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When you connect two batteries in series, why doesn't the middle short? The connection between the positive terminal on one battery and the negative terminal on the other battery has no load, so shouldn't it be shorted? But I don't notice heat or explosions or anything like that when I connect batteries in series.

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  • \$\begingroup\$ Well, (almost) all battery-powered devices with two or more batteries connect them in series (like your TV remote control, for instance); the current at the interconnection(s) is just the same as the one flowing through the load; if there is none, no current flows... \$\endgroup\$ – aschipfl Jun 26 '19 at 18:54
  • \$\begingroup\$ Voltage is a relative quantity, not an absolute one. \$\endgroup\$ – Hearth Jun 26 '19 at 19:00
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Your problem is there are two definitions for the phrase a short:

A) A short is high current flowing through a wire unintentionally.

B) A short is two places connected by a low-resistance interconnection (as a wire).

Both are correct but as much as A implies B, this does not mean B implies A. Using the same term a short for both makes you think they are equal, which they are not.


When you connect the plus from one battery to the minus of the other, you have a short of the second kind. However, there is no current flowing, as this requires a circuit —a closed loop— so obviously, B does not imply A.

As soon you connect the plus from the other battery to the minus of the first also, there is a closed loop, and your short of the second kind grows into a short of the first.


EDIT: There was some confusion about the heat production as well, which I explained in the comments before.

Where the batteries are connected bumper-to-bumper, you have a very low resistance \$R_b\$ of maybe 0.01 Ω. The outer wire resistance \$R_w\$ be 0.1Ω instead because the wire is a longer and thinner conductor than the bumper-to-bumper contact.

Because \$P=I^2\cdot R\$ for each individual connection, and \$I\$ being identical in both "resistors" as they are part of the same loop, this means there is ten times more heat produced in the wire than in the bumper-to-bumper connection.

In addition, the batteries are lumps of metal and able to take a lot of heat without a significant temperature rise.

Both is why the bumper-to-bumper connection stays considerably cool during an outer short circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I'd rather call B an (inter-)connection... \$\endgroup\$ – aschipfl Jun 26 '19 at 18:57
  • \$\begingroup\$ @Janka so then when the "short of the second kind grows into a short of the first", why is there still no high heat? \$\endgroup\$ – Serendipitous Epiphany Jun 26 '19 at 19:13
  • \$\begingroup\$ Because the bumper-to-bumper connection of the two batteries typically has even lower resistance than the outer wire connection. The current is the same in the whole circuit but the vast majority of the heat is produced in the wire due to its higher resistance. Yes, there are even grades of shorts. In addition, the batteries are lumps of metal able to take a lot of produced heat from the bumper-to-bumper connection without a significant temperature rise. \$\endgroup\$ – Janka Jun 26 '19 at 20:16
  • \$\begingroup\$ @Janka Do you mean the bumper-to-bumper connection has higher resistance and the outer wire has lower resistance? I thought higher resistance would mean less heat. \$\endgroup\$ – Serendipitous Epiphany Jun 26 '19 at 20:20
  • \$\begingroup\$ No, the bumper-to-bumper connection has lower resistance. The wire has higher resistance because it's thin and long in relation to the bumper-to-bumper connection. The individual heat produced in a conductor is equal to the current squared times the individual resistance. The current is the same in both connections as this is one loop. Higher overall resistance in the loop meant less heat at any place in the loop, because the current is voltage divided by overall resistance in the loop, and your voltage is set by the batteries. \$\endgroup\$ – Janka Jun 26 '19 at 20:24
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When you connect two batteries in series, why doesn't the middle short? The connection between the positive terminal on one battery and the negative terminal on the other battery has no load, so shouldn't it be shorted?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A short-circuit is usually defined as an unintended bypass of the intended circuit. Here the intended circuit is the loop consisting of the battery and lamp. The short-circuit is bypassing the lamp and providing a "short-cut" between the terminals of the battery. If the resistance is low then high currents will flow.

schematic

simulate this circuit

Figure 2. (a) No short-circuit occurs when the batteries are properly connected in series. (b) Attempted series-connection of two grounded batteries would result in a short-circuit as the current could flow through the ground connection as indicated by the red arrow. BAT3 is short-circuited while BAT4 is not.

But I don't notice heat or explosions or anything like that when I connect batteries in series.

That's because no current is flowing as in Figure 2a.


From the comments:

How come in Figure 2a there is no current flowing between the positive terminal of BAT1 and the negative terminal of BAT2.

Because there is no closed circuit.

I thought BAT1's positive terminal has a deficiency of electrons while BAT2's negative terminal has an excess of electrons.

No, the net charge in a battery is zero! There is no surplus or missing charge at either terminal.

So shouldn't the excess electrons in BAT2 flow towards the positive terminal of BAT1, thus creating a current?

Current can only flow when there is a circuit. As shown in Figure 1 the circuit is open so no current can flow.


I thought a battery separates itself into two sides, one positive and one negative, with a barrier in between.

If that were true then batteries suspended by a thread around their middles would all line up due to electrostatic attraction. (They don't.)

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  • \$\begingroup\$ How come in Figure 2a there is no current flowing between the positive terminal of BAT1 and the Negative terminal of BAT2. I thought BAT1's positive terminal has a deficiency of electrons while BAT2's negative terminal has an excess of electrons. So shouldn't the excess electrons in BAT2 flow towards the positive terminal of BAT1, thus creating a current? \$\endgroup\$ – Serendipitous Epiphany Jun 26 '19 at 19:21
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Jun 26 '19 at 19:27
  • \$\begingroup\$ Addressing "Because there is no closed circuit". Well what if we closed it by connecting BAT2+ and BAT1- and adding a load in between. \$\endgroup\$ – Serendipitous Epiphany Jun 26 '19 at 19:30
  • \$\begingroup\$ Addressing "No the net charge in a battery is zero! There is no surplus or missing charge at either terminal." Why is this? I thought a battery separates itself into two sides, one positive and one negative, with a barrier in between. \$\endgroup\$ – Serendipitous Epiphany Jun 26 '19 at 19:31
  • \$\begingroup\$ I don't understand your comment about "adding a load in between". You can add a schematic into your question to clarify. For your other comment see if the section on Chemical Reactions in Batteries by Explain That Stuff is of any help. \$\endgroup\$ – Transistor Jun 26 '19 at 20:04
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A short (of the type that you mean, that causes high current and thus heat) is caused by a high voltage being directly connected to a low voltage.

The positive terminal of a battery is higher voltage with respect to its own negative terminal. But that doesn't give it any particular relationship to a second battery.

So, if you take a 5V battery and ground the negative end, the positive end will be at 5V wrt ground. Then you take a second 5V battery and put its negative end on the positive end of the first battery, and its positive end will be at 10V wrt ground. But the positive end of battery1 and the negative end of battery2 are at the same voltage--and there's nothing trying to force different voltages--so no sparks.

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A battery looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

A battery is like an ideal voltage source. An ideal voltage source produces the same voltage no matter what. It doesn't matter if you sink 1 million Amps out of it, it will still give you the same voltage. If you put two ideal voltage sources in series, the voltages add. If you revers the polarity of one, it will fight the other one and the voltage will be subtracted.

A simmilar case applies to batteries. The ways that a battery is not like an ideal source is:

  • It doesn't have unlimited current sourcing abilities
  • The voltage drops the more current you pull from it (in the short term) it has series resistance
  • The batterys voltage will drop the more energy you pull from it.
  • It has internal resistance that is high (if you leave it there for a long time it will slowly degrade the battery.

So if you put two of the same batteries in series, it will double the voltage.

If you do this in the real world, it is important to match the batteries (which matches the series resistance, otherwise you can have more voltage running through one battery than the other and it will damage it or explode.

The middle doesn't 'short' because the voltage of the battery is pushing against the voltage of the other battery.

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  • \$\begingroup\$ The next-to-last statement is not quite true, the current is the same when the batteries are in series; or did I miss the point you're trying to make? \$\endgroup\$ – aschipfl Jun 26 '19 at 19:01
  • \$\begingroup\$ Yeah, sorry, its the voltage that changes in that case \$\endgroup\$ – Voltage Spike Jun 26 '19 at 19:12
  • \$\begingroup\$ Yes, but still, a weaker battery can't damage a stronger one, it's just the overall voltage that might be not enough for the load, and an (almost) empty one will limit the overall current since it's internal (serial) resistance heavily increased... \$\endgroup\$ – aschipfl Jun 26 '19 at 19:22
  • \$\begingroup\$ It can and it does. have you ever put an old AA battery in with a new one? The old battery will leak. When you make battery packs, it's important that the cells are matched especially in lion batteries or the series resistance can cause the battery to explode. A stronger battery can damage a weaker battery \$\endgroup\$ – Voltage Spike Jun 26 '19 at 19:37
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Im no pro but its easier than it looks bro. Please follow through. When you take one battery, you connect its positive with its negative together, it "explodes". But if you take two batteries, connect positive of the one with negative of the other, nothing happens. Why?

In the first case, you created a closed loop, a complete circuit. In second, the circuit is not complete because the two batteries have no relation with each other as you have not completed the circuit yet (i.e one terminal of the source connected to other terminal through some load/resistance). It will be complete once you connect the other two terminals, again ending up in an "explosion".

By definition, you are shortening the two middle terminals. You don't get heat or explosion simply because the circuit is not complete. I hope this was a little helpful. Stay Safe

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