# B Open circuit voltage of an old battery

1. Nov 30, 2018

### hokhani

Why the voltage of an exhausted battery, in open circuit state, is almost identical with that of new battery?
The internal resistance of an exhausted battery is very large and so, the large amount of energy consumed by the source (for overcoming the electric filed in the battery) would waste in such a way that it seems the source can not put enough charge on the terminals of battery.

2. Nov 30, 2018

### profbuxton

Since the battery is in open circuit state, there is no current through the battery so the full o/c volts appear at the terminals. As soon as a load is placed across the terminals a current flows and voltage will "drop" due to battery "internal resistance" so a portion of voltage will be divided across load and a portion across "internal resistance".

3. Nov 30, 2018

### hokhani

The internal resistance always exists in the exhausted battery, no matter it be in open or closed circuit. Part of the energy consumed by the source, to put the charges on the terminals, is exhausted by the internal resistance. So, how the full volts should appear at the terminals?

4. Nov 30, 2018

### profbuxton

There is NO energy "consumed" when the battery is open circuit because there is no current flowing so full battery volts(whatever it may be) appears at terminals.Have you measured a battery?. For instance, note that you can have a battery with a faulty internal connection(i:e high resistance) which you can charge to full volts (i:e 12volts) and with no load will measure 12 volts but when you put a load on the battery the volts on the terminals will drop due the internal resistance(faulty connection). This is very apparent on car batteries when they become faulty.
Look up information on "ideal batteries " and how they are modeled.
Note that "faulty connection" can be that the battery elements have deteriorated and are end of useful life.

5. Nov 30, 2018

### Staff: Mentor

Yes, the internal resistance always exists. Use Ohm’s law:

$V=IR$

Since the battery is exhausted $R$ is large, and since it is an open circuit $I$ is zero. So what is $V$?

Exactly how much energy is consumed?

$P=IV$

Again, $I$ is 0 and you just found $V$ so calculate $P$.

Last edited: Nov 30, 2018
6. Nov 30, 2018

### profbuxton

Example: Recently put new battery in my car. On switch on battery volts is 12volts(from car dash meter). When I start car battery volts drops to 10volts. I have measured starting motor current to be 195 amps(6 cylinder motor). what is the battery internal resistance?
Note when old battery was failing on starting, battery volts dropped to 8volts. What would be the internal resistance?(assume same current draw). Note also that when not supplying current battery volts was still 12 volts.

7. Nov 30, 2018

### Drakkith

Staff Emeritus
An exhausted batter often outputs a voltage, it just can't sustain that voltage with any appreciable amount of current flow. Set up a simple circuit consisting of a voltage source and two resistors. One will serve as the internal resistance of the battery, the other will be our test resistor to see how voltage behaves in this circuit.

Let's say our battery outputs 10 volts. Set resistor A, representing the battery's internal resistance, to something very high. Let's say 1 giga-ohm. Set resistor B to something very small, such as 1 ohm. Calculate the voltage drop across both resistors. You'll find that nearly all of the voltage drop is across the resistor A. This is why if you test a nearly dead battery the current is extremely low, and not even a short circuit between the terminals generates any appreciable current. The internal resistance of the battery is simply too high.

Now ramp up the resistance of resistor B to 1000 ohms, 1 mega-ohm, 1000 mega-ohms, etc, calculating the voltage drop across each resistor as you go along. You'll find that more and more of the voltage is lost across resistor B. As resistor B approaches infinite ohms, corresponding to an open circuit, you'll find that all of the voltage drop occurs at this resistor. Hence, you'll measure the full voltage of the battery across it even though it can't support anything but a minuscule amount of current.

A new battery, fully charged, has a much lower internal resistance, so unless you do something like short the terminals together you'll find that most of the voltage drop is across the rest of the circuit.

8. Nov 30, 2018

### Mister T

The chemistry is the same, at that's what determines the voltage.

9. Dec 1, 2018

### hokhani

Right, I guess you mean that what voltmeter measures, in the o/c state, is exactly the difference between the chemical potentials of the two terminals? Isn't it?

10. Dec 1, 2018

### CH WILSON

If you are talking about a lead acid battery you can in fact run it down to ) volts or darn close. Leave your headlights on for a day or 2 then stick your digital voltmeter on it.You will probably measure a tiny fraction of a volt.

Discharged this far the battery won't even start to seriously take a charge for some time because all the electrolyte's acid has all been converted into lead sulfate in the plates and the water left in the battery doesn't conduct electricity very well. Given some time the trickle of current converts some of the sulfate into acid, the conductivity of the electrolyte starts to rise and charge current rises with it. Eventually the battery is recharged but its lifetime has probably been compromised.

11. Dec 7, 2018

### ruko

There is current flow in a voltmeter, certainly not much but some. A teeny tiny bit, micro amps I would guess. So it really isn't an open circuit with a volt meter across it.

Last edited by a moderator: Dec 8, 2018
12. Dec 8, 2018

### Staff: Mentor

This is true but negligible in this case.

13. Dec 8, 2018