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Question about voltage...

Discussion in 'Electrical Engineering' started by Default User, Mar 26, 2008.

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  1. Default User

    Default User Guest

    Hi,

    I've got a question and I hope I explain it right. Lets say you measure the
    voltage between two points, and you have a value, which is the difference
    between the two points. Lets say this value is 100 volts ac.

    Does this tell you anything about the ability of those two points to deliver
    current across them? For example, is there any way to know if shorting
    those two points would yield massive current, or virtually no current?

    I've got a couple examples. A Honda eu2000 inverter generator has an
    inverter that produces 120vac. It does not tie its hot or common to the
    ground in the plug by default (not bonded). If you measure between the hot
    and common you will get 120vac. If you measure between the hot or common
    and ground you will get 60vac. But, if you connect a wire from the hot or
    common to ground very little current will flow just a few ma. Not all
    inverters would support this type of bonding and it will destroy many MSW
    based ones. But, my question is, it read 60vac, but in reality it might
    have well been 0vac.

    I was working on a car yesterday, and it really didn't make any sense to me.
    I was working on a solenoid valve that had 2 wires going to it. Wire #1
    when I measure resistance to chassis ground is grounded, but when I switch
    to volts I get 11.5 volts. This was a test without the engine running,
    battery actually had around 12.7 volts at the time. Does this make any
    sense? How can it be grounded and yet have 11.5 volts between it and
    ground? Is this the same sort of issue as the generator example?

    Is there a name for voltages like this that read something, but are
    misleading?

    Thanks,

    Alan
    www.sadevelopment.com
     
  2. You have to find out the inner resistance of the voltage source. This
    can be accomplished in measuring the voltage between both points with a
    high imdedance voltmeter.
    While driving a known load (resistance with known value) and measuring
    the reduced voltage you can derive the inner resistance of the source
    and which maximum current it could deliver.

    Example: In the idle state (no load) you measure 120 volts. connecting a
    load with 50 ohms to the load reduces the voltage to 100 volts. the
    current through the resistor is i=100v/50 ohms = 2amps.
    these 2 amps lead to an inner loss of 20 volts, so the inner resistance
    is r=20volts / 2 amps = 10 ohms.
    The short circuit current would be 120 volts / 10 ohms = 12 amps.

    This only applies to typical DC current sources. A more complex inner
    resistance and waveform of the generator would lead to completely
    different results. The leakage current you measured is a result of
    alternating current and the capacitance which acts as an AC resistance
    between the generators coil and ground.

    - Udo
     
  3. Guest

    | I've got a question and I hope I explain it right. Lets say you measure the
    | voltage between two points, and you have a value, which is the difference
    | between the two points. Lets say this value is 100 volts ac.
    |
    | Does this tell you anything about the ability of those two points to deliver
    | current across them? For example, is there any way to know if shorting
    | those two points would yield massive current, or virtually no current?

    It is possible for a test device to determine the impedance of the source
    system fairly accurately. This is done by measuring the current going
    through a known resistance for 2 or more different resistances. The
    current will be greater through the lower resistance. You can then figure
    out what the additional unknown resistance is from these two currents and
    two know resistances in each test. You do not need to know the voltage
    of the source (you'll actually get that as part of the tests).


    | I've got a couple examples. A Honda eu2000 inverter generator has an
    | inverter that produces 120vac. It does not tie its hot or common to the
    | ground in the plug by default (not bonded). If you measure between the hot
    | and common you will get 120vac. If you measure between the hot or common
    | and ground you will get 60vac. But, if you connect a wire from the hot or
    | common to ground very little current will flow just a few ma. Not all
    | inverters would support this type of bonding and it will destroy many MSW
    | based ones. But, my question is, it read 60vac, but in reality it might
    | have well been 0vac.

    There is a very high impedance in the circuit when you get 60 volts. That
    impedance is the capacitive coupling between the wires.


    | I was working on a car yesterday, and it really didn't make any sense to me.
    | I was working on a solenoid valve that had 2 wires going to it. Wire #1
    | when I measure resistance to chassis ground is grounded, but when I switch
    | to volts I get 11.5 volts. This was a test without the engine running,
    | battery actually had around 12.7 volts at the time. Does this make any
    | sense? How can it be grounded and yet have 11.5 volts between it and
    | ground? Is this the same sort of issue as the generator example?

    The exact details of this are unclear for your case. But this can happen
    even with DC, when the meter involved is a very high impedance meter such
    as today's digital ones.


    | Is there a name for voltages like this that read something, but are
    | misleading?

    Phantom voltage.

    Use an older style volt meter that uses a resistor which draws more current.
    Compare the readings from both this meter and a digital one.

    Maybe someone makes a digital voltmeter that includes a current flow test
    to verify the true voltage or even the circuit impedance.
     
  4. James Sweet

    James Sweet Guest


    It tells you virtually nothing on its own. You need a load, in this case a
    low wattage light bulb of sufficient voltage rating to handle what you
    measure. If it lights up, then you have at least that much current
    available.
     
  5. I have a consumer-grade DMM that has both the usual high-impedance DC
    volts scales and a set of "battery test" settings as well. In battery
    test mode there's a load resistor across the test leads to draw some current.

    The meter has 4 "ranges" for batteries, labelled 1.5 V, 6 V, 9 V, and 12
    V. The meter circuitry is always working in the 20 V range for any of
    these (it doesn't switch to 2 V full scale even for the nominal 1.5 V
    setting), but the load resistors vary.

    The manual doesn't document what the load resistors actually are, and
    you can't use them on any range other than 20 VDC, but it's a nod in the
    right direction.

    (Meter brand is "Equus", bought at Canadian Tire).

    Dave
     
  6. Default User

    Default User Guest

    Hi,
    It was a Fluke 29.

    Thanks,

    Alan
     
  7. Default User

    Default User Guest

    Hi Everyone,

    Thanks for all the info everyone, it gives me a ton to figure out!

    I appreciate it!

    Alan
     
  8. Don Kelly

    Don Kelly Guest

    --

    Don Kelly
    remove the X to answer
    ----------------------------
    -------------
    a)No- the current depends on the load resistance plus the internal
    resistance of the source. The maximum would be determined by the open
    circuit voltage divided by the internal resistance- don't try to measure
    this resistance with an ohmmeter!

    b)In the ungrounded Honda generator there is also capacitive coupling to
    ground- and the natural capacitive divider can result in this voltage.
    Since the impedances are large, the current that can be drawn is small.
    Insulation leakage can do the same.

    c)You are apparently trying to measure resistance, using an ohmmeter, in an
    energised device. The ohmmeter is in parallel with the device and the
    battery source is bucking or boosting the little battery in your meter so
    that the meter current has nothing to do with the resistance of the device.
    An example is a meter which has an internal voltage of 3V and an internal
    resistance of 1Kohm. then the meter will read 0 for a current of 3ma. If
    the meter is connected to an isolated 1k load, it will have a current of 1.5
    ma and the ohmmeter scale is calibrated to indicate this. Note that in an
    analog ohmmeter the ohms scale is reversed compared to the normal voltage or
    current scale so that infinite resistance corresponds to 0 current or full
    voltage.
    Now when connected to a device supplied by a 12V battery, the effective
    voltage will be either 9V or 15V leading to a current of 9 or 15ma trying to
    drive it offscale below the 0 resitance mark. Not good for the meter and the
    readings will be meaningless. So- never use the ohmmeter range to measure
    the resistance of an energised device. If you can measure current and
    voltage- do that. Sice many devices draw a current that is too high for the
    meter and for other reasons it is better to put a small known resistance (R)
    in series with the device, measure the voltage across this resistance (Vr)
    and that across the device and calculate Rdevice =Vdevice/Vr.
     
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