0
$\begingroup$

enter image description here

In the beginning the first battery is fully charged, and the second is fully discharged. As the circuit is connected the discharging battery recharges a dead second battery and light is emitted from the bulb, using up energy. Because the amount of electrode material moved in the batteries is proportional to the current, that also determine the amount of electrochemical potential energy stored. After some amount of discharging the voltages will come to equilibrium with the batteries both half charged. But then you would still have half the chemical potential energy in each = full potential! So you could take one of the batteries and flip it in its place (reverse its polarity so their voltages don't cancel out) and run the lightbulb for as long as you'd have run it with the one battery. See the problem? 1 chemical energy turned into about 1.5 of itself... what did I break? lol

Improve this question
$\endgroup$

4 Answers 4

Reset to default
2
$\begingroup$

As the circuit is connected the discharging battery recharges a dead second battery and light is emitted from the bulb, using up energy.

You state this, but the circuit will not perform in this way. Imagine two large weights connected by a pulley. Assuming they have the same masses, the higher one will not spontaneously descend to raise the other. Similarly, even without the light in place, the full battery will not push current into the discharged one.

With the light in place, it is even less likely for that to happen. For the weight analogy, it would be like adding a brake to the system. You could push them to another position, but only by adding extra energy.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is not entirely correct. The charged battery has a higher resting voltage than the discharged one, even if the difference may be small. So some amount of current will flow, until the resting voltages are equal. I just tried it with two NiMH cells and even with both almost fully charged with a very small difference in voltage I got 3.1 mA of current by putting them in parallel. Didn't have an empty cell handy unfortunately. $\endgroup$
    – Sunday
    Nov 14 at 12:14
0
$\begingroup$

I'm not exactly sure on the details, but on a higher level the answer will have to do with conversion of energy. As the first battery is discharging, chemical energy converts to an electromotive force, which must be converted back into chemical energy in the second battery. This cannot be used in the light bulb until the polarity switches

Improve this answer
$\endgroup$
0
$\begingroup$

No battery is perfect, it has internal resistance, which when added to the R of the bulb will not have the potential to charge the other battery to half. And when the are in series the not half charged battery will limit the total current flow.

Yes it's interesting. Think about the fresh battery (just 2 atoms) one atom wants the electron and the other is happy to give it up but they need a current path, the other dead battery (also 2 atoms) has 1 atom that has happily taken the electron and the other that happy gave it away. Adding the second battery 2 the first with a perfect conductor will likely produce a situation where it is only probabilistic that an electron flows. Now if the first battery had a zillion atoms then "obviously" current would flow but this force is more driven by entropy rather than a potential difference. Adding a resistor in the circuit just slows down an already pretty slow process, none of these electrons have much energy, its like applying a very weak voltage to the light bulb, it won't glow. Some electrons will lose energy in the bulb but some will also gain, it will be equilibrium.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Wouldn't adding resistance to the circuit just slow down the rate of discharge of the battery? Slowing down how fast something happens doesn't mean it doesn't happen. But also even assuming less of the capacity of the battery is used, we still have a problem of x% cap turning into 1.5x when the polarity flips $\endgroup$
    – user273872
    Dec 15, 2018 at 4:00
  • 1
    $\begingroup$ I think the correct reasoning won't rely on non ideal components because you shouldn't be able to break conservation even with ideal energy storage $\endgroup$
    – user273872
    Dec 15, 2018 at 4:09
  • $\begingroup$ Anyway the bulb resistance is functional. An ideal bulb isn't even conceivable and in all cases the radiated energy is lost. Answer is that by @BowlOfRed $\endgroup$
    – Alchimista
    Dec 15, 2018 at 9:17
  • $\begingroup$ Added to my answer above $\endgroup$
    – PhysicsDave
    Dec 15, 2018 at 15:31
0
$\begingroup$

In your diagram you have a current going through both batteries, the charged one and the uncharged one. Please notice that the direction of the current flow for charging has to be the opposite of the one for discharging, because the chemical reaction has to be reversed.

Take a lead acid battery:

leadac1

in a circuit with a lamp, the resistor. The direction of the current is from the positive to the negative because of the chemical reaction

When recharging an empty battery, the direction of the current is from the negative to the positive.

leadac charg

Thus in your circuit the empty battery will be just a small (hopefully) extra resistance, it cannot be charged.

So if you have them in series, there is no charging, if you have them in parallel as I think your drawing shows:

so you could take one of the batteries and flip it in its place (reverse its polarity so their voltages don't cancel out) and run the lightbulb for as long as you'd have run it with the one battery.

It seems that having batteries in parallel which are not of the same voltage is a very bad idea , according to this link.. The charging will be violent if not controlled and the too high currents will damage the batteries.

The only way to prevent these subtle series-parallel string interactions is by not using parallel strings at all and using separate charge controllers and inverters for individual series strings.

As far as energy conservation goes the high currents will turn a lot of the energy to heat, the batteries will not be half charged because of these losses.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.