# Confused boy.

Discussion in 'General Electronics Discussion' started by RedGoblinz, Jun 18, 2012.

1. ### RedGoblinz

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Jun 18, 2012
I've got a few questions in my head, and I hope you guys don't mind if i post it all here, and I hope you guys can answer it all.

1. I bought a cheap analog multimeter, it reads my nimh battery as 2v, my lithium ion 7.4v battery as 9v. Tried changing the batteries and cleaning the probes, still the same. Though it reads my wall socket correctly at 240v.

2. The needle is supposed to be on all the zeros right?

3. One 1.2v 1600mAh nimh cell, one 1 ohm resistor, one clock. Connected in parallel. Calculating the current with ohms law would give us 1200 mA. I wanna ask, what does this value mean? The resisted value or the value that should be passing?

4. If it is resisted, how come higher resistance comes up with a lower current? Example 1.2v with 100 ohms, is 12 mA. Doesn't that a high resistance resistor resisted less current...?

5. In the parallel circuit, the clock was removed, multimeter plugged in, measurement came out about 25mA, which is 900000 mAh...... Okay what the hell am I doing...

6. Forgot to ask, when measuring current in a circuit, can I just pluck the probes into the
circuit without redoing the whole thing so that the probes are PART of the circuit?

7. While measuring the current for (5), the needle stayed at 25 even if I set the dial to DC mA 500, it's the same with 50, lol, my meter is failing. Only at 0.5 does the needle go to max.

8. What the heck is going on?

2. ### Harald KappModeratorModerator

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Nov 17, 2011
1) you measure the batteries in DC mode, the wall socket in AC mode, do you?
2) I don't understand this part. Do you mean that the needle points to 0 when no voltage is applied? If so: right.
3) What is the clock doing in this setup? If you want to know the current forced by a 1.2 V source through a 1 Ohm resistor your calculation is right.
4) Where is the problem? 1.2 V / 1 Ohm = 1200 mA, 1.2 V / 100 Ohm = 12 mA. Therefore the higher resistance has lower current. Why do you question that?
5) See 3: What is the clock doing here? Maybe you mena smething different, but not clock? Could you post a schematic of your setup?
And no, 25 mA of current doesn't tell you anything about the battery's capacity in mAh. What is it you want to measure?
6) Measuring current by way of probe tips requires you to open the circuit, insert probes connected to a multimeter thereby re-closing the circuit. The consequence is that in order to remove the probes (or the multimeter) you'll have to close the circuit by a piece of wire. Otherwise you have an open circuit and no current can flow.
7) I'm not sure about what you're doing. A schematic could help to understand. Anyhow: if you switch the range while the setup remains the same, the indication of the multimeter should change. Unless it's crap.

Harald

3. ### RedGoblinz

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Jun 18, 2012

1. Yes, dc mode.
2. Yeah that's what I meant.

3. This circuit is supposed to estimate the capacity of the battery. So the calculated value is the current that should be flowing through the entire circuit? If I measure the circuit with the resistor and battery in it, should I be getting 0.33 mA? Is the forced current (1200) used up no matter what? What turns the resistor hot? If a short is done to the battery 1600mAh), a wire from positive to negative, what's the supposed current?

4. I thought the value calculated is the resisted current, not forced current.

5. I want to measure how much current is flowing through the circuit.

6. What happens if I don't open up the circuit?

7. As I said I somehow got 25mA in question 5, I got 25 in the 50mA mode, I decided to change to 500mA, but the needle did not move.

8. I read somewhere that when measuring anything, the probe should only touch the measured object, and not anything else (ie plastic), is that true?
Thanks

4. ### Harald KappModeratorModerator

11,799
2,749
Nov 17, 2011
1), 2) o.k.
3) No, you can't measure the capacity by measuring the current alone. In order to measure the capacity you would have to discharge the battery with a known constant current while monitoring the voltage. If the voltage reaches the "empty" level (which depends on the type of battery, temperature etc.), you can calculate the capacity by C=I*t. The capacity thus measured will vary depending on the current you use.
Using a very small current will cause a long time for measurement. During this long time leackage and internal discharge will contribute to the voltage drop, thus giving you less capacity.
Using a very high current will cause the battery to heat up, incresing its internal resistance which in turn also leads to faster voltage drop.
It is best to use the same current as intended for the later ue to measure capacity.
See here http://batteryuniversity.com/learn/article/how_to_measure_capacity for a detailed explanation.
4) I have never heard the term "resisted current" before. A battery is a voltage source (you can measure the voltage without any load current flowing). The current is the result of this voltage applied across a resistor. In your case: 1.2V / 100 Ohm = 12 mA.
5) Which circuit? So far your circuit consists of the battery and the resistor plus a multimeter. And the current is as described above.
6) if you don't open the circuit you cannot insert the multimeter for measuring the current. Or, if the multimeter is already inserted, you cannot remove the multimeter. Only option: short circuit the multimeter, then remove it and leave the short circuit wire in place. Or use an expensive current porobe (I won't go into the details of this).
7) Since you haven't provided a schematic I still do not understand your circuit completely, especially the function of the "clock". Are you telling us that the current changed from 25 mA to 250 mA by switching the range of the multimeter only? How did you connect the multimeter to the battery? I do hope in series, not in parallel!
8) Always a good idea to keep tzhe probes away from anything but the object under inspection. However, if you use low voltages (you do) there is usually no problem if the probe touches something else. Just make sure that you do not accidentally short circuit anything.

Harald

5. ### RedGoblinz

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Jun 18, 2012
1. So, what's wrong with my meter, lol.

3. Yes, that's what I intend to do. Discharge the battery and calculate the time. So what happens to the 1200mA forced current? If a short is done to a 1600mAh 1.2v battery, what's the maximum current?

4. Resisted current meaning the current that is being stopped by the resistor and turning into heat (?), maybe I got this whole thing wrong...

5. -

6. What does short circuiting the multimeter mean?

8. Cuz I can't hold the probes to the wires firmly, is it fine if I use plastic clips?

6. ### davennModerator

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1,970
Sep 5, 2009
1) ... more likely beginner user error

3) what 1200mA "forced current' ? no such thing as forced current
that is not a way to test a battery, rather its a VERY GOOD way to destroy a battery
possibly even have it explode in your face ...DONT do it

4) resistors dont resist current, they limit current. I suggest you do some google searching on Ohms law and understand how resistors in series operate

6) Harald was referring to shorting out the probes of the multimeter so that its no longer in circuit.
8) yes thats OK

.....
OK some hints and tips.....

To measure Amps, you must break into the circuit and put the meter in series
DO NOT put an ammeter ( or multimeter in Amps mode) directly across a battery or other power source.
Its likely to have a very short life, probably ending in smoke and sparks

To measure Volts, you measure parallel across the circuit

Ohms law is V (Volts) = I (Amps) x R ( Resistance)

therefore in the above circuit if we want to find how much current is flowing we rearrange the formula to give Volts / Resistance = Amps
therefore 10V / 10Ohms = 1 Amp of current
you can measure the current in either of the 2 positions shown and the meter will read 1 A
the current flowing in a series DC circuit is the same everywhere

OK
see how you go digesting that

Dave

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7. ### RedGoblinz

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Jun 18, 2012
1. What am I doing wrong? ._.

3. Well, if that's not the way, I have no other ways of discharging my battery. I've done a few and non has exploded. Forced current was stated by Harald.

Last edited: Jun 18, 2012
8. ### RedGoblinz

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Jun 18, 2012
What happens to the current in a circuit?

The results from Ohm's law calculation, 1200, is it in mA or mAh? If its mA it would mean per second...right? My goodness im so confused...

I just broke into a circuit and tested a 1.2v batt with the 1 Ohm resistor, first thing i noticed is the resistor did not heat up like crazy the way it does if the multimeter wasn't there. Whats up?

Last edited: Jun 18, 2012
9. ### dssteven

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May 9, 2012
But for further explination, from wikipedia via a simple google search I found this:
"In practical terms, the ampere is a measure of the amount of electric charge passing a point in an electric circuit per unit time with 6.241 × 1018 electrons, or one coulomb per second constituting one ampere"

Battery capacity is rated in mAh (milliAmp*hour) or Ah (Ampere*hour). A 10Ah (Ampere*hour) battery will hold a current draw of 10 Amps for 1 hour or 1 amp for 10 hours. (Amps * hours = the unit Ah)

If you tested a 1.2v battery with a 1 ohm resistor, the current in the circuit would be 1.2A or 1200 mA. In a 1600mAh battery as you seem to have, the battery would supply a current of 1.2A to the 1 ohm resistor for:

1200mA*t(hours) = 1600 mAh (where t is the amount of hours it can supply that current) so t = 1.33 hours, about an hour and 20 minutes.

for a 100 ohm resistor, it would draw 12mA (1.2V/100 ohms - note this is in mA not mAh) and would therefore draw 12mA*t = 1600 mAh = 133.33 hours

Davenn's diagram explains very well how to measure voltage and current in a circuit like the one you've explained.

10. ### Harald KappModeratorModerator

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Nov 17, 2011
Re-read my post thoroughly. I said
which is just another way of expressing Ohms law.

Determining the capacity of a battery and charging a battery in the proper way is not a simple feat. Read the link to the batteryuniversity I gave you.
Therefore and besides, I think davenn is right: before you deal with an already complex problem you should understand the basic principles of an electric circuit (Ohms law, Kirchhoffs laws). This will help you a lot along the way.

Harald

11. ### CocaCola

3,635
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Apr 7, 2012
Seriously? Put a load on the battery and let it drain it, this can be a resistor, a light bulb, anything really, just don't short it with little to no resistance/load, that is just foolish...

Count your luck and don't push it...

12. ### RedGoblinz

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Jun 18, 2012
dssteven - What happens if the battery only has 400mAh left and is put into the 1200mA circuit? Will it still drain to 0mA?

Harald - My apologies for the misreading.

CocaCola - I am putting a load on the battery, a resistor. I'm not just shorting the battery.

13. ### CocaCola

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Apr 7, 2012
On paper in a perfect world it will last 20 minutes, in the real world more likely a few minutes, maybe a little more maybe a little less... As long as there is a load on the battery it will continue to drain until it's flat dead, that is assuming there is no fail safe that prevents further draining as is the norm on Lipo cells that generally have a built in circuit that will cut off the drain well in advance of flat dead so that no damage happens to the cells...

Even a full wire short has a resistance, you have limit the drain to a safe level or you risk damage or worse, again most Lipo cells have a built in circuit that prevents excess drain levels... A 1 ohm resistor IMO might as well be a short in many cases...

Last edited: Jun 19, 2012
14. ### RedGoblinz

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Jun 18, 2012
So, back to the reason i got a multimeter, measuring voltage. What am i doing wrong? I keep getting overly high voltages.

Right, is 4.7 Ohms fine? (Or so it says 4.7...)

15. ### CocaCola

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Apr 7, 2012
You determine the safe or recommended drain rate of the battery in question and use Ohm's Law to determine the appropriate minimum value resistor to accomplish this...

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Jun 18, 2012
17. ### CocaCola

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Apr 7, 2012
That is applicable for those batteries, every battery type and manufacture differs...

Taking those values, that says .25C is the recommend discharge...

1C = total rated mAh drain in one hour... So 1C of a 1600mAh battery is 1600mA

If you had a 1.2v 1600mAh battery you would do the math...

1.2 / (1.6*.25) = Ω

1.2 / 0.4 = Ω

And in this case the answer is 3Ω or larger...

18. ### RedGoblinz

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Jun 18, 2012
Alright!
I have 2 last questions, so far. I hope you can try to imagine this, a circuit.
____ _____
Battery |____|R|_____|Clock

If the resistor was 2.5 ohms and the battery is 1.2v, can the clock be powered? If so, powered until what happens before it stops (ie fully drained, insufficient voltage)?

19. ### CocaCola

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Apr 7, 2012
Possibly, depends on the power requirements of the clock...

What happens to what? The clock? It will probably continue to drain the battery trying to run even though it isn't running, or it might do other things, depends on how the clock is designed...

I believe that was addressed earlier, likely operator error... That or the batteries are really being overcharged to a level that will likely cause them damage...

Take some pictures of your measuring technique...

20. ### RedGoblinz

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Jun 18, 2012
Im guessing i cant find out myself what the requirements are?
I've had a clock's needle not move but is able to give out the ringing tone, should it be that way?

The way i measure the voltage is exactly like how this guy did it at 4:20