If two or more cells of unequal voltages are connected in parallel (with the same terminal on the same side), is there a formula that gives the net potential difference?
Also, by Kirchoff's loop law, if we go round in the loop (the internal one, not the external one), we gain $V_1$ going in the direction of current and lose $V_2$ going against it. This implies $$V_1-V_2=0$$ but $V_1\neq V_2$. What does this mean?
The voltage of a cell is caused by a chemical process, and the details of this process determine the behavior of the cell when a current flows. Simplistically, that behavior can be described as the cell having an internal (series) resistance - although the real behavior is more complex (an internal resistance would not explain that a battery goes flat after a certain amount of current has flowed, for example).
Once you have a series resistor in your model, it is easy to see how to compute the voltage obtained when the cells are in parallel: you assume an output voltage $V$ and calculate the net current flowing into our out of each of the cells.
One complicating factor is that, depending on the battery chemistry, the "internal resistance" may be different for the cell when it is charging vs when it is discharging. But with all these simplifying assumptions in place, we would calculate the net (no-load) voltage of cells with no-load voltage $V_i$ and internal resistance $R_i$ with:
$$\sum{\frac{V-V_i}{R_i}}=0\\ V = \frac{\sum\frac{V_i}{R_i}}{\sum\frac{1}{R_i}}$$
As a simple sanity check of this equation, if you had two cells with an internal impedance of 1 Ohm each, the expression for their voltage would be $V = \frac12 (V_1+V_2)$ which intuitively makes sense.
This means that you are creating a short circuit. Suppose you connect the + of cell 1 to the - of cell 2. So far everything is OK. When you close the loop by connecting the + of cell 2 to the - of cell 1, you will be applying a potential difference $V_2-V_1$ to the wire connecting both. This wire has a very low resistance, so the current will be very high. Your equations do not seem to make sense, but they will again make sense if you take into account the resistance of the connecting wires.
I did not mention the internal resistance of the cells, but also this internal resistance (think of it as a resistance in series with an ideal cell) can be the factor that will determine the max current, depending on which is the smallest, internal cell resistance or wire resistance.
Easyest anser is a battery with more volt than the other will drain until they are equle voltage unless you put a diode from low battery + to full battery +. Sorry for spelling errors
What does this mean?
It means that your model isn't valid.
Only in the context of ideal circuit theory does a voltage source produce a voltage across independent of the current through.
In this context, connecting two (ideal) voltage sources in parallel leads to a contraction, e.g.,
$$1 = 2$$
But physical voltage sources cannot produce unlimited current and maintain the voltage across their terminals.
The most straightforward extension that produces a reasonable result is to model the cell with a Thevenin equivalent; an ideal voltage source in series with a resistor of resistance
$$R_{TH} = \frac{V_{OC}}{I_{SC}}$$
where
Then, given two cells, the voltage across the parallel combination in this simple model is given by
$$V_{||} = \frac{V_{OC_1}R_{TH,2} + V_{OC_2}R_{TH,1}}{R_{TH,1} + R_{TH,2}}$$
Now, while this gives a reasonable result, it is still a simple model and doesn't take into account that, for example, one or both cells may rapidly heat up to the point of rapid unplanned disassembly.
In summary, one should be careful to not apply a mathematical model outside of the context in which it is valid.
You cannot connect ideal cells of unequal emf in parallel. The potential difference between the common ends would then have two or more different values at the same time, which is impossible. See Combination of ideal voltage sources.
All real cells have a non-zero internal resistance. This allows their terminal PD to differ from their emf. Terminal PD is the emf less the drop in potential across the internal resistance. For real cells in parallel the current through each cell adjusts so that the terminal PD has the same value in each branch.
Your mistake in your application of Kirchhoff's Voltage Law is that you have ignored the PD across the internal resistance in each cell. Or perhaps you have assumed that the cells have no internal resistance.
Thevenin's Theorem states that any network of resistors and cells is equivalent to a single resistor in series with a single cell. If the network consists of cells in parallel, each with an emf $V_i$ and an internal resistance $R_i$, then the voltage (ie emf) and resistance of the equivalent series resistor and ideal cell are given by
$$\frac{1}{R_{Th}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$$
$$\frac{V_{Th}}{R_{Th}}=\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}+...$$
