Answer
2 volts
Work Step by Step
We can apply Kirchoff's loop rule to this problem, recall that that is $\Sigma_i^n \Delta V_i = 0$ over a closed circuit loop. Let $\Delta V_?$ be the unknown potential difference.
Starting at the battery, this gives us:
$\varepsilon + \Delta V_?$ - 8 - 6 = 0
You may ask why are we subtracting the potential difference across the resistors since the picture has then listed as positive values? This is because when we traverse the loop and go through a resistor from the positive to negative end, we must negate the potential difference. Continuing to solve for $\Delta V_?$ gives us
$\Delta V_? = 14 - \varepsilon = 14-12 = 2$ volts
