Can someone help me with this 3 phase WYE calculation? - voltage divider.pdf (0/1)

Anyone want to take a stab at this one?
This is basically a voltage divide array of identical 2% tolerance
resistors that are arranged in a WYE configuration with 4 series
conected 300Kohm resistors in each leg.
I connected the input to a 3 phase 4 wire supply of 213V - 215V
along points a-a-a in any combination I measure 153V-155V
along points b-b-b (center) in any combination I measure 100V- 101V
along points c-c-c in any combination I measure 51V
across each resistor there is 29V-30V dropped
from any input line to neutral I measure 123V to 124V
A little high, this is from a 120V/208V 3 phase 4 wire system.
I would expect that since point b-b-b is exactly half way between
line (average 214V) and neutral (average 0V) I would measure line/2
or 107 volts phase to phase here.
I dont however, I measure 100V which confuses me. I am really rusty on
my 3 phase calculations and hope someone can shed some light on this.
I am trying to create a voltage divider that will give me exactly half
the line voltage.
Thanks.
Frank White
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Frank White
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Some puzzles do indeed attract my attention. The puzzle here is to figure out what the puzzle is. Do you have a schematic?
Bill -- Fermez le Bush--about two years to go.
Reply to
Salmon Egg
stab at this one?
Leave L-L values out of it. Measure one L-N voltage, then measure the divider voltages to neutral on the same phase. Come back with those numbers. Also, what type of meter are you using?
Ben Miller
Reply to
Ben Miller
I assume aaa is 1/4 the way from line to neutral, etc. but this is not specified. However you do not mention the voltmeter that you used. What is the impedance of the meter? If it is not very large with respect to the resistors, it can affect the readings. My estimate (after too much wine) is that the meter may be about 2.5MOhm (based on ignoring the meter and finding the Thevenin model of each side and then connecting the meter between). Secondly, with each resistor +/-2% the readings can be +/-8%
Reply to
Don Kelly
I did not measure L-N voltages. I did however measure the drop across each resistor at 30 - 31 volts. I made the assumption that the voltage ratings would be cumulative along the string towards N. When I add up the four 30V drops, this equals 120V when the actual voltage line to N is closer to 124V. I don't know how accurate my meter is at different voltages, but maybe this is a case of meter error. It is not a true RMS meter, but these are relative measurements in a purely resistive circuit.
The way I see it , I should have 3/4 x V-line across A-A-A and 1/2 x V-line across B-B-B and 1/4 x V-line across C-C-C
Is this assumption accurate or am I missing something. I can try some more measurements from line to N along the resistors, but I wont have my RMS meter back until after next week. Maybe this will measure differently, but this is more of a mathmatical question than anything. I have attached a schematic of the array
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Reply to
Frank White
Meter error is part of it, but meter input impedance is the other part. If it is a digital meter, then it is probably 10 MegOhm, and would have a minor effect. However, an analog meter at 3000 ohm/.volt on a 250 volt range would be 750 kOhm, which would have a huge effect on your divider ratios. Yours may be somewhere in between. If you can give us the actual AC sensitivity and range for your meter, we could tell you for sure.
You are playing with something connected to a three-phase high energy system, that could create a catastrophic failure, including arc flash, if something goes wrong. I will again recommend, as I did in your other thread, that you consider potential transformers instead. They provide accurate ratio for a wide range of instrument impedance, and they isolate your instrumentation from the power system which enhances safety.
Ben Miller
Reply to
Ben Miller
Input impedence is 10Mohm, but the sensitivity and range are not listed. Resolution at this scale is 0.1 mv.
The resistors are 300Meg so I suspect that introducing 10Meg to the circuit might cause some error as well. Where I expect to read 107V , I read 100V. Fo this I am using proper fused leads for safety. I have not shopped for CT's but suspect they are expensive and definitely bulky.
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Frank White

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