# Measuring amps on 9 volt battery

Discussion in 'Electronic Basics' started by George, Oct 28, 2004.

1. ### GeorgeGuest

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

2. ### Jason RosinskiGuest

There is an internal source resistance to the battery. It cannot supply current faster than a
certain maximum rate. In this case it looks to be ~250Ohms. Think of it as a resister in series
with the battery internals.

3. ### GeorgeGuest

I'm not understanding this.
Are you saying there is a Resistor within the battery?

The "decreasing value" behaviour is not described in the MM manual.
It does say "The maximum input limit for DC/AC current measurement is 400 mA."
As shown below, I calculate with a 150 ohm R, the MM should read 60 mA.

I probably should question Radio Shack.

4. ### Guest

Heee hee hee hee!!

Let us know what they say <g>

John

5. ### Tom BiasiGuest

There is a 'resistance' within the battery.

The battery is not a perfect device, it can not supply voltage with

You seem to know this already. What does limit the battery's ability to

The main factor is the internal resistance. You are testing the battery at
levels beyond its capability and are seeing the output drop as you heat up
the battery and increase the internal resistance even more.

Regards,

Tom

7. ### Bob MastaGuest

Others have pointed out about internal battery resistance.
I'd like to point out that if you put the ammeter leads directly
across the battery, the current may easily exceed the limits of the
meter if the battery is *not* old and nearly dead. In that
case you will pop a little fuse which is typically hard to get
at in cheap meters, and is usually some bastard size (physical
size and electrical size) you don't have on hand.

So, as a good rule of thumb, *never* put ammeter leads across a
battery or other power source. Always insert the ammeter in series
with a branch of the circuit being tested.

Saves wear and tear on meters, and on your nerves!

Best regards,

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

8. ### Tom BiasiGuest

At the level the poster appears to be I should have mentioned that Bob, glad
that you did.

9. ### GeorgeGuest

Thanks Tom and others for your info.
I guess this poster (me) is at a beginner level.
Learning OHM's law, relationship between each value and trying to
confirm with a 9V battery, resistors and MM.

Using letter R to represent ohms, showing Law and actual MM readings
at decreasing levels of magnitude:

Law: .06 mA = 8.2 V / 118,500 R
MM: .06 mA

Law: .556 mA = 8.2 V / 14/730 R
MM: .556 mA

Law: 3.79 mA = 8.2 V / 2162 R
MM: 3.75 mA (99% of Law)

Law: 55.40 mA = 8.2 V / 148 R
MM: 52.1 mA (94% of Law) ***

Law: 83.25 mA = 8.2 V / 98.5 R
MM: 76.5 mA (92% of Law) ***

Law: 138 mA = 8.2 V / 59.4 R
MM: 120 mA (87% of Law) ***

Law: 332 mA = 8.2 V / 24.7 R
MM: 253 mA (76% of Law) ***

*** At these levels, the MM showed initial 473 mA then the number
shown here then decreasing values. At the 24.7 R test, the rate of
decrease was in larger increments than at 148 R.

Did not proceed with lower R values because MM manual says it can read
max 400 mA.
In bottom two levels, I learned how to put resistors in Parallel and
calculate the net resistance. (98.7 and 148.7 for 59.4 R; 4 x 98.7
for 24.7 R)

Q1. Is the apparent error percentage due to my MM or would all meters
show these variances?

Q2. Same question but regarding the "decreasing value" behavior.

Q3. If the meter could read larger amounts of current, and I used
lower amounts of R, the Law predicts higher current.
Law: 4.1 A = 8.2 V / 2 R
Can an 8.2 V battery produce 4.1 A?
Is there an additional math factor needed in the Law to filter out
"impossible" results?

Q4. Continuing to lower R to Almost zero (metal of the MM lead wires),
if the MM can read up to 400 mA, why isn't it safe to measure the
current of a 9 V battery? Is there another way to find available
current?

10. ### John FieldsGuest

---
Since you haven't mentioned anything to the contrary, it seems you're
_assuming_ that the battery voltage will remain constant at 8.2V
regardless of the load on it. It won't.

You also seem to be having trouble with the concept of the internal
resistance of the battery. Consider it a variable resistance in
series with the battery, located in the same housing as the battery,
but which you can't physically get to.

But, you _can_ measure it. Consider:

+----------+
| |
+---|----+ |
| | | |
| [R1] | |
| | | [R2]
| |+ | |
| [BAT] | |
| | | |
+---|----+ |
| |
+----------+

Where R1 is the internal resistance of the battery, BAT is the stuff
in the battery making the voltage, and the box around them indicates
that they are inseparable. But, let's go ahead and separate them
anyway, just for fun, like this:

+-----+---->E1
| |
| [R1]
|+ |
[BAT] +---->E2
| |
| [R2]
| |
+-----+---->

Now, let's assume that the stuff in the battery makes BAT a true
voltage source which can supply an infinite current into a zero ohm
load. If that's true, then the current BAT can supply into R1 and R2
will only be limited by the sum of the resistances according to Ohm's
Law:

E
I = ---
R

which, in this case, becomes

E1
I = ---------
R1 + R2

Using your data, we can say that in all cases E1 will be 8.2V, and if
we retabulate the data for convenience we get:

VOLTS mA OHMS
E1 I R2
-------|-------|--------

8.2 0.06 118500

8.2 0.556 14730

8.2 3.75 2162

8.2 52.1 148

8.2 76.5 98.5

8.2 120 59.4

8.2 253 24.7

If we rearrange I = E/R to solve for resistance we get:

E
R = ---
I

and if we plug in the first set of values you got we get:

E 8.2V
E = --- = ------ = 136667 ohms
I 60µA

Notice that the resistance you used to get that 60µA was 118500 ohms,
some 18167 ohms short of what Ohm's law says it should be, and since
Ohm's law doesn't lie, where did that resistance come from?

From inside the battery, and it's the resistance we're calling R1

Continuing in the same vein and filling out the rest of the table, we
wind up with:

VOLTS mA OHMS OHMS OHMS
E1 I RT R2 R1
-------|------|--------|--------|-------

8.2 0.06 136667 118500 18167

8.2 0.556 14748 14730 18

8.2 3.75 2186 2162 24

8.2 52.1 157 148 9

8.2 76.5 107 98.5 8.5

8.2 120 68.3 59.4 8.9

8.2 253 32.4 24.7 7.7

Notice that R1 is the internal resistance of the battery, and notice
---
---
A1. Different ammeters would show the same variances if they all had
the same internal resistances.
---
---
---
---
A3A. If it has a low enough internal resistance and the resistance of
the load is also low enough, yes.
---
---
A3B. No, just proper reasoning.
---

---
A4A. The reason you shouldn't connect an ammeter directly across a
battery is because the internal resistance of the battery may be very
low, which will allow a large current to flow through the meter. Then,
if the meter isn't protected by a fuse, it could be damaged.
---

11. ### GeorgeGuest

Thanks John,

This is the method I've been using:
1. Measure all the components separately (no load).
2. Using Ohm's Law, calculate the expected result.
3. Connect the circuit
4. Measure with a MM.
I did the first and last tests again and measured the battery under
In the first test (highest R), the battery under load showed same 8.2
V.
In the last test (lowest R), it showed 6.15 V.

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact. (Example, say my company makes a product. We buy a
component widget. The widget company may have various "internal
resistance" to make it, but when it arrives at our company, all we
care is that meets our specs.)

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

Also note that my MM is an inexpensive model. In the first test I
reported that it showed .06 mA. But in my second test I noticed it
also showed .05 and .07. My point is that some of my data may not be
accurate enough.

This is a great learning experience.

12. ### Peter BennettGuest

When you use Ohm's Law to calculate the current through a resistor,
you must use the voltage across the resistor at the time the current
is flowing - that is the only voltage the resistor knows about.

The internal resistance of the battery is indeed inside the battery,
but it is still in series with the battery terminals, so it will cause
the battery terminal voltage to drop when current is drawn from the
battery.
Yes it does: "discharge history, etc.".

The battery operation depends on chemical reactions. When you draw
current from the battery, "used" chemicals may accumulate on the
electrodes, and obstruct further reactions. At low currents, these
used chemicals can diffuse back through the electrolyte, so they have
little effect on the battery operation. At high currents, the used
chemicals are created faster than they can diffuse, so they have a
greater detrimental effect on the battery operation.
A lead-acid battery (car battery) will have a very low internal
resistance. A small battery, like the tiny cells in a 9 volt battery,
will have a much greater internal resistance. Both the cell size and
chemistry affect the internal resistance.
The specs for most digital meters say "+/- 1 count". If my
only .01 mA, can legitimately read .05, .06 or .07 mA.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

13. ### John FieldsGuest

---
It refers to the current which will be forced to flow through a
resistance with a known voltage across it. _Measure_ the voltage
across the resistance, then divide that voltage by the resistance of
the resistor and you'll have the current flowing through the resistor.

If you don't, and you make the assumption that a 9V battery will stay
at 9V regardless of whether 1 milliamp or half an amp is being taken
from it you'll be wrong.
---
---
You still don't understand. It's not after the fact, it's part of the
deal.

The internal resistance is the same as a resistor that you can't get
rid of that's in series with the battery, and since the current in
series resistances is the same in all the resistances, the internal
resistance will drop some voltage when the load is drawing current,
and that voltage won't be available for the load to get the current it
needs.

Let's say you have a 9 volt battery with a 1 ohm internal resistance
hooked up to some load. It'll look like this:

+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| |
+--------------------------------+

Now let's say the load is 9 ohms. That'll make the circuit look like
this:

+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| [9R]
| |
+--------------------------------+

Since resistors in series add, and the current in series resistances
is the same in all the resistances, we have 9 volts and 10 ohms, so
the current flowing in the circuit will be

E 9V
I = --- = ---- = 0.9A
R 10R

Now, if your 9 ohm load is a 9 watt light bulb and you expect to get a
certain amount of light from it when you connect it across the 9 volt
battery, you're in for a surprise, since you'll only be able to get
0.9A out of the battery because of its internal resistance.

14. ### GeorgeGuest

snip

If I measure the battery at 8.2 V, can you not assume that is what is
realized in the circuit?
If you're taking an exam and you calculate an answer based on the
be correct?

Not understanding this yet. If you measure the V at the battery's
external posts, does this not include the internal resistance? Does
the internal resistance only get activated when a load is connected?
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".

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

Do you know how much internal resistance there is on a new 9 V
battery?
Why would it be higher than a 12 V car battery?
I've sent an inquiry to a 9 V manufacturer; it's not stated on their
data sheets.
I'm not convinced a 9 V battery could have 18167 ohms.

15. ### Peter BennettGuest

No - the actual battery voltage once you connect a load will depend on
the current drawn by the load, and the age of the battery.
Exams are not real life - on exams you can probably assume that you
are given an ideal voltage source, which will provide the specified
voltage, regardless of how much current you try to draw.
It would definitely be much higher than a car battery, but I don't
know by how much - and I don't want to risk damaging my meter (or
killing a battery) trying to find out.

A car battery has very large plates, so that it can deliver the 200
amps or so that are required to operate the starter without the
voltage dropping too much.

That sounds very much too high for a fresh battery.

I'd guess you might be able to get 250 mA or so through a short
circuit - that would be about 36 ohms. However, if you attempt to
draw 250 mA from a 9 volt battery, the chemicals near the plates will
be rapidly depleted, so the internal resistance will rise, and the
current and voltage will drop. If you let the battery rest for a
while, it will recover somewhat, so you will be able to draw a fairly
high (but probably not so high) current for another short time.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

16. ### GeorgeGuest

Thanks John,
With your info and suggestions, I'll be learning more.

I'm sure all of this will sink in eventually; I'm one of those "slow"
learners, needing to know more details. (another example - I feel the
need to know what materials are utilized in the battery to provide 1.5
volts and how all those amps are stuffed in there, and how they get
out. I force myself to accept it as a "black box" and proceed with my
projects.)

Perhaps my basic question about measuring amps will prove to be
unimportant;
it's just that I saw unusual patterns and behavior; did not know if it
was due to my MM or something else.
The question has brought up others, like how to measure the battery's
amp capacity (how many amps are left?). I guess this cannot be done
with a cheap MM; just use the related Voltage capacity, since it

It does seem important to know the basics before moving on to more
complicated circuits.

My career has been in computers, but at 64 I'm finally getting to
learn the electronics side. One of my first electronic projects is a
simple robot, to follow a line. "Robot Building for Beginners" by
David Cook.

Think of it this way. If I needed to program a robot to calculate
exactly how many amps will flow from a 9 v batttery in seven simple
circuits of only resistors at successive levels (120,000, 12,000,
1200, 120, 12, 1.2, and .12 ohms) what would be the steps?
Pretend the robot is in an assembly line. It can measure each
component with an expensive/precise MM before making the simple
circuit.

17. ### John FieldsGuest

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

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

That's also borne out if you do:

E 9V
I = ------- = ----- = 0.42857...A
R1+R2 21R
---
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.
---
---
Read what follows "AHA!!! There it is...", previous.
---
---
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.

18. ### Guest

You guys actually have the patience to do these great "schematic" sketches in
than it appears?

They are handy, and given the right font do a great job. Just seems rather
labor intensive to me.

Hey, if its by hand, and takes as long as I think, kudos to all who do it as
its a big help.

John

19. ### John FieldsGuest

---
For general info:
http://www.duracell.com/oem/default.asp

For 9V alkaline:
http://www.duracell.com/oem/Pdf/MX1604.pdf
---
You can't even really do that, since remaining capacity varies with
load and if you don't have a valid Voltage Decay VS Load profile it's
all pretty much guesswork. Educated guesswork, but guesswork
nonetheless.
---
---

start: measure r

120k: bne 120kR 12k
r = 1.2E5
bra calci

12k: bne 12kR 1200
r = 1.2E4
bra calci

1200: bne 1200R 120
r = 1.2E3
bra calci

120: bne 120R 12
r = 1.2E2
bra calci

12: bne 12R 1r2
r = 1.2
bra calci

1r2: bne 1.2R 0r12
r = 0.12
bra calci

0r12: bne err1
r = 0.12

calci: connect resistor to 9V battery
measure battery voltage
divide battery voltage by resistance
quotient = i
write i to RAM for later use
bra end

err1: resistor out of range
write error flag to RAM

end: bra end
---

20. ### John FieldsGuest

---
Seems most of us use a program I always forget the name of and the URL
to, but I prefer to do it "by hand" since I think it makes for a more
compact drawing.