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.
The battery is not a perfect device, it can not supply voltage with unlimited current to a load.
You seem to know this already. What does limit the battery's ability to supply current to a load?
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.
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 dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis
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?
--- 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.
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:
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:
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
Notice that R1 is the internal resistance of the battery, and notice that it changes with load.
---
--- A1. Different ammeters would show the same variances if they all had the same internal resistances.
---
--- A2. Same answer
---
--- 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.
---
--- A4B. Measure the [momentary] short-circuit current using a shunt.
Measure with a MM. I did the first and last tests again and measured the battery under load. 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. Question: Does the Law refer to unloaded or loaded voltage source?
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.
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 meter,which will read to .0001 mA,reads .0600 mA, yours, reading to 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
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--
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.
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 unloaded battery (before connecting to the circuit), will the answer 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 the load, as John stated.
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.
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
Vancouver Power Squadron: http://vancouver.powersquadron.ca
Thanks John, I was just reporting more info, no offense intended. 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 degrades at about the same rate.
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.
You guys actually have the patience to do these great "schematic" sketches in your mail reader, or is there a secret I don't know about that makes it easier 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. :-)
--
For general info:
http://www.duracell.com/oem/default.asp
For 9V alkaline:
http://www.duracell.com/oem/Pdf/MX1604.pdf
>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
>degrades at about the same rate.
--
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.
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