If I measure the battery at 8.2 V, can you not assume that is what is

realized in the circuit?

---

Only if you measure the battery voltage when the battery is under

load.

---

If you're taking an exam and you calculate an answer based on the

unloaded battery (before connecting to the circuit), will the answer

be correct?

---

No, but if the the purpose of the exam is just to see whether or not

you know how to use Ohm's law, it won't matter. However, if the

purpose of the exam is to see whether you know what internal

resistance is about, you'll flunk!

---

Not understanding this yet. If you measure the V at the battery's

external posts, does this not include the internal resistance?

---

AHA!!! There it is...

Yes. it does, but you don't see it because the resistance of the meter

is so much higher than the internal resistance of the battery.

Taking a look at this circuit,

+--------------------------+

| |_

| +-[9V SOURCE+]--[Rint]--->_|+9V

| | |

| | |

| | |_

| +------------------------>_|-9V

| |

+--------------------------+

the 9v source and Rint represent the battery and its internal

resistance, and the box around them represents the case; just like a

regular 9V battery is put together.

Now, let's say that Rint is one ohm and that you've got your voltmeter

hooked from +9V to -9V, like this:

+--------------------------+

| |_

| +-[9V SOURCE+]--[Rint]--->_|<----+

| | | |

| | | [METER]

| | |_ |

| +------------------------>_|<----+

| |

+--------------------------+

Let's also say that the internal resistance of your voltmeter is 10

megohms.

Then, for convenience, we can redraw the circuit to look like this:

+-----+---->E1

| |

| [R1]

+| |

9V +---->E2

| |

| [R2]

| |

+-----+---->0V

Where R1 is the internal resistance of the battery, R2 is the internal

resistance of the meter, E1 is the voltage from the 9V source, and E2

is the voltage across the meter.

Now, if we calculate the current being drawn from the battery, we can

say:

E 9V

I = ------- = ------------- = 0.00000089999991A = 899.9991nA

R1+R2 10,000,001R

and if R2 is equal to 10 megohms, the voltage the meter will see will

not be 9.0V, it'll be 8.9999991V, because the other 0.0000009V will

have been dropped across R1, the battery's internal resistance.

Considering that R1R2 is a voltage divider, there's a much more

convenient way to find out what the voltage across the meter will be,

and that's to use

E1R2

E2 = -------

R1+R2

For the 10 Megohm case you'll get:

E1R2 9V * 10000000R

E2 = ------- = ----------------- = 8.9999991V

R1+R2 1R + 10000000R

which is just what we got before.

Now to see how the internal resistance of the battery affects the

battery output voltage, let's connect a 20000 ohms-per-volt analog

multimeter across the battery and see what happens. Let's say the

meter has a 10 VDC range. That would make its internal resistance

200000 ohms if we selected the 10VDC range and, with a 1 ohm internal

resistance for the battery we wind up with a reading of:

E1R2 9V * 200000

E2 = ------- = --------------- = 8.999955V

R1+R2 1R + 2000000R

See the voltage falling? That's because more voltage is being dropped

across the battery's internal resistance as it's required to supply

more current.

If we use a voltmeter with a 1000 ohm iron vane movement to measure

the battery voltage, it'll draw 9mA from the battery when it's

connected and it'll read:

E1R2 9V * 1000R

E2 = ------- = ------------ = 8.991001V

R1+R2 1R + 1000R

The reading dropped again because of the higher current flowing

through the battery's internal resistance, and the higher the load

current goes, the higher the drop across the internal resistance will

be.

So far, with the voltmeter loads, the drop has been small, but it

starts to matter when you start drawing significant current from the

battery. for example, assume you have a 20 ohm load and that because

you've figured out that

E 9V

I = --- = ----- = 0.45A

R 20R

you expect the 9V battery to deliver 450mA into the load.

Well, if you look at

E1R2 9V * 20R

E2 = ------- = ------------ = 8.57V

R1+R2 1R + 20R

you'll find that, because of the battery's internal resistance you can

only get 8.57V across the 20 ohm resistor, which is only going to

allow about 429mA to flow through the load.

That's also borne out if you do:

E 9V

I = ------- = ----- = 0.42857...A

R1+R2 21R

Does the internal resistance only get activated when a load is connected?

---

No, it's always there but it doesn't matter until you start taking

current out of the battery. Kind of like even though the electricity

from the power company is always there, on the other side of the

switch, you don't get charged for it until you turn on the switch and

start using it.

---

I'd like to see a circuit diagram of the innards, showing the path

from one post to the other, thru the resistance.

Evidently I missed "Batteries 101".

---

Read what follows "AHA!!! There it is...", previous.

---

Several web sites I've read indicate the internal resistance only

increases with age; a higher discharge history would increase the

resistance faster.

I didn't see indications that it's a variable resistance changing with

the load, as John stated.

---

Never mind what you read, find out for yourself what the deal is.

Make your measurements again measuring the battery voltage and the

current through the load for each measurement you make, then calculate

the various internal resistances you get for the battery with

different loads and post what you find.