-1
$\begingroup$

While playing with some 9 volt batteries in a series circuit, I noticed that the voltage of the battery as measured by my voltmeter was increasing at a constant rate. After placing a bulb in my circuit, the voltmeter when placed from terminal to terminal read around 8.8 or 8.9 volts. Then after replacing the bulb with a resistor my voltmeter read 9.0 volts, then, soon after, it read 9.02 volts, then 9.04, 9.06, 9.08, and so on. I was able to reproduce these results multiple times. Why do the batteries seem to "gain voltage" at all?

Edit Will be adding the further information within the next 24 hours

Improve this question
$\endgroup$
2
  • $\begingroup$ What type of voltmeter? Brand. Model. $\endgroup$
    – StainlessSteelRat
    Commented Mar 31, 2017 at 18:21
  • $\begingroup$ This is not a common occurrence. Such trouble-shooting is very difficult without having access to your equipment. We are totally dependent on you providing all relevant information. How fast does the voltage rise? Where does it stop? Were the batteries stored in a cold place and used in a warm place? Can you post a photo or video of your apparatus? $\endgroup$
    – sammy gerbil
    Commented Mar 31, 2017 at 21:16

1 Answer 1

Reset to default
2
$\begingroup$

This is a guess answer as we do not know enough as per @sammy gerbil comment. Most likely you loaded the battery (batteries) with your bulb and of course the voltage dropped to a figure below the open circuit voltage owing to the internal resistance of the battery/ies. Thus 8.8 - 8.9V measured compared to higher o/c volts (not stated). When you replaced the bulb with a resistor the loading was probably less so the internal volt drop was less so the battery terminal volts were higher.Hence measured 9.0V. Now we come to the rising voltage. You probably used a small transistor radio type 9v battery and the initial loading of the bulb may have been more than it was designed for or you my have started with a battery which had a fair bit of its life expired. In such cases after removal of a "heavy" load (the bulb) and replacing with a "light" load (the resistor) there will be a degree of "recovery" in the battery (it's a chemical process) and as it changes chemically inside the voltage will rise a little as experienced, 9.02, 9.04v etc.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This is almost certainly the right answer. An easy way to test this, if you don't care about the battery is to short the battery for a second or two and watching the potential difference across the terminals with a normal load afterwards: you will see it recover over time. Do not do this with any battery which has a low internal resistance... $\endgroup$
    – user107153
    Commented Apr 1, 2017 at 6:31
  • 2
    $\begingroup$ I would advise NOT shorting any battery, even a small 9v battery of a few cubic cm in size can pack a s/c current of many amps and with thin wire you can burn your flesh (been there done that) if doing it with your fingers. A correctly sleceted resistor for value and wattage is better. e.g. for a small 9V battery a 27R resistor gives you 333mA and 3 Watts. $\endgroup$
    – BetterBuildings
    Commented Apr 1, 2017 at 21:57
  • $\begingroup$ I think that's good advice: as a child I used to like making sparks with batteries. I once did this using an aluminium bar across the terminals of a car battery. The bar welded itself to the terminals and I was about to grab it to pull it off when I noticed it was glowing dull red. I knocked it off with a brick, I think, and ever since then I have had a deep respect for car batteries. I suspect I destroyed the battery in the few seconds it took, unfortunately. $\endgroup$
    – user107153
    Commented Apr 1, 2017 at 23:13

Your Answer

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

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