Confusing ohm readings

[mivey]

I have seen this before as well.

If the light was on for a period of time and the element heated up to temperature then you could shut it off shortly and should be able to measure the resistance that would work for your theoretical calculations.

If the element was already up to temperature and you switched the bulb back on would there be any current transient since there would be no change in temperature?

If you could hold the temperature, it would certainly reduce the inrush. The remaining small inrush would be the charging current for the wire.

But it cools down quickly. If the filament were at temperature, it would be "on".

 

[JWCELECTRIC]

I had an emergency panel change over the weekend. FPE panel, FPE service (to be changed shortly.)

The old panel developed some sort of issue (that I never fully understood), basically it developed an internal short, blew a hole in the bussing and knocked out the utility transformer fuse.

As I was hooking up the old circuits in the new panel, I tested for continuity and found a home run that had 4 ohms. I did a little math and concluded that the circuit would be pulling 27A when I energized it. When I fired it up, it was only pulling 4A.

What gives?

What circuit was it on? Lighting, Heat, Motor load?

 

[gar]

090428-2049 EST

mull982:

See my inrush waveforms at:
http://www.beta-a2.com/EE-photos.html

All photos are accessed by vertical scrolling. Both tungsten filament and transformer inrush are shown. Note: the peak voltage of a 120 V sine wave is about 170 V and at 10 ohms this would be a peak inrush of 17 A. Also see how rapidly the filament temperature rises, and how much less the peak inrush is when when turn on is at a voltage zero crossing.

.

 

[George Stolz]

What circuit was it on? Lighting, Heat, Motor load?

Lighting in the hallway I was in, for starters. I didn't investigate much beyond that, the 4A reading eased my mind.

 

[mull982]

090428-2049 EST

mull982:

See my inrush waveforms at:
http://www.beta-a2.com/EE-photos.html

All photos are accessed by vertical scrolling. Both tungsten filament and transformer inrush are shown. Note: the peak voltage of a 120 V sine wave is about 170 V and at 10 ohms this would be a peak inrush of 17 A. Also see how rapidly the filament temperature rises, and how much less the peak inrush is when when turn on is at a voltage zero crossing.

.

These were very interesting plots to look at. It looks like the transient associated with the resistance changing in a light bulb only lasts for a couple of ms. I guess this shows that the filament temp rises and increases to full resistance in this couple of ms as well. Does thie filament temperature and resistance decrease as fast when power is removed from the element?

Very interesting to see how the turn on point of the voltage waveform affectes the current transient magnitude. What is the explanation for this?

With the transformers, I couldn't figure out why the initial transient is not a symmetrical current waveform until steady state is reached. I understand the concept of the residual magnetism being present, but dont see how this would cause this unsymmetrical waveform. Is there a DC component of some sort that causes this?

 

[jghrist]

I have seen this before as well.

If the light was on for a period of time and the element heated up to temperature then you could shut it off shortly and should be able to measure the resistance that would work for your theoretical calculations.

If the element was already up to temperature and you switched the bulb back on would there be any current transient since there would be no change in temperature?

I don't know what kind of light bulbs you use, but mine go dark (cold) pretty quick after I turn the switch off. Too quick to get an ohmmeter on.

 

[gar]

090430-2003 EST

mull982:

I guess this shows that the filament temp rises and increases to full resistance in this couple of ms as well.

Quite fast. In the Henry Ford Museum is a very large incandescent bulb on display. I do not remember the wattage. Maybe 10,000 or larger. This would have a much longer turn on time constant. On an automotive head light I can get useful modulation at 20 Hz. A small pilot light would be faster.

Does thie filament temperature and resistance decrease as fast when power is removed from the element?

No. When you turn on the lamp you are putting power in very rapidly thru the wires into the bulb. When you turn the lamp off all the stored energy must be lost thru radiation. Thus, if you apply full voltage excitation to the lamp as a step function (from 0 to full voltage in near zero time), then it heats faster than when you turn it off.

Very interesting to see how the turn on point of the voltage waveform affectes the current transient magnitude. What is the explanation for this?

When you turn on at the voltage peak the voltage goes from 0 to 170 V and maximum current occurs and maximum instantaneous power. When you turn on at a zero crossing there is no voltage and no current. The voltage builds gradually and the filament is partially heated by the time you reach the peak voltage and thus resistance is greater than room temperature and therefore a lower peak current.

With the transformers, I couldn't figure out why the initial transient is not a symmetrical current waveform until steady state is reached. I understand the concept of the residual magnetism being present, but dont see how this would cause this unsymmetrical waveform.

This is because the magnetic flux in the magnetic core has a non-zero initial value. The ferromagnetic core has a hysteresis curve and in this case of a large inrush current the voltage polarity is such as to force the core more into saturation and that takes a lot of current. Following this the subsequent cycles follow a gradually changing unbalanced hysteresis curve until it works toward a balanced equilibrium curve.

Is there a DC component of some sort that causes this?

Yes there is a resulting DC component. It is not the cause. There is a gradually diminishing DC component until steady state is reached.

.

 

[mull982]

090430-2003 EST

mull982:

When you turn on at the voltage peak the voltage goes from 0 to 170 V and maximum current occurs and maximum instantaneous power. When you turn on at a zero crossing there is no voltage and no current. The voltage builds gradually and the filament is partially heated by the time you reach the peak voltage and thus resistance is greater than room temperature and therefore a lower peak current.

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Gar

Thanks alot for the great responses to my questions.

As far as the turn on voltage explanation above, I guess this could be represened for all turn on transients that can be represened by dV/dT. The higher on the voltage waveform the faster the voltage will be changing from turn on and a faster dV/dt will occur.

One example I can think of is the initial current through a capacitor effected by turn on voltage.

 

[gar]

090501-0754 EST

mull982:

dv/dt is not the cause.

Consider a series circuit of a battery, switch, resistor, and inductor. Assume the initial condition of no current in the inductor. Close the switch. An instantaneous voltage rise occurs to the RL series circuit, and to the inductor, but no instantaneous change in current because current can not instantaneously change in an inductor.

.

 

[mivey]

but no instantaneous change in current.

For a normal circuit, there will be some initial charging current for the conductor.

The formula we use in distribution systems is:
Vl-l / sqrt(3) / Xc

I've never calculated how much for a small circuit, but for a 3-phase, 25kV distribution line we get:
4/0 ACSR: 0.0913 amps/phase/mile
1/0 ACSR: 0.0858 amps/phase/mile
1/0 CU: 0.0845 amps/phase/mile
#6 CU: 0.0740 amps/phase/mile

So the same could be calculated for a building cable:
The charging current is:
2*pi*f*C*Vln * 10^-12
where:
f = frequency in hertz
C = cable capacitance in picofarads per ft
Vln = Voltage line-to-neutral

I could not find a table of cable capacitance value so I picked some reasonable values. The cable testing guys probably have some better data as they have to deal with this when testing cables. To avoid high AC charging currents, they will use DC tests. When using DC tests, they have to deal with the total leakage current which is made up of leakage current, charging current, and absorption current.

Anyway,
using 120 volts, 7 pf/ft, and a 100 ft run, you get 3.17 mA of charging current.

Using 120 volts, 50 pf/ft, and a 100 ft run, you get 22.6 mA of charging current.


For DC, the charging current in uA/kft is given by:
V/R*2.718281^(−t/RC)
where:
V = Voltage line-to-ground
R=dc resistance of cable in megohms for 1000 feet
t=time in seconds
C=capacitance of circuit in microfarads per 1000 feet

 

[gar]

090501-1157 EST

40 year old Romex is about 20 pfd/ft between the current conducting wires.

The inrush current to a capacitor with zero initial charge is Vsource at the instant of switch closure divided by Rsource.

As the source resistance approaches zero the current approaches infinity. But one can not create a real world circuit like this. If nothing else there is inductance in the wiring.

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