full-wave diode bridge calculation.

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minerman

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Columbus, Ohio, United States
On the image below. If V2 = 40kV, and you place a 10-ohm resistor between the opposite side of the diode bridge and GND, how do you determine the voltage across that 10-ohm resistor? I figured that it would be -40kV, but it does not seem that way.

Thanks!




Image89.gif
 
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LMAO

Senior Member
Location
Texas
minerman said:
On the image below. If V2 = 40kV, and you place a 10-ohm resistor between the opposite side of the diode bridge and GND, how do you determine the voltage across that 10-ohm resistor? I figured that it would be -40kV, but it does not seem that way.

Thanks!




View attachment 6218
Click to expand...

isn't V2 the voltage across the resistor?? if not, then what is it?
 
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hurk27

Senior Member
Location
Portage, Indiana NEC: 2008
If you are replaceing the direct connection to ground with a high resistance to ground you still will not have any current flowing, a transformer is isolating, and there is no referance to ground until you make it, except capacitive coupling, much like a high inpedance grounded service, only after another fault has gone to ground will you see any current on the resistor, so is your question more of finding the correct vqlue of the resistor, that would bleed off capicitive voltage, but withstand a fault to ground on the positive side of the supply .5 amps of bled off current at 40kvdc would give you 20k ohm resistor, that would have to disapate 20kw in power.
 
Besoeker

Besoeker

Senior Member
Location
UK
minerman said:
Actually I found an answer to my question. Sorry about being unclear.
Click to expand...
Indeterminate since just a single point is connected to ground.
But it would hold the negative end of the bridge at or about ground potential.

Some sort of precipitator circuit?
 
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