Calculating the voltage provided by batteries that have different voltages and are connected in parallel

Question

I'm trying to calculate the reading on a voltmeter connected to the following set of batteries, where each battery is of $1.5$ volts. These batteries are NOT rechargeable, so I'd really appreciate it if you could also shed light on potential risks.

The two batteries in series is equivalent to a 3-volt battery, so I guess the current would flow from point A to point B. This is where my question arises. I know there is always internal resistance inside a battery, but I'm not sure if we have to consider the electromotive force of the third battery, since it is not discharging as a battery normally does. I still don't know how to theoretically determine the reading on the voltmeter, but here are experimental results to support you fine people:

Can somebody please tell me the physics on which the reading of $2$ volts is based? Also, I'd like to know the potential risk of continuing the experiment. Thank you.

Reference: https://haha90.phy.ntnu.edu.tw/

Answer 1

The resistance of the voltmeter is very high, so the current in the circuit, $I$, flows only through the batteries. You have two equations for the potential difference between the top and bottom of the circuit, $U$: $$ U = 2{\cal{E}}-2Ir $$ and $$ U = {\cal{E}}+Ir, $$ where $r$ is the resistance of the batteries. Now it is clear that $$ I = \frac{\cal{E}}{3r} $$ and $$ U = \frac43\cal{E}. $$ The voltmeter shows $U$, and for ${\cal{E}}=1.5$V you will have $U=2$V. So the physics here is that batteries have their own resistance, which is much less than the resistance of the voltmeter.

I find it difficult to assess the risk that continuing the experiment entails for you. For myself, I would consider that there is no risk.

Answer 2

While the other answers dealt with the electrical aspects of the experiment, I will add somewhat of a chemistry touch:

These 1.5V ZnMnO2 cells come in two flavours:

• Alkaline cells. The electrolyte is water solution of potassium hydroxide.
• Salt cells. The electrolyte is a water solution of ammonia chloride.

Both of these chemistries (esp. the salt cells) are PARTIALLY rechargeable. This means that, with some luck, you can recharge a partially discharged cell to somewhat higher state of a charge. On the other hand, the commercially available cells are not designed to be charged, so one could expect neither a high cycle life (my own experiments some 35y ago yielded ~15 cycles between ~30% and ~80%) nor much of a safety.

What can go wrong?

1. Gas and pressure buildup. Both types use water-based electrolyte and the parasitic chemical reactions when you attempt to charge such a cell (esp. when the only limit of the current is its own internal resistance) tend to decompose the water and create hydrogen and/or oxygen with a tiny scent of chlorine and ammonia from the salt cells.

Cells are generally made gastight in order to prevent the electrolyte leaks (making the cells safer) and to limit the water vapor loss (improving the shelf life). Cell walls also have a finite tensile strength, so a pressure buildup will lead to a rupture and a disappointing explosion. The explosion will be disappointing because the walls have intentionally weak places - again for improving safety. What you will get is a "Pfffff" type of explosion.

Anyway, the event can be somewhat dangerous to the eyes, the skin or the clothes.

At least wear a safety goggles if you want to see the whole spectacle.

2. A lot of heat, leading to a steam pressure buildup in the two discharged cells. See the above for the effects of the increased pressure.

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Some higher-quality cells have safety features that will switch off the cell on heat or pressure buildup, thus preventing the Pfffff event.

Other, lower-quality cells may happen not to be this much gastight, so instead of a small explosion you could hear the hissing of the escaping gases.

Some, even lower-quality cells may start a fire.

Your mileage may vary.

The electrolyte that may leak is not immediately dangerous, but still not good when in contact with any part of human body. Wash with water.

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Be sure not to repeat the experiment with any type of lithium battery or with any type of battery exceeding 20g in mass.

Answer 3

Here is the circuit diagram with all emfs equal and all internal resistances equal but labelled differently for ease of identification.

Two things to remember:
The current through a resistor always flows from the node at the higher potential to the node at the lower potential.
The positive terminal of a battery is higher than the negative terminal.

Consider loop $ABCDEFA$.

$A\to B$ - potential change $-ir_1$
$B\to C$ - potential change $+\mathcal E_1$
$C\to D$ - potential change $-ir_2$
$D\to E$ - potential change $+\mathcal E_2$
$E\to F$ - potential change $-\mathcal E_3$
$F\to A$ - potential change $-ir_3$

For the whole loop
$-ir_1+\mathcal E_1-ir_2+\mathcal E_2-\mathcal E_3-ir_3=0\Rightarrow \mathcal E -3ir=0\Rightarrow i=\dfrac{\mathcal E}{3r}$

$A \to F \to E$
$+ir_3 + \mathcal E_3 = \left(\dfrac{\mathcal E}{3r}\right )r+\mathcal E=\dfrac{4\mathcal E}{3}$.

$A \to B\to C\to D \to E$
$-ir_1+\mathcal E_1-ir_2+\mathcal E_2=2\mathcal E -2ir = 2\mathcal E -2\left(\dfrac{\mathcal E}{3r}\right )r = \dfrac{4\mathcal E}{3}$.

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With $\mathcal E \approx 1.5\,\rm V$ and $r\approx 0.1\,\Omega$ the current is $\dfrac {1.5}{3\times 0.1}=5\,\rm A$ a high drain for an alkaline cell which it would not sustain for a long period of time.
Thus although the cells would heat up they probably would not explode because they would drain frairly quickly.

Answer 4

I'm not sure if we have to consider the electromotive force of the third battery, since it is not discharging as a battery normally does.

You need to consider the emfs of all three batteries along with their internal resistances as current has to be flowing in battery 3, otherwise the voltmeter reading would not exceed the emf of battery 3. This assumes the input resistance of the voltmeter is very high compared to the internal resistances of the batteries (which it ought to be if the proper voltmeter is used), such that there will be negligible current to the voltmeter.

To determine the current in the three battery loop, apply Kirchhoff's voltage law to the loop. With the current known, you can then calculate the voltage across the terminals of battery 3 which would equal the emf of battery 3 plus the voltage across its internal resistance. That should approximate the voltmeter reading.

I'm not sure what type of "risk" you are referring to.

Hope this helps.