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As electric current will flow in path with zero resistance when a zero resistance path and finite resistance path are in parallel combinations. In a Wheatstone bridge, the resistors have a finite resistance and the resistance of the galvanometer can be considered as either negligible or finite.

So,

  • Either all the current must flow through the galvanometer

  • Or some current must flow through the galvanometer as is it in a parallel connection with the resistors But, why doesn't the electrons flow through the galvanometer?

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    $\begingroup$ Low resistance does not alone mean current flowing. You still need a potential difference (voltage) to create the current. $\endgroup$ – Steeven Apr 20 '18 at 16:39
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Using this image from Wikipedia:

By Rhdv - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=2888809

Consider $R_1 = R_2 = R_3 = 2\Omega$ and the unknown $R_\text{x} = 1 \Omega$, but we don't know that yet.

Also consider that the galvanometer is not yet in place.

Now there will be more current through B than through D. Also the potential of D will be in the middle of the potential difference of the batterie's terminals while B will be at a lower potential than D because the voltage divider $R_3$-$R_\text{x}$ is not symmetric. Current will flow through the galvanometer from B to D.

This tells us to decrease $R_2$. At half the initial resistance, the whole thing becomes symmetric because now D is at the same potential as B. There will be no current between two nodes at the same potential, no matter the resitance of the galvanometer.

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As electric current will flow in path with zero resistance when a zero resistance path and finite resistance path are in parallel combinations. ... But, why doesn't the electrons flow through the galvanometer?

First, that galvanometer isn't in parallel with any of the resistors in the Wheatstone bridge so your logic here doesn't work.

Second, if there is a current through the galvanometer, the bridge cannot be balanced. You can conclude this without resorting to any equations as follows (refer to the diagram):

enter image description here

To say that the bridge is balanced is to say that the open-circuit bridge voltage $V_{out}$ is zero which implies both

(1) the voltage across $R_1$ equals the voltage across $R_3$

(2) the voltage across $R_2$ equals the voltage across $R_4$

Carefully note that (1) and (2) hold just when there is zero current from $C$ to $D$ (there must be zero current since there is an open-circuit between those nodes).

Now, replace the open-circuit with an ideal wire (short-circuit) between nodes $C$ and $D$ and see that now

  • $R_1$ is in parallel with $R_3$
  • $R_2$ is in parallel with $R_4$

Since parallel connected circuit elements have identical voltage across, it follows that (1) and (2) still hold and so conclude that, since a circuit solution is unique, the current through the wire and the open-circuit must be equal, i.e., the current through the wire is zero.

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