0
\$\begingroup\$

I've got huge three-phase transformer, primaries are connected in Y configuration, secondaries in Delta, then goes to 6 diode bridge rectifier. Output is about 200 amp 52 VDC.

In short, how many ohm resistor do I need to reduce current from 200 amp to 100 amp while dead-shorted ?

\$\endgroup\$
4
  • \$\begingroup\$ Are you sure a fuse or breaker wouldn't be better? With that kind of power dissipation, you might end up with a fuse regardless of what you intend. Why do you need to reduce the current? \$\endgroup\$
    – Adam Haun
    Jul 29, 2015 at 3:17
  • \$\begingroup\$ It's meant to be welder. 200 amps is way too much. It may be more I am not sure, but it's in this range. 3/16" electrode gets hot and melts quickly. Another transformer I got, has adjustable shunt. It same configuration just a bit bigger. Uses 85 mm2 welding cable. \$\endgroup\$
    – Wade J.
    Jul 29, 2015 at 4:26
  • \$\begingroup\$ Does the transformer have any taps? Set the primary for a higher voltage. Can you reduce the primary supply voltage? It's a transformer, you can change the applied primary voltage down to zero without a problem. You could even put an SCR bridge feeding the primary as a variable voltage source, essentially a big three phase light dimmer. \$\endgroup\$
    – R Drast
    Jul 29, 2015 at 10:45
  • \$\begingroup\$ No taps, unfortunately. I am wondering if SCR bridge would do I don't know if they work with transformers. \$\endgroup\$
    – Wade J.
    Jul 29, 2015 at 18:41

1 Answer 1

2
\$\begingroup\$

Simple Ohms law can answer this. R = V / I

52V / 100A = .52Ω

It would also explain why this would be a pretty bad idea.

P = VI

52V * 100A = 5200 W. 5.2 KILOWATTS. Of pure waste heat. And a few hundred dollars.

\$\endgroup\$
3
  • \$\begingroup\$ .52Ω hmm that's interesting. And it wouldn't be much waste if I use nichrome or tungster spring. Anyway thanks for the answer ! \$\endgroup\$
    – Wade J.
    Jul 29, 2015 at 2:13
  • \$\begingroup\$ @wade only if the wire can withstand that much power and heat. \$\endgroup\$
    – Passerby
    Jul 29, 2015 at 3:18
  • \$\begingroup\$ @Passerby That might work, but probably not. The chances that the secondary winding is exactly the right size to deliver 200A (Final, DC) at core saturation is about zero. More likely, the secondary resistance is much lower than V/I, and the transformer is rated for core-saturation. And it is wasteful, no matter what wire you use, it is pure heat. \$\endgroup\$
    – R Drast
    Jul 29, 2015 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.