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Thread: How do you measure amps?
12-30-2008, 06:14 AM #1
How do you measure amps?
If I put an amprobe on the hot leg of a 120v ac line I get a reading of some finite amount of electrons traveling thru the wire. If I put the amprobe on each of the hot legs of a 240v ac circuit I get two numbers. Is the amperage of the 240v load the sum of the two numbers or the average? If I get up on the shop roof and measure the amperage on each of the 240v single phase legs am I using the total or the average? These questions keep me up at night.
12-30-2008, 06:52 AM #2
They are subtractive. You'll find the answer to your puzzle on the neutral.
12-30-2008, 09:50 AM #3
For any 240 Volt load such as a dryer, hot water tank, or range the current in both hot leags will be the same and no current flows in the neutral.
For 120 Volt loads such as receptacles and lights the current in each hot leg will differ depending on the balance of loads. For 120 volt loads the neutral carries the current of each hot leg and they subtract because they are of opposite phase.
If you consider the volt-amps or watts(assuming all loads are resistive) if will help clear things up a little. The utility company watt-hour meter measures volts x amps for each hot leg and adds them, the current in the neutral is of no concern to the meter.
Normally the only concern for measuring amps in each wire is to determine adequacy of the wire size and protective devices such as fuses or circuit breakers.
12-30-2008, 10:35 AM #4
For a three wire 240 volt circuit, L1 - L2 = N where the neutral current N might
well be zero.
But the short answer is, each wire is carrying the current the meter says it is,
as read on the dial. This of course is for resistive loads only.
Defintion of an amp is one coulomb of electrons flowing through a surface in the
wire in a second. The typical analogy is "gallons per second" of water flowing
through a pipe.
Electrons not equal to water of course but the idea is that you are "counting"
the items going on by.
12-30-2008, 11:09 AM #5
If it is a mixed 240/120 load, the neutral carries the difference
If I get up on the shop roof and measure the amperage on each of the 240v single phase legs am I using the total or the average? These questions keep me up at night.
How that divides up into loads is another matter.......
12-30-2008, 03:28 PM #6
12-30-2008, 07:23 PM #7
"Does this work even if both hot service legs are suppling
Imagine a case where you have a 100 amp load running
off of 240 volts. No neutral current. L1 = L2 = 100 amps.
Then imagine a similar load that draws 100 amps of 120
volt power. Now L1 = 100 amps, L2 = zero, and N = 100 amps.
Put a duplicate of that 120 volt item on the other side,
now L1 = 100, L2 = 100 amps, and N again is zero.
If you could magically change that second load to be,
say, 120 amps, then L1 = 100, L2 = 120 amps, and N = 20
If you have one 100 watt light bulb running on one side
of your service, the neutral current is one amp. If you
put an indentical light bulb on the other side, now the
neutral current is zero.
12-30-2008, 07:28 PM #8
ARRRGH! You measure voltage ACROSS two points and current THROUGH a single point. I don't mean to sound pedantic, but this tells you something about what you're trying to measure. If you want to measure volts, you put the meter in parallel. If you want to measure amps, then you have to break the circuit and put your meter in series and then hook it to the load, whatever it is. Since we're talking about AC current, the meter should read the RMS current being drawn through the meter.
You can get one of those meters with the split doughnut on the end and measure current by means of the magnetic field generated by the flow if you don't want to break your circuit.
12-30-2008, 08:14 PM #9
He's using an amp-clamp. He says so in the first post.
12-30-2008, 08:58 PM #10
Actually, we need to make sure that his service is in fact single phase ( meaning 2 hot legs and a neutral ) and not 2 legs of a 3 phase service.
In the single phase system, the legs are 180 degree apart, while a 3 phase system has 120 deg between hotlegs.
Using the neutral for current measurement is useless in either, as it is unlikely to show the actual current useage unless there is one and only one hotleg being used.
As Jim has pointed out, in a single phase system 1A on L1 and 1A on L2 will produce 0A reading on the neutral.
In a 3 phase, the meter would read 1A. However 1A on L1 only would also produce 1A on the neutral. ( again, true 3 phase )
SO anyway, if you want to find out the actual current useage, just measure all the hotlegs and sum them together.
It will likely not give you WATT reading, rather VAR (Volt-Amp-Reactive), as your load is typically inductive in nature ( coils rather than capacitance )
Watt reading is also useless for most, as the power co. will bill you on Voltage x Amp (VAR) amount.
OTOH if you are a large enough user, you may want to worry about the Watt vs. VAR amount, which could easily add up to a tidey sum. With active or passive monitoring, you can reduce your current useage therefore your monthly bill by switching in parallel capacitance to offset the inductive loads.
12-30-2008, 09:28 PM #11
Now that last answer has me confused. If a device is running on 240 volts, would its current consumption not be read on a single leg? If you're running 120 volt devices on both hot legs of the 240, you'd have to sum them to get total usage, but if you simply summed the readings on the shop roof for both types of loads, wouldn't you be doubling the readings for any 240 volt devices?
12-30-2008, 09:41 PM #12
For a three phase system, at any given instant, the current in from all phases pretty much equals the current out, that is, there is no charge buildup. They can be 120 out of phase, which works if you have all three phase currents the same magnitude. Not really sure how this would affect the RMS readings using an amp clamp, but it seems to me that even if individual currents add to zero at any time, t, or
i1(t) + i2(t) + i3(t) = 0
it doesn't necessarily mean that the rms values of each current over time (which I think you are getting with an ampclamp). That is,
RMS(i1) + RMS(i2) + RMS(i3) doesn't have to equal 0.
Last edited by bosleyjr; 12-30-2008 at 09:53 PM. Reason: clarity
12-30-2008, 09:41 PM #13
two wires of a 3 phase service is ONE PHASE.
it is 180 deg from one wire to the other and no nonsense.
The phasing is only relevant when you have the third wire. It supplies current that creates the various 3 phase currents you can measure. But with two wires, and reversing voltage, the total ends up with a single phase differential voltage and a "common mode" voltage that drops out in any transformer.
The most common way to get your house voltage is phase to neutral, but it could also be obtained phase-to-phase. The transformer would be more expensive though, because the insulation would have to be better.
Don't worry about Semour-Dumore's post. Both parts are basically wrong.
The power company bills you at your house, for WATT_HOURS in the US. No VAR compensation can help you as things stand now at your house. That may change eventually, as the powerco discovers another revenue stream in terms of power factor penalties. First they have to get it past the public service commission, then deploy the meters.
At a business, you get lots of other charges, and VAR compensation may be helpful.
12-30-2008, 10:51 PM #14
The RMS value between L1 and L2 on a 3 phase system wil NEVER!!! be as high as the RMS value on a single phase (180deg) L1-L2 voltage.
Put a damm scope on it and see.
The sum of the currents in both legs will be the total current useage. Period.
You think it's a single transformer providing L1 and neutral?
While I dunno about yours, all I've seen is a center tapped transformer with the center tap used as the neutral, L1 and L2 is derived from the coil ends. That my friend is 180 out of phase, which provides X Volts between L1 and N, X Volts between L2 and N and 2X volts between L1 and L2.
On a 3 phase system L1 and L2 will only get you Sqrt3X volts.
Sure, a single phase load doesn't give a hoot about the phasing between L1 and L2, but the current on each leg will be 1/2 of the equivalent voltage L1-N arrangement. This is also why the neutral is not factored into the meter. Hence yes, you add the L1 L2 measurements together.
This one should explain why you add the currents:
12-30-2008, 11:14 PM #15
Indeed, all of my meters say KWh. I did not know that and assumed that they base their readings on Voltage x Current, meaning they are VArh meters.
Certainly is different than what I was thought.
Does that mean that meters installed on business customers are different than residential meters?
12-31-2008, 12:20 AM #16
The short answer is, yes residential and business services are metered DIFFERENTLY.
In most cases. In most cases residences are NOT charged for anything aside from
in-phase current. If you draw a large reactive current you are not charged for that.
Most business, and certainly those that use polyphase power, ARE charged for both
reactive current and real current. A further difference is that business typically
have a "peak demand" meter that charges a continous rate, proportional to the
largest peak current draw in some fixed time period, like 24 hours or so. Residences
rarely have that.
"If a device is running on 240 volts, would its current consumption not be read on a single leg? "
Yes. The current in each leg is the same. The wires have to carry that current. But the
power company effectively figures out how to charge according to the number of energy
units (say, joules or BTU if you wish) the service provides, and the loads dissipate.
You get charged the same if you run a 1000 watt heater from 120 volts, or if you
ran it off a 240 volt connection.
For the sake of this discussion, I've been fiddling things a bit, the fiddles are:
1) your electrical service is center-tapped 240 volts, with 120 volts between each
hot leg and the neutral.
2) your loads are purely resistive, not reactive. So your 'power factor' is close to one.
Some of the discussion cropping up right now, involves deviation from one or both
of those boundary conditions, that I have arbitrarilly chosen, to simplify the discussion.
12-31-2008, 12:25 AM #17
Certainly the 3 legs voltage is 120 degrees out of phase with each other.
When you use 2 of the 3 phases the load is going to see it as single phase and each leg 180 degrees out. The current is going in and out of each leg with a 180 degree phase difference. Think about it. The current has to be 180 out of phase with 2 legs. Add a 3rd leg- then the current has a path to travel and be 120 degrees out. Yes you are right that if you measure the voltage in each leg relative to each other they are 120 out.
I hope I make sense here.
12-31-2008, 12:40 AM #18
I was taught that 3 phase has each cycle peak 120 deg from each other.
How could you have 180 deg between two legs of a three phase system when they are 120 deg apart?
12-31-2008, 12:44 AM #19
Yes, I see where thge confusion came about.
I was trying to refer to Jim's post where he is describing the current readings on a single phase system with only L-N loads. That is where the differences are in the meter readings on the neutral.
Phase loads (L-N) show different Neutral current measurements when the supply is single phase L1-L2-N or true 3 phase L1-L2-L3-N.
In pure line loads ( L1-L2) the load sees 180.
12-31-2008, 12:50 AM #20
Simple. The load never sees the neutral, therefore it never references to it.
Now, if there is no neutral reference, all the load sees is the potential difference between L1 and L2. If you do a sum of the sinewave you'll get another sinewave with a Sqrt3 peak. That's all the load sees.
Basically you can picture it by sitting on one of the waves, ride it and observe where the other one is with respect to you at each moment.