$Kirchhoff's\space law$ can prove useful here.

*Case 1.* Consider a closed circuit with two batteries, of emf $V_1$ and $V_2$, and a resistor of resistance R (all in series). A circuit with no resistance will catch fire. The resistor is added to prevent short circuiting.

According to Kirchhoff's law,

*(i)* If the $+ve$ terminal of a battery is connected to the $-ve$ of the other :

$V_1 + V_2 -IR=0$

In this arrangement of batteries, their emfs will just add up. No problem encountered here.

*(ii)* If the $+ve$ terminal of a battery is connected to the $+ve$ of the other :

$V_1 - V_2 -IR = 0$

In this arrangement of batteries, the emfs will subtract. The battery with larger emf will decide the current's direction. The current will be directed from the $-ve$ to the $+ve$ terminal of the battery with larger emf.

If batteries of same emf are used then their emfs will simply cancel out, resulting in zero current.

*Case 2.* In a circuit with parallel batteries as shown by the 2nd picture in your question, you can find out the direction of the current and the net emf just like it was found in *Case 1*. You just have to apply Kirchhoff's law correctly. Also, do not forget to insert a resistor.

Kirchhoff's law will prove to be a very useful tool. You just have choose a closed loop and use the sign conventions appropriately.

FYI, for kirchhoff's law to work you can assume any direction of current in the closed loop, either clockwise or anti-clockwise. If you assumed the correct direction of the current, you will get a positive value of current, otherwise negative implying the current is flowing in the direction opposite to the assumed one.

**tl;dr** Batteries do not create electric fields to move charges. They move charges, which creates electric fields.

a battery [...] gives out some electric field that moves through the circuit and gives a force on electrons in conductor to produce current.

This description is, if not completely wrong, at least misleading. A battery is not a source of *electric field*, it is a source of *electric potential*. Imagine a battery with terminals in the shape of a pair of parallel conductive plates with an air gap in between (this is a capacitor): as you move the plates toward each other, the field strength (between the plates) increases, and as you pull them apart, the field strength decreases. There is no upper limit to how strong the field can be (well, until it reaches the breakdown voltage and begins to arc), and no lower limit either -- the strength of the field is not determined by the battery. So the battery itself does not directly create an electric field between its terminals.

Moreover, electric field does not "move through" a circuit; charges do. In a simple DC setup like a battery driving current through a resistor, all electric fields are stable over time -- the *charges* move through the circuit, but the field itself does not. Saying that electric field moves through a circuit is a bit like saying that gravity moves through a rollercoaster. Moving electric fields do come into play with AC circuits and devices with moving parts, such as electric motors; but even a non-moving electric field causes charges to move through a conductor (this is, after all, essentially the definition of "conductor").

So how does current work? Batteries are a source of electric *potential*, which is measured in volts. Potential is a kind of pressure and in a typical battery this pressure is caused by chemical reactions inside the battery pumping electrons from one terminal (+) to the opposite one (-). The potential difference between the terminals does create an electric field. In the experiment above where the terminals of the battery are parallel flat plates separated by a distance $d$, you could calculate the field strength between them as simply $E = {V \over d}$. But the field is just a way of observing the potential difference between the terminals: it's not the reason the charges are moving in the first place (which is, again, the chemical reactions happening inside the battery).

Pressure is relative, and electric potential is no different. To be strictly accurate, what a battery provides is an *electric potential difference* ("voltage") between its two terminals. This is basically a measurement of how strong an electron-pump the battery has inside it. This is why connecting two identical batteries in parallel does not add their voltages: because they both provide *the same* potential difference between the (+) and (-) terminals, there is essentially no difference between two batteries connected in parallel and two isolated, disconnected batteries. The electric potential between the (+) terminals and (-) terminals is the same, and both batteries "agree" on it, so there is no reason for current to flow between them. (Connecting two *mismatched* batteries in parallel *will* cause current to flow, in a direction determined by which battery has higher voltage.)

What connecting two batteries in parallel *does* do is change how the system behaves when under load. If you connect a load, say a 1kΩ resistor, across the terminals of a single 1.5-volt battery, the current through the resistor will be 1.5 mA, all of which is supplied by the same battery. If you connect the same load across the terminals of two 1.5-volt batteries connected in parallel, the current through the resistor will still be 1.5 mA, but now each battery only has to supply 0.75 mA of current. This means that each individual battery is under less load than before, because the electrochemical pumps inside it only have to move half as many electrons to maintain the same voltage. These batteries may last longer and behave better under a variety of loading conditions. (This assumes ideal conditions and perfectly matched batteries. In real-world scenarios, batteries are never perfectly matched, and so you may need a load-balancing circuit to protect the batteries from each other.)

## Best Answer

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.