nov 2 '19 at 20:00
• 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... – vicatcu Nov 2 '19 at 20:06
• 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 $\small V _x =\frac{R_2}{R_1+R_2}V_{s}$ – Chu Nov 2 '19 at 20:13
• 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,$$2\,V_\text{S}=2\,V_\text{X}+V_\text{X}+2\,V_\text{X}=5\,V_\text{X}\implies 2\,V_\text{S}=5\,V_\text{X}$$ – jonk Nov 3 '19 at 5:02