Hard to admit but watch out, very basic and elementary question ahead :o .

Let me expose my problem with an example and maybe someone could teach me where I am wrong?
To my understanding any current meter return an “instantaneous” value. You wire a multimeter on a circuit and have a direct current reading. A 10k resistor will pass about 1.2mA of current if connected to a 12volts supply. Now, if I record this over and hour, I will read 1.2mA for every single reading (it could be 1 record per hour or in theory an infinite amount of records per hour). To estimate current consumption, one should sum all the instantaneous value and divided by the recording time. In my example and for one hour of 1 sec interval recording, my resistor would pass on average 1.2mA X 3600 sec / 3600 sec or 1.2 mA/hr! How come a resistor that pass every single instant 1.2mA will only pass 1.2 mA over an hour? On the other side, if I don’t divide by the monitoring time and only take the sum of the instantaneous data, I can theoritically reach infinite amount of current over the hour and my resistor clearly don’t pass this much electron. How much current per hour my resistor really pass?

You don't calculate or measure current consumption per hour or other unit of time like days, weeks, months or years. If you want power consumption then it becomes watts as the unit of measure for power. You can have watt-seconds, watt-hours How to calculate my energy consumption

Device Wattage (watts) x Hours Used Per Day = Watt-hours (Wh) per Day.

Energy consumption is measured in watts per unit of time. The Power in a simple DC circuit would be Voltage * Current or 12 Volts * 0.0012 Amp = 0.0144 Watt (14.4 mW).

any current meter return an "instantaneous" value.

FALSE
DMMs measure rms current.
Instantaneous current is something completely different.
Instantaneous current is current at an exact point in time down to the uS.
Which a switch is closed connecting a battery to a load, the instantaneous current may be very high during the
first few uS and then decrease as the voltage across the load increases.
This kind of measurement can only be captured on a storage oscilloscope and will appear as a peak.
2uS later it would be different, hence the need to 'capture' it with a screen snapshot.
You need to use the correct terms. Unless you have a storage scope the only thing we can really discuss is rms
current.

Ron_Blain:
Energy consumption is measured in watts per unit of time. The Power in a simple DC circuit would be Voltage * Current or 12 Volts * 0.0012 Amp = 0.0144 Watt (14.4 mW).

Since the 0.0012 amp is a RMS current of unknown period (one shot DMM reading), I end up with 14.4 mW power consumption for the same unknow period. If I assume that the 0.0012 amp was measure for 1 sec then the resistor need 14.4mW / sec. Correct?
Going further, 14.4 mW X 3600 sec. X 24 hrs will give about 1244 W/day.
If I power this resistor with a 12v battery, I would need a 100 Amp batterie!!! since 1244 W / 12 v = ±100 amp. Not possible. So I still miss something.
To raschemmel,
yes, I oversimplified it. Thank for this clarification. Do we know the averaging time that a dmm take to return a RMS current value? In DC circuit.

cdb101:
How come a resistor that pass every single instant 1.2mA will only pass 1.2 mA over an hour?

Because if you take 3600 readings and then divide the total by 3600 what you end up with is the AVERAGE current. And if something carries 1.2mA at every instant then the average is obviously going to be 1.2mA.

A few points here just for clarification. There are a few types of meters for measuring current, a lower cost meter is likely Average responding RMS Indicating and for that reason it will only read accurately measuring a true sine wave. A better quality meter will be RMS Responding RMS Indicating which will measure the true RMS value of a wave shape be it sine, square or anything else.

[RMS Current may be defined as](http://"Alternating Currents | Boundless Physics root mean square (abbreviated,the square root of two."):
"The root mean square (abbreviated RMS or rms ) is a statistical measure of the magnitude of a varying quantity. We use the root mean square to express the average current or voltage in an AC system. The RMS current and voltage (for sinusoidal systems) are the peak current and voltage over the square root of two".

"In physics, the RMS current value can also be defined as the " "value of the direct current that dissipates the same power in a resistor.".

So 12 VAC RMS has the same heating effect on a resistor as 12 VDC.

If I apply 12 VDC to a 10,000 Ohm resistive load the power will be 1.2 mW or 0.0012 Watt. Now from my first reply:

Device Wattage = 14.4 mW * 24 Hours used per day = 0.3456 Wh per day

Different DMMs have different sample time for the signal they are sampling but it matters not.

Ron_Blain:
If I apply 12 VDC to a 10,000 Ohm resistive load the power will be 1.2 mW or 0.0012 Watt. Now from my first reply:

Device Wattage = 14.4 mW * 24 Hours used per day = 0.3456 Wh per day

Device Wattage = 14.4 mW * 24 hours used per day = 0.3456 Wh per day.
Is the "h" in "Wh" above is a typo? or then this is power consumption on a hour basis for a day?
Then how to size a battery for using this resistive load for a full day? Normally, I would divide the wattage by the voltage and this would be 0.3456 Wh/day / 12 volts = 0.0288 amp.
This value is the total amount of current that the device will dissipate for the full day? A perfectly efficient and theoritical 30 mA battery would do the job for 1 day?
I ask because if I do the same calculation but for 24 seconds utilisation per day, I also obtain 0.0288 amp. :
Device Wattage = 14.4 mW * 24 seconds used per day = 0.3456 Ws per day.
The only difference would be the wattage basis (here second) above (hours). If it was not a typo.

Is this correct,
Case 1 (24 hours) = 0.0288 amp * 24 hours / hours in a day (24) == 0.0288 amp (28.8 mA)
Case 2 (24 seconds) = 0.0288 amp * 24 second / seconds in a day (86400) == 0.000008 amp (8 uA)
Case 2 would require 8 uA per day. Does this make sense?

You're still confusing 2 different things. There's current measured in A or mA. Then there is battery CAPACITY and that's measured in Ah or mAh where the h is not a typo it means HOURS. So you'd never see a 30mA battery but you might see a 30mAh battery. And that can deliver 30mA for 1 hour or 10mA for 3 hours etc.

The h on the end of units is important. You can't keep confusing capacity and current. Because you are your case1, case2 calaculations don't make any sense.

Is this correct,
Case 1 (24 hours) = 0.0288 amp * 24 hours / hours in a day (24) == 0.0288 amp (28.8 mA)
Case 2 (24 seconds) = 0.0288 amp * 24 second / seconds in a day (86400) == 0.000008 amp (8 uA)
Case 2 would require 8 uA per day. Does this make sense?

slipstick:
You're still confusing 2 different things. There's current measured in A or mA. Then there is battery CAPACITY and that's measured in Ah or mAh where the h is not a typo it means HOURS. So you'd never see a 30mA battery but you might see a 30mAh battery. And that can deliver 30mA for 1 hour or 10mA for 3 hours etc.

The h on the end of units is important. You can't keep confusing capacity and current. Because you are your case1, case2 calaculations don't make any sense.

Steve

Ok, my case 1 / case 2 calculations aim to convert everything on a daily basis while I understand from your reply that everything should be on a hourly basis at least to size my batteries.
Device wattage = 14.4 mW
Case 1 (24 hrs. per day) = 14.4 mW * 24 hours = 0.3456 Wh === a battery of 0.0288 Ah capacity will do the job for 1 day.
Case 2 (24 sec. per day) = 14.4 mW * (24/3600) = 0.096 Wh === a battery of 0.008 Ah capacity will do the job for 1 day.

Device wattage = 14.4 mW
Case 1 (24 hrs. per day) = 14.4 mW * 24 hours = 0.3456 Wh === a battery of 0.0288 Ah capacity will do the job for 1 day.
Case 2 (24 sec. per day) = 14.4 mW * (24/3600) = 0.096 Wh === a battery of 0.008 Ah capacity will do the job for 1 day.

Why are you using wattage ?
Use mAh's and then convert it by multiplying it by the battery voltage later.
Batteries are not rated by wattage so it makes more sense to use current (mAhs)
You should start with current instead of starting with wattage.

raschemmel:
Why are you using wattage ?
Use mAh’s and then convert it by multiplying it by the battery voltage later.
Batteries are not rated by wattage so it makes more sense to use current (mAhs)
You should start with current instead of starting with wattage.

Well? This is the point of my initial post. I did start with current and plan to convert to watt if required later in the chain. My problem is that my current measurement are not related to any time. It is rms value from my DMM. So it is not mAh. It is mA. I don’t know how to transfrom this into mAh. I was pointed toward using watt.
My original question and the title of my post is “how to go from instant current data to current per hour.”

If you want to size a battery then look at it this way. Batteries are rated a few different ways but lets just look at mAH and AH. Milli-amp hours and Amp-hours.

“The rated capacity of any battery expresses the average amount of current it releases over a period of time under normal use. This means that a battery with a rating of 200 Ah can deliver 20 amps of power at a constant rate for 10 hours”.

There is more to this. Using a 12 volt SLA (Sealed Lead Acid) as an example with a 10 AH rating. The reality is the actual battery voltage is about 12.9 volts fully charged and when that drops to about 11.4 volts it is considered discharged. In theory the battery will deliver 10 Amps for one hour, 1 Amp for 10 Hours or 0.5 Amp for 20 hours. With that thinking the same battery should deliver 100 Amps for six min and good luck with that. Also the numbers only apply to a new battery in good shape. With age batteries lose their ability.

Note your load current and maybe add a little like 20%. Now look for a battery that meets your needs for a known period of time. Consider things like battery chemistry and how it gets recharged. With all things considered choose your battery. Do not make this more complicated than it is. Allow headroom in your calculations.

Thank you for the explanation. Once I have a clear Wh or Ah value, I can transform my data to scale a batterie if required.
I guess I have the answer to my questions so if I still don't get it, it is because I ask the wrong questions.
I will try one last way.

In a DC circuit with a resistive load, my DMM read 10 mA. I can start a timer on my phone and read this for 1 minute, 10 minutes, 1 hour and so on. I will read 10 mA during the whole test. Base on the last reply, if I stop measuring after 1 minute, I should conclude that my circuit need 10 mA / minute. If I stop after 1 hour, I should conclude the circuit need 10 mA / hour. etc.

If I convert all these cases to mA per hour I will have different results.
600 mA / hour (when I stop after 1 minute), - an hour is 60 minutes,
10 mA / hour If I stop after 1 hour.

Base on the Device_Wattage example from Ron, I should do the following:
A- 10 mA * 12v = 0.012W
At this point there is still no unit of time involve since I am using the DMM reading and it is not scale to time.
Next is:
B- 0.012W * 5 minutes (I don't choose hour on purpose to try understanding) = 0.06W / 5 minutes.
This would be the power required by my device to run 5 minutes.
Then if I want the data for 1 hour I would do:
C- 5 minutes * 0.012W + 55 minutes * 0W = 0.06W / hour.
Finally I can obtain Ah and scale my batterie using :
D- 0.06Wh / 12 volt = 0.005 Ah.
please tell me this is it!

How come a resistor that pass every single instant 1.2mA will only pass 1.2 mA over an hour?

Amperage is a rate of current flow.

If it was water flow it could be gallons per hour. There is a common water analogy where voltage is water pressure and current is water flow. A skinny pipe is high resistance and you get less water flow. A fat pipe has low resistance and you get more water flow.

The main difference is if you cut a wire you get infinite resistance and no current flows. If you cut a pipe you get zero resistance and water flows-out everywhere. And with zero water-resistance, nothing bad happens (except maybe a flood ) but with zero electrical-resistance (a short circuit) you generally get excess current and something burns-up, or you blow a fuse or circuit breaker, etc.

If you know the current you can actually calculate the number of electrons per second, but it's a HUGE incomprehensible number! (Look up "coulomb" if you're interested.)

If your load is 10 mA and the load duration is 1 min then you have 10 mA/Min but normally it would not be expressed that way. A common unit of measure is mAH So 10 mA for 1 min would be 10/60 = 0.1666 mAH. We have 60 min in an hour and your load was only on for 1 min. Now if I measure a current of 10 mA and measure for 1 min but my load is on for 1 hour then I get 0.166 ^-3 * 60 = 10 mAH. That's all there is to it. What is the load current and how long is the load applied? Stop trying to read more into this than there is.

Ron_Blain:
If your load is 10 mA and the load duration is 1 min then you have 10 mA/Min but normally it would not be expressed that way. A common unit of measure is mAH So 10 mA for 1 min would be 10/60 = 0.1666 mAH. We have 60 min in an hour and your load was only on for 1 min. Now if I measure a current of 10 mA and measure for 1 min but my load is on for 1 hour then I get 0.166 ^-3 * 60 = 10 mAH. That's all there is to it. What is the load current and how long is the load applied? Stop trying to read more into this than there is.

Ron

I get it. This is clear with this example. Thank's a lot.