Resolution a problem with voltage divider
I built this in CircuitLab.



Consider the sliding contact of the potentiometer to divide the 2kΩ resistance into two parts, the part "above" the sliding contact and the part "below" the sliding contact.

Set the position of the sliding contact so that the ratio of the resistance above the contact to the resistance below the contact is 8.3:3. This gives resistances of \$ 2kΩ \times \frac{8.3}{3.0+8.3} = 1,464Ω\$ and \$ 2kΩ \times \frac{3.0}{3.0+8.3} = 536Ω\$.

When simulated, the current through R2 is 13 mA and the current through R6 is 23 uA. That is, the current through R6 is a thousand times less than the current through R2, i.e. effectively zero.

Note that \$ R2/R3 = R4/R5 \$ so that the voltages at "NODE 2" and "NODE 3" are identical, so that there is no potential difference between "NODE 2" and "NODE 3" and the current must necessarily be zero.