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.