Battery amp-hours don't add up??

rick_od · 11-01-2006, 12:34 PM

I have a hand held Volt-Ohm Meter with an amp probe. I hooked it up around the coach battery cable and went around the coach and turned everything on & off one item at a time and recorded the amp draw. I then turned on everything from my list and recorded the total amp draw. With everything on at once, the amp draw was 25 amps. When I add up everything individually on my list it adds up to 55 amps.
Anybody know why the difference??? If one lamp draws 2 amps, why don't 2 lamps draw 4 amps??? For me they draw 2.8 amps.
Thanks

rick_od · 11-01-2006, 12:34 PM

I have a hand held Volt-Ohm Meter with an amp probe. I hooked it up around the coach battery cable and went around the coach and turned everything on & off one item at a time and recorded the amp draw. I then turned on everything from my list and recorded the total amp draw. With everything on at once, the amp draw was 25 amps. When I add up everything individually on my list it adds up to 55 amps.
Anybody know why the difference??? If one lamp draws 2 amps, why don't 2 lamps draw 4 amps??? For me they draw 2.8 amps.
Thanks

Moder2 · 11-01-2006, 12:59 PM

I am not electrical engineer, and really don't have a good answer, but my thought on the subject is that it takes a certain amount of power to get to the appliance, and those that are "branched" off the same line, would be mearly drawing from the branch when both were on. In other words, there is loss in the wiring. Lets say two lights, A & B are on the same string. When only A is on, and draws 2 amps, and only B is on it draws 2 amps. But, when A and B are both on the current is already going to A, so that part of the route is already incurring the loss, and you only need the power only has to go the extra bit to B and of course, the draw of B.

If you are comparing draws that are you are certain are on seperate circuits, I have no answer.

I could be dead wrong on this, and am curious to see a more qaulified answer...

John_Canfield · 11-01-2006, 03:26 PM

It could be that manufacturers rate the current requirements for various equipment for the total inrush current or surge current at startup.

I wouldn't be at all concerned that calculated vs. actual current has that much varience.

I'd also want to know the accuracy of your VOM.

rick_od · 11-01-2006, 05:08 PM

The VOM is made by Fluke. No arguments as to its quality.
What I am trying to figure out is this...
Furnace = 4.5 amps
Bedroom Lights = 1.7 amps
Livingroom Lights = 1.7 amps
TV & VCR through the inverter = 3.5 amps
Each of these was measured one at a time.
If I turn them all on at the same time you would think that it should = 11.4 amps....it doesnt. When tested with all of these items on it = 8.5 amps.
I reduced the size of my example considerably. My actual list contains 18 items and differs by 30 amps.

John_Canfield · 11-02-2006, 04:22 AM

Fluke is one of the best - I use an old model 77. You really have me scratching my head and the only factor that comes to mind is that your Fluke is not as accurate at lower current readings as it is at higher ones. I have never been concerned with precision current measurements and I have no idea what the Fluke specs are for current.

You could test the meter by taking a known resistance and apply a known voltage to it and compare measured vs. actual. Set up the test so that you measure 0.5 amps, 1.0 amps 5.0 amps amd 10.0 amps. You would have to be creative in finding a suitable resistor for 5 and 10 amps due to the power dissipated with 12 volts, or use a low voltage (<6 volts) to keep the heat down.

If it is not instrument accuracy issues, I don't have a clue

tomsm · 11-02-2006, 04:57 AM

Are you monitoring battery voltage during your tests? If you have a weak battery/cell, the voltage would drop with the higher load and thereby reduce the amps.

Clay L · 11-02-2006, 05:19 AM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by tomsm:
Are you monitoring battery voltage during your tests? If you have a weak battery/cell, the voltage would drop with the higher load and thereby reduce the amps. </div></BLOCKQUOTE>

Very good point!!
If the battery voltage is constant and the amps don't add up then the only answer is measurement error as John said.

tomsm · 11-02-2006, 09:25 AM

Another thing you might check is the connections on both ends of your battery cables. A slightly loose/corroded connection can work fine with a light load, but become more resistive as the load increases.

rick_od · 11-02-2006, 11:43 AM

Tomsm, I think your on to something here. I have a '03 Sightseer that I bought used. I think that the coach batteries are the origional. I have the stock converter. I do not leave it plugged in all the time in fear of cooking the batteries. I have it hooked to a timer that turns on for 2 hours every night. I think that the batteries were either not completly charged or are too old to hold a full charge. When I was doing my test, and I turned on everything, the batteries died. Which means that the battery voltage must have dropped when everything was turned on and therefore explains why I have different amp readings.
Bravo!!!

John_Canfield · 11-02-2006, 12:01 PM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by tomsm:
Another thing you might check is the connections on both ends of your battery cables. A slightly loose/corroded connection can work fine with a light load, but become more resistive as the load increases. </div></BLOCKQUOTE>Ooh Tom - you might the cigar! Rick needs to monitor the voltage as the current increases, or make the tests with the charger/converter running.

JRP · 11-02-2006, 12:04 PM

I think moder2 had the most valid point. you're measuring the individual current flows thru the entire circuit all the way back at the battery, which includes portions of shared cable common to all the branch circuits. Since cable has resistance youre individual measurements include extra resistance from the resistance of those shared cable lengths (branch circuit plus home run) in all 18 individual measurements, which drives those amp readings down by following ohms law I=E/R
Then when you measure with every thing on at once, the resistence of that shared cable length counts once, instead of 18 times, giving your higher reading.
Those differences in individual vs total circuit resistance account for most of the difference you see.
lets say one of your light circuits had 12 vdc / 6 ohms = 2 amps , but if 2 of those 6 ohms of resistence were from the common/shared home run cables, then the actual light circuit was only 4 ohms and 12 vdc / 4 = 3 amp

Its not the fault of your Fluke meter, its just the method youre using to measure, doesnt really isolate each circuit path from their shared common path.

Jim

Clay L · 11-03-2006, 05:48 AM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by JRP:
I think moder2 had the most valid point. you're measuring the individual current flows thru the entire circuit all the way back at the battery, which includes portions of shared cable common to all the branch circuits. Since cable has resistance youre individual measurements include extra resistance from the resistance of those shared cable lengths (branch circuit plus home run) in all 18 individual measurements, which drives those amp readings down by following ohms law I=E/R
Then when you measure with every thing on at once, the resistence of that shared cable length counts once, instead of 18 times, giving your higher reading.
Those differences in individual vs total circuit resistance account for most of the difference you see.
lets say one of your light circuits had 12 vdc / 6 ohms = 2 amps , but if 2 of those 6 ohms of resistence were from the common/shared home run cables, then the actual light circuit was only 4 ohms and 12 vdc / 4 = 3 amp

Its not the fault of your Fluke meter, its just the method youre using to measure, doesnt really isolate each circuit path from their shared common path.

Jim </div></BLOCKQUOTE>

That sort of violates Kirchoff's Current Law.
You might want to take a look at: http://cnx.org/content/m0015/latest/

Moder2 · 11-03-2006, 11:14 AM

Ok, I amm confused now.

I understand how the voltage varing could effect the current readings, and am curious to here the results after he tests them hooked to Shore Power.

I do not understand how KCL could be violated here? All the current going in is still coming out of the nodes. Could you please explain in further detail? I would like to gain the knowledge of this topic.

Thanks!

John

rick_od · 11-03-2006, 11:22 AM

Just for the record...I will not be amp testing with the coach hooked to shore power. I origionally put the amp clamp on the battery cable at the battery. If I am connected to shore power, the only reading I will get at this location is what the converter is putting back into the battery to charge it. The only other way to do it while hooked to shore power is to disect my dc load center and amp clamp it from there. And thats assuming that I know exactly which fuse goes to what load. Too much hastle. I need new batteries. When I replace them, I will retest and let everyone know the results.

Clay L · 11-03-2006, 12:00 PM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Moder2:
Ok, I amm confused now.

I understand how the voltage varing could effect the current readings, and am curious to here the results after he tests them hooked to Shore Power.

I do not understand how KCL could be violated here? All the current going in is still coming out of the nodes. Could you please explain in further detail? I would like to gain the knowledge of this topic.

Thanks!

John </div></BLOCKQUOTE>

Basically KCL says all the currents are additive. (Current flowing in different directions are still considered additive but one will have a minus sign and the result will be one minus the other). This means that all the currents flowing in the same line will just add - in that line - no matter where they enter or leave the entire circuit. With one current source all the currents will have the same sign and will add.
If he is measuring at the battery, all currents flowing out of the battery will add and be equal to the total current leaving (or entering) the battery.

This something that is hard for me to explain with just words and I am afraid I am not doing a good job.
It would easy enough if we could sit down and write the loop equations.

Maybe someone with more word skills will drop in and clarify things.

NR4A · 11-03-2006, 01:23 PM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I have a hand held Volt-Ohm Meter with an amp probe </div></BLOCKQUOTE>

If I may add my $0.02

Yes, Fluke makes great test equipment.

The "amp probe" is what's nagging me on this question, however.

An inline ammeter rated for 30 or 40A full scale might offer a little more accuracy and resolution than a probe.

IMHO

John_Canfield · 11-03-2006, 02:06 PM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Clay L:

That sort of violates Kirchoff's Current Law.
You might want to take a look at: http://cnx.org/content/m0015/latest/ </div></BLOCKQUOTE>Kirchoff's law - oh my gosh.. I haven't thought about that for 40 years. My electronics theory education is mostly a distant memory

rick_od · 11-03-2006, 03:40 PM

quote:
I have a hand held Volt-Ohm Meter with an amp probe

If I may add my $0.02

Yes, Fluke makes great test equipment.

The "amp probe" is what's nagging me on this question, however.

An inline ammeter rated for 30 or 40A full scale might offer a little more accuracy and resolution than a probe.

IMHO

Sorry,

It is a fluke VOM with a amp clamp on one end. You clamp it around the wire to be tested and it gives you the amps. I don't know if an inline amp test would be more accurate.

JRP · 11-03-2006, 03:57 PM

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Kirchoff's law </div></BLOCKQUOTE>

My previous explaination was based on looking at each of his 18 measurements as 18 different series loop circuits using total volts, total current & total resistence; Kirchoff's law has nothing to do with that approach, its purely Ohms Law.
However, if you wish, we could apply Kirchoff's law to the condition where all 18 branch loads are turned on. In that case its Kirchoff's law that explains the total of all the individual 18 branch circuits must equal the total current flowing in the main battery circuit. Reaching the same conclusion, he is not actually measuring the individual branch circuit currents, since his measurement method includes the resistence load of the main battery cable in series with the branch load.
If he were to use a clamp on amp meter on each actual branch circuit wire, then he would get the real branch circuit currents and they would all add up to the total current.
Jim