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A few beginners questions

commander768

Jul 18, 2014
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I did another test. I inserted a battery, under load it was 0.86V and 1.35A, then i measured resistance without battery ( used a wire to connect those terminals where battery would be ) and i got a reading of 1 Ohm. It doesnt make sense since from voltage and amperage readings i should be getting about 0.6Ohm
 

Gryd3

Jun 25, 2014
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I did another test. I inserted a battery, under load it was 0.86V and 1.35A, then i measured resistance without battery ( used a wire to connect those terminals where battery would be ) and i got a reading of 1 Ohm. It doesnt make sense since from voltage and amperage readings i should be getting about 0.6Ohm
Remember that batteries have an internal resistance as well which will throw your measurements off if you don't account for it.
 
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BobK

Jan 5, 2010
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Well, that is the correct way to measure.

You really need two meters so that you are reading both the voltage and the current at the same time. It is not possible to get such different current through the same resistor at the same voltage.

Bob
 

BobK

Jan 5, 2010
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I did another test. I inserted a battery, under load it was 0.86V and 1.35A, then i measured resistance without battery ( used a wire to connect those terminals where battery would be ) and i got a reading of 1 Ohm. It doesnt make sense since from voltage and amperage readings i should be getting about 0.6Ohm
Your meter might not be good at measuring resistance below 1 Ohm. Also, your resistor is obviously heating up (as evidenced by the burn marks). This will change the resistance.

Bob
 

commander768

Jul 18, 2014
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If without battery it is 1 Ohm and with battery 0.6 Ohm, then it would mean that battery somehow produced -0.40Ohm, i am not sure if that is possible. Also i am possitive that 2 batteries produced different Amperage at same Voltage. I measure amperage and right after that i measure voltage. Those 2 values dont change in a few seconds. And i tried all of that with batteries many many times ... Could it be that those batteries are damaged that badly to produce those numbers ? Otherwise i will start thinking that my multimeter is bad ..
 

Gryd3

Jun 25, 2014
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Otherwise i will start thinking that my multimeter is bad ..
According to Ohm's law. You cannot change the value of current without changing the value of resistance, or voltage.
You are over-looking something. Heated wire will change it's resistance. Batteries have an internal resistance (think of it like an extra resistor hidden inside the battery) that become more apparent when under heavy load. Your meter will have an internal resistance as well.
You are apparently dealing with a 1Ω, and at that range, your meter may introduce enough resistance to skew the amperage reading. Your meter 'should' have a very high internal resistance when measuring voltage, so that is most likely your most reliable test result. (Assuming you have reliably measured the resistance of your home-made resistor)
 

commander768

Jul 18, 2014
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Thank you for all your replies,

The first time i noticed something strange btw was before i even measured amperage. I was testing different batteries and i kinda knew that if a battery shows 0.7V under load for about a minute, that wire will get very hot. And then if it stays like that for 10 more minutes, the wire will still be very hot. But some batteries stayed at 0.7V for 10 minutes but wire was completely cold, which would show that not enough current is running through it. This was indeed very strange. It was almost like if the current would be running through multimeter instead of through the wire ...

Anyway i am not looking for exact voltage and amperage in this case. I simply had 24 old batteries and i wanted to pick 12 best of them. I did this by puting them under load ( on picture ) and then measuring voltage drop over time. This way it was easy to find which batteries are better.
 

Gryd3

Jun 25, 2014
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Thank you for all your replies,

The first time i noticed something strange btw was before i even measured amperage. I was testing different batteries and i kinda knew that if a battery shows 0.7V under load for about a minute, that wire will get very hot. And then if it stays like that for 10 more minutes, the wire will still be very hot. But some batteries stayed at 0.7V for 10 minutes but wire was completely cold,
Let me give you some links:
Remember. Batteries and Meters have resistance too!
Using such a low value resistor when you do current tests is a bad idea if you don't account for your meter's resistance.

http://www.allaboutcircuits.com/vol_1/chpt_8/3.html
http://www.allaboutcircuits.com/vol_1/chpt_8/5.html

I am hoping you read these, and I look forward to seeing something you have learned referenced in one of your future posts.
 

BobK

Jan 5, 2010
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It was almost like if the current would be running through multimeter instead of through the wire ...
Most likely, it was. Because you actually had the meter set to read current instead of voltage.

Does your meter have a different socket for the + lead when measuring current? If it does, and the lead was in the current socket, then the multimeter would behave that way even if you had it set to measure voltage, because that input has a very low resistance shunt to ground.

I would like to see you try to repeat this experiment with two meters if you have them, or do it again with one meter being very careful of where the leads are plugged in and what the multimeter is set to read.

Bob
 

commander768

Jul 18, 2014
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Gryd3: Interesting reading .. obviously my readings were off as far as the absolute A and V goes .. That would explain why measured and calculated values didnt match. But it is not what i am looking for, i am comparing batteries to find out which ones are good and which ones are bad. And if we say that all batteries have the same internal resistance, same voltage reading should produce the same amperage ( if i measure first voltage and right after that i measure amperage, in few seconds values cannot change that much ... ). The only rational explanation i can think of is heat of the resistor. I will have to use different resistor for future testing, one that doesnt heat and change its rating .. as for using bigger resistance, this will be harder to do. When i compare batteries i want as big as possible amperage to flow, otherwise this tests take a very long time.

BobK: My multimeter has a different socket for measurnig current ( i am using the unfused 10A socket ). I am sure that i used the right socket and the right setting. Getting a second multimeter to doublecheck would be a good idea. Right now like i said i am thinking how the heat of the resistor affects the values, and if this could be the reason for what my multimeter showed: different amperage at same voltage ( for same type of batteries ) and also in 1 case completely cold wire in othercase searing hot wire, again, same voltage of same type of batteries.

Again thank you for all the answers, this experiment is really interesting and also i learned many new things on the way
 

BobK

Jan 5, 2010
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The measurements you are getting are not correct. They are impossible. Stop looking for reasons for them until you are sure you are measuring correctly. I will bet that if you carefully check the meter setting and position of the probes, you will not be able to reproduce your incorrect results. It is not possible to have 0.7V across the resistor with one battery and have the resistor cold, and 0.7v from another battery and have it hot. In the first case, there was no current going through the resistor, as I said, I will bet that you had the probe in the 10 amp position, which would cause it to do exactly what you said it did. I know this because I have done it myself!

Please try the measurements again taking picture of the whole setup for each measurement.

Bob
 

Gryd3

Jun 25, 2014
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And if we say that all batteries have the same internal resistance,
The internal resistance of a battery will vary greatly. You can determine the internal resistance of a battery, but the result will only be valid while you are dealing with a battery of a similar charge and condition. I have also found numerous charts online that show the relationship of resistance / foot and the effects of different temperature. I would suggest you take a look, as the heating may not be contributing to what you are seeing.

I would also like to stress the importance of using a larger value resistor, unless you specifically require the same or similar current draw in normal operation. Using such a low value resistor will cause incorrect readings. Even if you account for the internal resistance in the meter, a change of 1/10th of an Ω will through your reading off. Once you have a reliable load to discharge your battery into, you can very easily use the voltage drop across it to determine current.
You should also check the data sheets for the batteries, as discharging a battery at a rate that is higher than designed for can cause damage.
 

commander768

Jul 18, 2014
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This are Ni-Cd batteries from battery drill so they can handle a lot more then 2A current that i am drawing from them :)

I will try to get a resistor with more Ohms to get better readings, also i will post info about 2 batteries that show interesting numbers
 

commander768

Jul 18, 2014
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A few days ago i got 2 old dewalt 14.4 battery packs. They were dead packs that haven't been used for many years and would be thrown away. I opened them and inside i found 24 4/5 subc 1.25Ah batteries. After some testing i found out that some batteries still hold voltage and some were at 0.0V ( even charging them would leave them at 0.0V ). After zapping them i managed to bring 4 batteries back to life while 4 i couldnt ( i will try zapping them more though :) ). Then i tested them. Most of the batteries survived my testing bed for 25 minutes or more ( running 1.2 - 1.5 amps, as you know i cant be completely sure of the value .. ). I will use those batteries for 2 of my old drills ( 1x 14.4 and 1x 4.8) as backups. I am learning more about zapping too .. zapping from 0.0V to >0V works, but for how long, i dont know. Zapping batteries with good capacity but bad voltage also helps a bit .. batteries that were 0.6V before zapping under load, were 0.8 or 0.9V after zapping ..but what is interesting is that this didnt seem to affect the time they hold. This would mean their capacity also improved ..

Btw i noticed something interesting. I measured battery voltage and amperage again and i got 0.8V and 1.44A. This would get us about 0.55 Ohms .. and a while ago i measured resistance of this testing bed and it was about 1.1Ohm. Now i know all those measurements are off because of such low resistance i am using but if you sum 0.55 Ohms and internal resistance of multimeter which is +-0.5 Ohm, you get the ~1 Ohm resistance of testing bed that i am getting. Add battery resistance and it is damn close. Just thought i would mention it :)

I noticed that certain smart chargers know when to cut of charging of NiCd batteries because when batteries are full, their voltage drops for about 0.1V. This is hard to notice but if you are charging 4 batteries or 12 batteries in series it is easy to notice it. How sure can i be about that ? For instance if i am charging with 0.22 amps for few hours and then suddenly i read 0.26 amps, does this mean that batteries are full or could it be some other factor ?
 

commander768

Jul 18, 2014
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Gryd3: about that 14.4V drill that had smoke coming out .. i am afraid that u are right about what happened. I tested it today and it had less torque then my little 4.8V drill. They have the same batteries so it cant be that. Drill will still be ok for some sanding, it has double the speed of my other drills so that is great. It just sucks that when i finaly got some batteries that can hold a charge for a while, the drill loses its torque.
 

Gryd3

Jun 25, 2014
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Sorry to hear.
You're going to have to baby your drill. The coils that have shorted out won't work again... so it will have a tendency to heat up and further damage the remainder of the coils. The drill is not completely dead though. Keep your eyes open for other dead drills, or pop your open and see if you can find some info on the motor. If you can find a replacement motor, just swap it out.
 

commander768

Jul 18, 2014
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It is not a big loss since that drill is a cheap fartools that i got for free ( with a dead battery ). But there are 2 things that i really like about it. First, it can spin up to 900 rpm, while my other drill goes only to 550 rpm. And second, the button is dynamic, meaning drill can do anywhere between 0 and 900 rpm, while this new cheap drill i got basicly has an on and off switch. Meaning it can drill either at 0 or 550 rpm. This can make screws fly of sometimes.
Ppl are selling their drills with dead batteries for like 5 euros ( ~3.5 dolars ) so maybe i will get 1 and just put those batteries i salvaged into a new pack and use with that drill.
 

commander768

Jul 18, 2014
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Interesting. I decided to test that drill a bit and i put it to 3/8 strength and it worked, i held the drill and clutch got into effect. If i tried 4/8 or higher, it didnt, drill simply stopped. Then i recharged it just out of curiosity and tested it again. Put it to 8/8 ( highest setting ) and i couldnt hold it. Since i didnt use that drill much but batteries obviously discharged anyway, it got me thinking that while drill maybe has less torque, it still has enough to hurt my fingers when i am trying to hold it, but more importantly, obviously batteries are self discharging very quickly.

This is kind of strange. First of all, i put battery pack out of the drill when storing for this exact reason, to make sure there is no connections that could discharge battery. Also i tested those 12 batteries and their charge is very close to eachother, under load and when not under load. To my understanding for that fast self discharge, 1 or more batteries must have a lot lower charge then the rest. Here this is not the case, so why is that battery pack doing that ? It was maybe 2 or 3 weeks and power went down considerably.
 
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