Does the resistance in the secondary circuit of a potentiometer circuit affect the balance length? I just learnt how potentiometer circuits work, and I was taught that the resistance in the secondary circuit does not change the balance length as no current flows through the secondary circuit and thus the only potential drop is through the EMF of the unknown battery in the secondary circuit. However my proffessor mentioned that if the resistance of bulb x was decreased in this particular arrangement, the balance length would shorter!
The reason he gave was "This is because with a lower resistance
bulb used for X, the current flowing in the lower circuit increases, the voltage drop
across the internal resistor increases, and hence the terminal potential difference
across the cell in the lower circuit decreases. " However I am struggling to understand why the concept of no current flowing through the circuit does not apply here, I still feel that decreasing the resistance of bulb x would not affect the balance length for the reason stated above. Thank you for clarifying this conceptual error of mine in potentiometer circuits!

 A: 
However I am struggling to understand why the concept of no current
  flowing through the circuit does not apply here

The condition required for there to be zero current through the ammeter connected to $C$ is this: all of the current through the bulb $Y$ is through the battery labeled with voltage $E$ (note that the battery is modeled as an ideal battery in series with an internal resistance).
If you'll think about this a little bit, you'll realize that if there's zero current through the ammeter, the ammeter can be removed with changing the value of any voltage across or current through in the circuit (if that isn't clear, stop here and think about it some more).
Now, since the (ideal) ammeter has zero volts across (regardless of the current through), it follows that with the ammeter in circuit, the voltage AC equals the voltage across the cell.
So, stipulate that the potentiometer is adjusted such that the ammeter reads zero and then the ammeter is removed. Given the above, it must be that the voltage AC is still equal to the voltage across the cell.
That is, the condition that there is zero current through the ammeter is the condition that the voltage AC equals the voltage across the cell when the ammeter is not connected.
Carefully note that this does not imply that there is no current through the bulbs!

if these bulbs are in parallel with the battery, musn't their
  potential drops match that of the battery in the secondary circuit?

Correct (if you mean the combination of the bulbs and not the individual bulbs), but that potential drop (voltage across) isn't constant, it is a function of the resistance of the bulbs, the internal resistance of the cell, and the voltage $E$. Assuming the ammeter has been removed, the voltage across the cell is given by:
$$V_{cell} = E\left(1 - \frac{r}{r + R_Y + R_X || R_Z}\right)$$
where
$$R_X || R_Z = \frac{1}{\frac{1}{R_X} + \frac{1}{R_Z}}$$
Only in the case that $r=0$ is this a constant.
A: The bulbs X Y and Z are not part of a normal potentiometer. Normally the battery E is balanced against AC and CB, but now AC is replaced by the combination of AC, X, Y, and Z. The internal resistance of the battery and meter don't matter, but these extra ones certainly do because they are in parallel with the battery instead of in series, and changing one of them will affect the balance point.
