Current being 0 in potentiometer Why does the current become 0 in potentiometer when the potential difference is equal to the emf of the battery? I know how a potentiometer works, this is the only part troubling me.
 A: The potential difference and the emf of the battery are trying to push current in opposite directions. When they are equal, neither is stronger than the other, so there is no current.
A: If two points are at the same potential and a conductor is connected between them no current will flow.
Consider an ideal potentiometer arrangement with a $2\,\rm V$ driver cell as shown below.  

The potential diagram for loop $ABCDA$ is shown on the left.
The potential rises from $0\, \rm V$ to $+2 \, \rm V$ along the potentiometer wire $AB$, does not change along the connecting wire $BC$, drops from $+2\, \rm V$ to $0 \, \rm V$ between the terminals $CD$ of the driver cell and then does not change does along the connecting wire $DA$.
For the other part of the circuit without a connection at $G$ has the potential not changing along the connecting wire $AE$, a rise in potential from  $0\, \rm V$ to $+1 \, \rm V$ across the terminals $EF$ of the other cell and then staying at a potential of $$+1 \, \rm V$ between $F$ and $G$ because the circuit is not complete an no current flows in that part of the circuit.  
Now make a connection with the potentiometer wire at $G$ such that the potential difference across the potentiometer wire is exactly the same as the emf of the cell.
There is still no potential difference between $F$ and $G$ and so no current flows through the galvanometer.  
Making the connection anywhere else along the potentiometer wire will produce a potential difference between $F$ and $G$ which will produce a current through the galvanometer.
A: If you know how a potentiometer works, then presumably you're happy about the current being zero when the 'unknown' pd is equal to the pd across a portion of the slide-wire. So your difficulty must be in understanding why, when a battery is is connected as the unknown pd, we can take this pd as equal to the battery's emf. Perhaps what you've not seen is that when the current is zero, there is no pd across the battery's internal resistance, so the terminal pd is equal to the emf. To give you a better answer, I think we need to know more about what it is that you don't understand.
