12
\$\begingroup\$

V = IR

Resistance stays same, and I know as a fact that I (or current) decreases (my stuff runs slower on old batteries).

So could a 9 Volt battery turn into a 1.5 volt one?

\$\endgroup\$
  • 2
    \$\begingroup\$ you won't get nearly as far down as 1.5 before you'd call it a dead 9v; a fresh 9V will actually run about 9.6V (o.c.), and by the time it's down to 9.0V it's noticeably 'tired', and it's pretty well spent below that. \$\endgroup\$ – JustJeff Sep 5 '11 at 21:01
  • \$\begingroup\$ Dave got some interesting videos about batteries: maybe you find this one interesting (there are some more details than you actually ask here): eevblog.com/2011/01/23/eevblog-140-battery-capacity-tutorial \$\endgroup\$ – PetPaulsen Sep 5 '11 at 21:31
12
\$\begingroup\$

Both effects occur as a battery is drained. The open circuit voltage goes down and the internal resistance goes up. Note that open circuit voltage is specifically measuring just the voltage the battery puts out with the internal resistance taken out of the equation. That is because there is no current thru that resistance, hence no voltage drop across it. Any decent voltmeter will have at least 10 MΩ input resistance, which is so way more than even a dead battery as to not matter.

All that said, different battery chemistries have different characteristics regarding both these parameters as they are drained. NiCd and NiMH have rather flat discharge curves after a short initial period. That means the open circuit voltage doesn't drop much for most of the discharge cycle even as the stored energy is getting steadily lower. These batteries then show a rather steep falloff in voltage as the last 10% or so of energy is drained. For a NiMH or NiCd therefore, it's tricky to determine a state of charge just from the voltage.

Other chemistries have a more linear discharge curve (voltage as a function of accumulated Coulombs drained at a fixed current). Old fashioned carbon-zinc cells are more like this. Usually, there is a significant temperature dependence too, both in terms of voltage and capacity.

Yes, batteries can get complicated.

\$\endgroup\$
  • \$\begingroup\$ Can you please make it clear what is the reason exactly for the voltage drop as the State of Charge decreases? Maybe due to internal resistance increases? \$\endgroup\$ – Tina J Mar 3 '15 at 7:17
8
\$\begingroup\$

Your 9V battery will indeed give a lower voltage reading when it's exhausted and that's not just because of higher internal resistance; you may read 6 or 7V even with a very high impedance DMM. I'm not sure you can go as low as 1.5V; the increased internal resistance makes that in the end you can hardly draw any energy from it anymore, so I expect that the voltage will go asymptotically to a somewhat higher voltage. Even so, a 9V depleted till 1.5V will never be able to supply the current a 1.5V battery can supply.

\$\endgroup\$
  • 1
    \$\begingroup\$ The problem I think the user needs help with is the concept of internal resistance increase and the 1.5 V is just an example dead battery attempting to show that voltage drop is the issue. \$\endgroup\$ – Kortuk Sep 5 '11 at 21:17
  • \$\begingroup\$ Although I'm not sure that a single cell feeding a reasonable resistive load could deplete itself in a reasonable time to the point that its open-circuit voltage would fall to essentially nothing, it's possible for some cells in a series-wired pack (which is all a "9-volt battery" is) to have their open-circuit voltage go negative. Indeed, I once had an AA cell which measured something like 0.2 volts negative even when driving a 20mA load. A 9-volt series-wired pack can easily have its open-circuit voltage fall below 1.5V, though as noted internal resistance does go up. \$\endgroup\$ – supercat Mar 19 '13 at 15:02
  • \$\begingroup\$ @stevenvh Can you please make it clear what is the reason exactly for the voltage drop as the State of Charge decreases? How internal resistance increases? \$\endgroup\$ – Tina J Mar 3 '15 at 7:16
4
\$\begingroup\$

Actually, resistance dramatically changes as the battery is used up. The voltage will go down with use, but in many applications the increased internal resistance will render the battery unusable long before the reduced voltage does.

\$\endgroup\$
1
\$\begingroup\$

As a battery runs down it's open circuit voltage will drop and it's internal resistance will go up. Unless the battery is nearly totally dead though the open circuit voltage will remain reasonably flat compared to the internal resistance which seems to drop quite linearly (I imagine different chemistries will vary though).
A 9V battery might start off with, say 5 ohms of internal resistance, reaching over 100 ohms when discharged (figures are rough guide, not researched exactly). If we took a moderately discharged 9V battery (internal resistance risen to 50 ohms) and read with a multimeter (a load of say 1 megaohm) we might read around 9V still, as the multimeter has almost no load on the circuit (e.g. 9 * 1000000/1000050 = 8.99V).
Under a 500 ohm load though it would drop to 9 * 500/(500 + 50) = 8.18V.
Maybe the open circuit voltage will end up at say 7.5V and the resistance 200 ohms (again these figures are just a rough example, google will no doubt know better)

So yes the voltage drops as the batteries get used up, and also the internal resistance rises. It's usually better to check a battery under a load to get a good idea of how flat it is.

\$\endgroup\$
0
\$\begingroup\$

The Voc or open circuit steady state voltage is very linear in decline with SOC as the battery is a fairly constant capacitance with a charge voltage. However the ESR rises sharply past 90% SOC and rises slowly below 50% SOC then rapidly below 10% somewhat like a bathtub curve . So the ESR and recent current with memory secondary charge capacitance with higher ESR greatly affects loaded battery voltage with SOC. The ESR increases the slope with a load current of V vs SOC at each end.

\$\endgroup\$
-1
\$\begingroup\$

As we know Dc circuits are rated in VA, product of the voltage and current i.e;if the voltage of the battery goes down during discharging process the battery has supply high current to match the required VA load, but has voltage dec the internal resistance of the battery increase so the battery is not able to give the required amount of currnet what the load is actual required,so the battery is found to be discharged.

\$\endgroup\$
-1
\$\begingroup\$

Wouldn't it be safe to use the analogy that the battery is like two cycliders joined at the bottom with a tube, one full of water the other empty. As you open up the circuit, the full cylinder tries to run into the empty one. For a while electron imbalance keeps the full cylinder dumping into the empty. After the sides start the equalize, the pressure of the water stream slows and it is like the tube (or internal resistance) allows very little water to cross, dropping the voltage. There remains some pressure (voltage). But the resistance of the circuit or components is too great for the voltage to be effective

\$\endgroup\$
-2
\$\begingroup\$

It's abusive to use V=IR in the way you've used! this is strictly applicable to a flowing electrical current flowing through a resistive load under the effect of a potential (voltage). Now define the resistive load, voltage and current you're applying Ohm's law on:

  • Is it: V is the voltage of the battery, R is the external resistance or load, and I is the current passing through. then this has nothing to do with the voltage of the battery being lower as being consumed.
  • Is it: V is the voltage of the battery, R as the internal resistance of the battery, and I as the current supplied by the battery to the external load? Applying Ohm's law here can tell us that the voltage read at the terminals of the battery gets lower if the current supplied by the battery increases.

As for the voltage of the battery getting lower as the state of charge getting lower (the more we consumed the battery), this is related to the change in the chemical materials that actually produce the voltage, that is electrodes dipped in electrolyte. That is, the electrode loss of extra free electrons.

The rate and behavior of how the voltage change with respect to the state of charge depends on the chemistry of the battery and not on any electrical law. As an example, here's a comparison between the shape of the voltage drop of Alkaline batteries compared to NiMh batteries as the batteries are being consumed(source):

voltage drop

note that when I mention "voltage of the battery", I mean the open circuit voltage, that is no current flowing through the battery. The internal resistance has no effect what so ever on this voltage.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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