# Why can't we add the emf of two batteries in a parallel circuit?

My teacher told me that two batteries' emf cannot be added in a parallel circuit but can be added in a series circuit. However I can't seem to fully grasp why and I think there may be some loopholes in my understanding of emf. I've attached two photos to illustrate my argument for why I think two batteries in an emf can be added in a parallel circuit. To exemplify my argument, I've drawn a circuit with two batteries of 9V and 3V in the first diagram. The two batteries' terminals are opposite to each other.

This diagram can then be redrawn like this (diagram 2). From my understanding of emf of a battery, it is the energy supplied per unit positive charge, by the battery, in moving the charge from the negative terminal to the positive terminal of the battery. It can therefore be thought of as the potential difference between the two terminals. In the case of the 9V battery drawn, I can take the negative terminal of the 9V battery to have a potential of 0V and the positive terminal to have a potential of 9V, hence the PD between the two terminals will be 9V-0V=9V. This can then be similarly applied to the 3V battery where its negative terminal has a potential of 0V and its positive terminal has a potential of 3V.

As shown in diagram 2, since the 9V and 3V batteries are connected in parallel at the point of the two thick dots drawn, hence they share a common potential difference at those two points. The potential at the left thick dot will have a potential of 9V+0V (drawn in blue) due to the 9V battery's positive terminal and 3V battery's negative terminal. Similarly, the potential at the right thick dot will have a potential of 0V+3V (drawn in blue) due to the two batteries' terminals. Therefore the potential difference between the two thick dots will be 9V-3V=6V. Therefore, showing that batteries emf can be added (or subtracted in this case since the two terminals are in opposite direction).

In the end, the circuit can thus be drawn as a battery with 6V like this. • Sep 10, 2018 at 15:55
• Have you learned Kirchhoff's Voltage Law (KVL)? What happens if you apply it around the upper loop in your circuit? How about if you first apply it in the loop containing the 3 V battery and the resistors, then apply it in the loop with the 9 V battery and resistors --- are the results self-consistent? Sep 10, 2018 at 16:35
• Bobby, you didn't truly draw a parallel circuit for your batteries. For two batteries wired in parallel, the positive side of one battery must be wired to the positive side of the other battery, and the negative side of one battery must be wired to the negative side of the other battery. Sep 10, 2018 at 17:42
• @DavidWhite, do you have a reference for this? As drawn, how are the two cells connected if not parallel connected? Anti-parallel? Sep 11, 2018 at 2:02
• @AlfredCentauri, I don't quite "get" what you are asking. If I wire the negative terminal of one battery to the positive terminal of a second battery, then I wire the positive terminal of the first battery to the negative terminal of the second battery, the batteries are obviously wired in series. Sep 11, 2018 at 2:30

The diagram you have drawn is an arrangement which has a loop with the emfs of two cells adding up and the resistance of that loop being zero (ideal cells and connecting wires) or the internal resistance of the cells and the resistance of the connecting wires (real components).

For the ideal component circuit the current through that loop would be infinite and extremely high if the cell resistances etc were low.
The cells would discharge very quickly and in the real world generate a lot of heat and probably explode or cause a fire.

So your teacher was right in this case.

• Bobby, just take your teacher's word for it, and quit trying to find the exception to the rule. Sep 11, 2018 at 16:28

You can think of the emf as force or pressure, if you connect the 9V to the 3V, the 9v will actually overpower the 3v battery and make it run in reverse (and overcharge or blowup). 2 batteries both 9v in parallel have the advantage of suppling 2x current which is better. Sometimes a newer battery is put in parallel with an older battery and that can lead to trouble.

The most straightforward way to see why your third drawing is not correct is to place a resistor in series with each of the cells and label them, e.g., $r_{9V}$ and $r_{3V}$ respectively. These resistors represent the internal resistance of each cell.

It's straightforward to solve for the voltage $V$ across the two resistor branch with superposition:

$$V = \frac{9\,\mathrm{V}}{1 + \frac{r_{9V}}{r_{3V}||5}} - \frac{3\,\mathrm{V}}{1 + \frac{r_{3V}}{r_{9V}||5}}$$

Now, stipulate that the internal resistances are much less than $R_1 + R_2 = 5\,\Omega$ and then

$$V \approx \frac{9\,\mathrm{V}}{1 + \frac{r_{9V}}{r_{3V}}} - \frac{3\,\mathrm{V}}{1 + \frac{r_{3V}}{r_{9V}}}$$

See that only if $r_{3V} = 3\cdot r_{9V}$ will it be the case that $V = 6\,\mathrm{V}$.

While you might try fixing the ratio $\frac{r_{3V}}{r_{9V}} = 3$ while letting both internal resistances go to zero and conclude that your 3rd drawing is correct for ideal cells, note that the cell current diverges in this limit.

So, for finite cell current, one or both cells must have non-zero internal resistance when connected together as in the first diagram.