Question about voltage...

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.

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

Default User schrieb:

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

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

| 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.

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.

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

Hi,

It was a Fluke 29.

Thanks,

Alan

Hi Everyone,

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

I appreciate it!

Alan