Answer by periblepsis for Simple circuit analysis question, voltage across an unknown component

Perhaps it may be easier to see if you select the negative probe as the default \$0\:\text{V}\$:

Then just place the voltmeter measurements where they belong (on the + side, because that's what the voltmeter is reporting.)

With the given zero-reference, the upper wire is \$+1.4\:\text{V}\$ and the lower wire is \$+0.8999\: \text{V}\$. So the unknown device sits between these two, with the difference being the voltage applied across it.

However, if you want KVL then start with this:

Then you find, working counter-clockwise as shown:

$$\begin{align*}0\:\text{V}-V_2+V_1-V_3&=0\:\text{V}\\\\\therefore V_3&=V_1-V_2\end{align*}$$

And this gets you the same result.