Electrical Engineering - Question about voltage...

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Question about voltage...

Default User

2008-03-26, 3:25 am

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

2008-03-26, 3:25 am

Default User wrote in message <47e9bea3$0$9837$c3e8da3@news.astraweb.com>...
>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
>
>
Here's a simple example that might help - you can jump start your car with
another 12v car battery, but if you try to jump start the car with a 12v
wall wart, the voltage output of the wal wart would drop to zero. Both are
12 volt sources but the difference between the two is that the wall wart has
a higher internal resistance. The current that the wall wart outputs has to
flow thru that internal resistance. If you are familiar with ohms law, it
states that voltage equals current times ohms. So a high current thru the
high resistance means that all of the wall warts voltage is being dropped
across it's own internal resistance. An automotive battery has a low
internal resistance so that it can output higher current. Usually you see
this sort of thing refered to as output impedance rather than internal
resistance. The output impedance of any voltage source determines how low
it's output voltage will drop when it has a load connected.

Salmon Egg

2008-03-26, 3:25 am

In article <47e9bea3$0$9837$c3e8da3@news.astraweb.com>,
"Default User" <nospam38925@forme.com> wrote:

> 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?

Just knowing this potential difference (voltage) is insufficient to know
wht current will be drawn through a short circuit. It can range from
virtually zero to destructive levels. Look up Thevenin's theorem you are
missing the internal impedance of Thevenin model.

Bill

Udo Piechottka

2008-03-26, 9:25 am

Default User schrieb:
>
> 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?

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

operator jay

2008-03-26, 9:25 am

"Default User" <nospam38925@forme.com> wrote in message
news:47e9bea3$0$9837$c3e8da3@news.astraweb.com...
> 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
>
>

Internal impedance of the source is important as well as having a
complete circuit. If you have two electrical systems, and one or both
of them is isolated and ungrounded, you can connect a wire between an
arbitrary point in the first system and an arbitrary point in the
second system, and experience little or no current flow.

The two arbitrary points may have originally been at different
potentials, and it may seem like putting a resistance (e.g. a very low
resistance 'short circuit') between them should cause a current
(I=V/R) to flow. However what would happen instead is that the two
arbitrary points would go to the same voltage, and the current through
the 'short' would be I=V/R=0/R=0. The voltages in one or both systems
may shift so that the voltage at both arbitrary points becomes the
same. If one system is grounded, then its voltages will likely not
change and the voltages in the isolated system will shift.

This is like the 'shift' in voltages you experienced with your
inverter and receptacle. The output of the inverter is isolated and
ungrounded. The voltages were originally 60vac to ground from hot and
60vac to ground from neutral. When you connected ground and neutral,
the voltage between them went to 0vac. The voltage from hot to ground
went to 120vac.

There are 'ungrounded delta' power systems in which a short circuit
from a hot to ground does not cause a large 'short circuit' current to
flow. There is no complete circuit. Instead, the voltages of the
system shift, similar to what is described above. Obviously, circuit
breakers will not trip on a single fault to ground when there is no
large current flow. One measures voltages between hots and ground in
such a system to detect a fault. 'Ground fault detection systems' do
this. Looking them up may give a better (clearer) explanation of what
I'm trying to say.

The above is a bit 'idealized'. There is probably normally SOME
current when shorting points, or shorting a phase to ground. Leakage
currents, capacitive and inductive couplings, etc. could cause that.
However the systems operate largely as described.

j

phil-news-nospam@ipal.net

2008-03-26, 9:25 am

On Tue, 25 Mar 2008 22:08:46 -0500 Default User <nospam38925@forme.com> wrote:

| 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.

--
|---------------------------------------/----------------------------------|
| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
| first name lower case at ipal.net / spamtrap-2008-03-26-0846@ipal.net |
|------------------------------------/-------------------------------------|

James Sweet

2008-03-26, 9:25 pm

Dave Martindale

2008-03-27, 3:25 am

phil-news-nospam@ipal.net writes:

>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.

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

Default User

2008-03-27, 3:25 am

Hi,

> 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.

It was a Fluke 29.

Thanks,

Alan

Default User

2008-03-27, 3:25 am

Hi Everyone,

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

I appreciate it!

Alan

Don Kelly

2008-03-27, 3:25 am

--

Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
"Default User" <nospam38925@forme.com> wrote in message
news:47e9bea3$0$9837$c3e8da3@news.astraweb.com...
> 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
-------------
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.
--

Don Kelly dhky@shawcross.ca
remove the X to answer

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