kwired
Electron manager
- Location
- NE Nebraska
Europe
Appliance shopping involves a vacation as well, should make many very happy wives:happyyes:
100 watt light
- Thread starter domnic
- Start date
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Appliance shopping involves a vacation as well, should make many very happy wives:happyyes:
mbrooke
Batteries Included
- Location
- United States
- Occupation
- Technician
The MGN might only look good to dispense on new rural lines. Older 3 phase lines already have an MGN, all that would be needed would be a double bushing unit. In urban areas it would not be an issue even on new construction since an LV MGN or ground has to exist no matter the conditions in order for 120 and 277 volt services to work/be safe.
However changing whats norm is the hard part.
Europe
Appliance shopping involves a vacation as well, should make many very happy wives:happyyes:
:lol:
If one Googles 230 volt appliances dozens of sights come up that actually run and ship within the US... guess there is a demand? :blink: Don't ask how I know:lol:
Are you talking about faults due to lightning surges or about power frequency faults?
Because the tungsten bulb presents a non-linear resistance, the voltage drop can not be calculated readily with any formula as in post #6 but only be measured.
broadgage
Senior Member
- Location
- London, England
The 100 watt lamp connected as described in the first post will light but not at full brightnes.
The actual voltage on the lamp cant be readily calculated for at least three reasons
Firstly we dont know the actual resistance of the lamp filament, which varies according to the temperature.
Secondly although the ground rod resistance is given, the current flows through TWO ground rods, and we are not given the resistance of the other ground connections at the transformer or service entrance.
Thirdly, to be really accurate we need to know the resistance of the wires, and what the ACTUAL voltage is, not just the nominal voltage.
The actual voltage on the lamp cant be readily calculated for at least three reasons
Firstly we dont know the actual resistance of the lamp filament, which varies according to the temperature.
Secondly although the ground rod resistance is given, the current flows through TWO ground rods, and we are not given the resistance of the other ground connections at the transformer or service entrance.
Thirdly, to be really accurate we need to know the resistance of the wires, and what the ACTUAL voltage is, not just the nominal voltage.
kwired
Electron manager
- Location
- NE Nebraska
But if the grounded conductor is also bonded to a MGN (which is very likely in most of North America) there is many more than just two ground rods, you have a parallel path to the majority of ground rods and other grounding electrodes on the entire continent. That ultimately makes the bulk of the resistance of the circuit in rod #1.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
140207-0017 EST
Actually you can write an equation for a standard tungsten filament bulb that will reasonably accurately predict power vs voltage.
For a nominal 120 V 100 W bulb this is approximately P = V^1.6/21.22 . A test gave the following values:
0.832 A, 99.8 W, 100.0, 120 V
0.564 A, 33.8 W, 32.9, 60 V
0.386 A, 11.6 W, 10.9, 30 V
0.270 A, 4.1 W, 3.6, 15 V
The first column is measured current.
Second is calculated from measured current and measured voltage.
Third is calculated from the equation using the measured voltage.
The fourth is the measured voltage.
The equation is quite close for a range from 90 to 130 V for an experiment with a different bulb at a different time. In this case the error was less than 1/2 W at 90 V and 3/4 W at 130 V.
Obviously the equation can be written in terms of resistance rather than power if you want.
.
Actually you can write an equation for a standard tungsten filament bulb that will reasonably accurately predict power vs voltage.
For a nominal 120 V 100 W bulb this is approximately P = V^1.6/21.22 . A test gave the following values:
0.832 A, 99.8 W, 100.0, 120 V
0.564 A, 33.8 W, 32.9, 60 V
0.386 A, 11.6 W, 10.9, 30 V
0.270 A, 4.1 W, 3.6, 15 V
The first column is measured current.
Second is calculated from measured current and measured voltage.
Third is calculated from the equation using the measured voltage.
The fourth is the measured voltage.
The equation is quite close for a range from 90 to 130 V for an experiment with a different bulb at a different time. In this case the error was less than 1/2 W at 90 V and 3/4 W at 130 V.
Obviously the equation can be written in terms of resistance rather than power if you want.
.
K8MHZ
Senior Member
- Occupation
- Electrician
So...am I the only one waiting for somebody to say
"OK, I just ran outside and did this, and here's the amps, voltage, and brightness of the lamp I used..."
Well, we have almost 2 feet of snow and ice on the ground and it's 11 degrees F.
Get back to me in a couple months.
Well in Dallas we have 2 inches of powder snow and 20 degrees this morning. Never seen it get so cold down here, especially this late in the season. Yesterday morning it was 11 degrees. :rant:
Not all is bad though, suppose to be 50 tomorrow which is still cold for Dallas in winter.
Mayimbe
Senior Member
- Location
- Horsham, UK
Tesla use to play that experiment to impress kids and a guy called T. Edison.
And he usually did it over an ice layer of 1 m, as it is shown in the movie "the prestige" with the fella from batman
And he usually did it over an ice layer of 1 m, as it is shown in the movie "the prestige" with the fella from batman
kwired
Electron manager
- Location
- NE Nebraska
Actual temp doesn't seem to matter lately, wind chill does. I was outside yesterday afternoon and thought it actually felt warm, but there was no wind and temp was still only maybe 15F at the most. I hate it when 15 F seems warm. Wind chill right now is around -15 haven't been outside yet and am not really all that excited about going out.
Probably a computer might have created the approximate equation P = V^1.6/21.22 after having input data of V-I characteristics of various filament bulbs for an operating voltage range of 90 to 130 V and an algorithm to fit the power values within 1/2 W at 90 V and 3/4 W at 130 V of the actual power values.
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