Calculating voltage through unknown resistor
Well, we have the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Using KCL, we can write:

$$
\begin{cases}
\text{I}_1=\text{I}_3+\text{I}_4\\
\\
\text{I}_4=\text{I}_2+\text{I}_5\\
\\
\text{I}_6=\text{I}_2+\text{I}_3\\
\\
\text{I}_1=\text{I}_5+\text{I}_6
\end{cases}\tag1
$$

Using KVL, we can write:

$$
\begin{cases}
\text{I}_3=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\
\\
\text{I}_5=\frac{\text{V}_2}{\text{R}_2}\\
\\
\text{I}_6=\frac{\text{V}_3-\text{V}_4}{\text{R}_4}\\
\\
\text{I}_6=\frac{\text{V}_4}{\text{R}_5}
\end{cases}\tag2
$$

Now, we can solve for the unknowns.



Using your values we get:

FullSimplify[
 Solve[{6 == I3 + I4, I4 == 10 + I5, I6 == 10 + I3, 6 == I5 + I6, 
   I3 == (V1 - V3)/1, I4 == (V1 - V2)/5, I5 == V2/2, 
   I6 == (V3 - V4)/R4, I6 == V4/3}, {I3, I4, I5, I6, V1, V2, V3, V4}]]


Which gives:

{{I3 -&gt; -((2 (4 + 5 R4))/(11 + R4)), I4 -&gt; (74 + 16 R4)/(11 + R4), 
  I5 -&gt; (6 (-6 + R4))/(11 + R4), I6 -&gt; 102/(11 + R4), 
  V1 -&gt; 92 - 714/(11 + R4), V2 -&gt; (12 (-6 + R4))/(11 + R4), 
  V3 -&gt; (102 (3 + R4))/(11 + R4), V4 -&gt; 306/(11 + R4)}}


Now, when we know that \$\text{I}_5=4\space\text{A}\$, we get:

$$\text{I}_5=\frac{6\left(\text{R}_4-6\right)}{11+\text{R}_4}=4\space\Longleftrightarrow\space\text{R}_4=40\space\Omega\tag3$$

So:

$$\text{V}_3-\text{V}_4=\frac{102\left(3+40\right)}{11+40}-\frac{306}{11+40}=80\space\text{V}\tag4$$