Measuring Wattage with Clamp Meter

Discussion in 'General Electronics Chat' started by seagull369, Jan 23, 2016.

  1. seagull369

    Thread Starter New Member

    Jan 23, 2016
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    Hi, everyone. Just signed up to the forum.

    I recently purchased a true sine power inverter (converts ~12VDC to ~115VAC) and would like to measure its wattage output whilst something is plugged into it. I know kill-a-watt type meters are good for doing this, but I don't have one. I do, however, an AC clamp meter. Can I get the amp reading from the meter then multiply it by the line voltage to get total the power? If so, should I use the voltage value when the inverter is under the load in question or in no load? I ask 'cause the values do tend to differ some.
     
  2. #12

    Expert

    Nov 30, 2010
    16,354
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    Measure under load. P=IE
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Welcome to the forums.

    The clamp on meter is fine for current. I assume you have some other meter for voltage.

    If you want to measure power you need to take both measurements under the same conditions. It makes no sense to measure voltage without a load but the current with a load.

    That measurement is actually called "apparent power", as the true power may be less for certain loads. A load such as a heater or incandescent light bulbs give you a better reading.

    (Edit to fix iPad typos)
     
    Last edited: Jan 24, 2016
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Note that the clamp on meter almost certainly is measuring the average current and converting it to the RMS value of a sine wave of equal value. This will not be correct for loads that draw a distorted (non-sinusoidal) current such as a non-power-factor-corrected power supply.

    ak
     
  5. RichardO

    Well-Known Member

    May 4, 2013
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    Using an amp meter and voltmeter to measure only works with non-reactive loads. This is because a reactive loads such as capacitance or inductance have the voltage and current out of phase. There is no way for the separate amp and volt meters to know the phase difference of the current and voltage being measured.

    A purely resistive load does not have a phase shift so that is the best load for your tests. An incandescent lamp is non-linear for voltage verses current but this non-linearity is probably small enough to be OK in your case.
     
  6. Mr.Meter

    New Member

    Jan 24, 2016
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    You will need a clamp on meter specifically designed to measure AC power. To do so, you would have the clamp on the conductor, and the voltage probes connected to line (+) and neutral (-) simultaneously.

    If you just measure the voltage and current and multiply the two, the product will be the VA which is total power.

    In order to measure W (useful power), you need to know the phase difference/angle between the voltage and current waveform. For an inductive load the current always lags the voltage. For a capacitive load, the current will lead the voltage. Most loads you will encounter are inductive.

    As an example, if you knew the current waveform was lagging the voltage waveform by 25 degrees (inductive) you could calculate the Power factor and get W:

    Variables:
    PF = Power Factor
    Phase Angle = θ

    Equation:
    θ = cos -1 (PF)
    PF = cos (θ)
    PF = cos (25)
    PF = 0.90

    Say you measured 120Vac and 10A. You would simply multiply the two to get VA:

    VA = 120Vac * 10A
    VA = 1200
    VA = 1.2 kVA

    Then you would multiply the phase angle by VA to get W:

    W = PF * kVA
    W = 0.9 * 1.2kW
    W = 1.08kW

    Long story short - You can only really measure VA using the tools you have right now. However, if you have a CT (current transformer), and an oscilloscope you can figure out the phase angle quite easily.
     
    Last edited: Jan 24, 2016
  7. seagull369

    Thread Starter New Member

    Jan 23, 2016
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    Thank you for the great replies, everyone.

    To answer your question, Ernie, yes, I was planning on using a separate RMS meter to measure the voltage.

    Is it fair to say if I ran a reactive load, the wattage value I obtained (via my proposed setup) would be an overestimation of power that was actually drawn?

    Do devices like the kill-a-watt meters do a better job at measuring the wattage drawn by reactive loads? I read a review online from a person who said the results it gives for those are NOT accurate.
     
  8. Mr.Meter

    New Member

    Jan 24, 2016
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    1
    Is it fair to say if I ran a reactive load, the wattage value I obtained (via my proposed setup) would be an overestimation of power that was actually drawn?

    - That is correct for an inductive load. If your load is mainly resistive, the VA reading will actually be quite close to the W value because the power factor will be close to 1. I design and inspect electrical metering systems for my work, and the lowest phase angle that I have ever seen in a commercial setting was about 5-6 degrees. That was on a service that was completely lighting loads with short conductor runs.

    Do devices like the kill-a-watt meters do a better job at measuring the wattage drawn by reactive loads? I read a review online from a person who said the results it gives for those are NOT accurate.

    - You need a meter rated to measure AC power (kW, KVA, kVAR, Power Factor) to accurately measure reactive loads. If the meter is rated to those standards, it should be able to measure any AC load accurately no matter what the reactance. The meter I use as my reference to do inspections is the Amprobe-330. Never had an issue with it.

    http://www.amprobe.com/Amprobe/usen/Clamp-Meters/amp-200-300-series/AMP-330.htm?PID=79189
     
  9. RichardO

    Well-Known Member

    May 4, 2013
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    Yes, since maximum power is drawn when voltage and current are in phase. (That is the assumption made in your measurement.)

    I would expect that the Kill-a-Watt or other similar power meters to be fairly accurate with reactive loads. Just keep in mind that inexpensive meters such as the Kill-a-Watt are not laboratory instruments. They do give reasonable results for their cost.
     
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