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Voltage is to be calculated through a path between points $A$ and $B$. The path travelled can either be from $A$ to $B$ or from $B$ to $A$. Don't consider KVL conventions as this is not a loop.

Does the direction affect the final voltage value?

If we go from positive terminal to negative terminal of the cell, is the voltage taken to be positive ? or does it depends on what convention we want to follow?

For Example, $A\:-(3)+ -(2)+ -(-1)+\: B$

Here + and - indicate the polarity of the cells $3\,V$, $2\,V$, $-1\,V$

If I take positive terminal to negative terminal flow as positive and negative terminal to positive terminal as negative, since the voltage is the algebraic sum

$$\begin{cases} V_{ab} = -3-2-(-1) = -4\,V \\ V_{ba} = -1+2+3 = 4\,V \end{cases}$$

Is this correct? And if a resistance is also connected between the path, what sign convention is the voltage drop IR given?

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For cells there can be no ambiguity about the potential differences across the cells.
When na ideal cell is labelled $3$ V then it means that the cell will maintain the positive terminal a potential which is $3$ V higher than the negative cell.

The first possible point of confusion is the labelling of the voltages.
For some $V_{\rm AB}$ means the change in potential in going from node $A$ to node $B$ whereas others will say that is the potential of node $A$ relative to node $B$.
I think that you are using the first convention?

Consider your arrangement of cells.

enter image description here

Below the circuit digram I have drawn a potential against position diagram. which clearly shows that the potential at node $A$ as $4$ V higher than the potential at node $B$.
This is written as $V_{\rm AB}= -1+2+3 = +4 $ V which means the potential of node $A$ relative to node $B$ or the potential difference between $B$ and $A$.
The potential of node $A$ relative to node $B$ is written as $V_{\rm AB}=+4 $ V.

Note that So going from $B$ to $A$ the potential rises and going from $A$ to $B$ the potential drops.
The current direction through the cells can be either from $B$ to $A$ which it often is, or $A$ to $B$ where you can think of the cell as being recharged or electrical energy is in some way being converted into other forms of energy within the cell.
The direction of the current through a cell does not change the potentials difference across the cell.

Although you said that you did not want to use KVL as there was no loop, there is in fact a loop although in this case the loop from $A$ to $B$ which has an infinite resistance.
So using KVL which is a really a convenient form of the law of conservation of energy we can walk around a complete circuit with a charge and have no net work done on the charge.
The direction of the walk does not matter.

Loop $BDCAB \Rightarrow V_{\rm BD} + V_{\rm DC}+V_{\rm CA}+V_{\rm AB} =0 \Rightarrow -1+2+3 + V_{\rm AB} =0$

and there you have $V_{\rm AB} = -4$V as before.

Going around the loop in the opposite direction gives the expected result $V_{\rm BA} = +4$V.

For a resistor the current always flows from the node at the higher potential to the node at the lower potential.
However there may a difficultly in assigning a current direction in a complex circuit.
This should present no difficulties if one is consistent with whatever convention is being used as is illustrated in the trivial example below.

enter image description here

Using KVL for the left hand circuit will result in a value for the current of $-2$ A whereas for the right hand circuit the current is $I=+2$ A.
So the guess for the current direction in the left hand circuit was incorrect and the current is actuallly two amps flowing anticlockwise with the left hand side of the resistor being at a higher potential than the right hand side so $V_{\rm R} = -4$V.

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