I need a restor that will reduce 10,000vdc to 12vdc?

Question

I am working on a progect and i need a resistor that will reduse the 10,000 vdc power from a dc generator to 12vdc and supply 200amps on a contenual bases

Answer 1

10000V - 12V = 9988V

R = 9988V / 200A = 49.94 ohms
P = 200A x 200A x 49.94 ohms = 1.9976 MegaWatts

Put this resistor in series with the 200A load and the output will be 12V.

Terribly inefficient way to do things. The load will be supplied 2400 watts. This is only 0.12% efficient!

This is not a realistic way to power things, but answers the question as asked. I suspect there is an error in what you are asking or it is simply a mathematical exercise. No one would ever use this technique in the real world. Do you realize how large a 2 megawatt generator is, not to mention fuel consumption !?

Answer 2

per David, you are talking Tesla numbers in total power.
Recently I had to specify power for a production line that was made in Europe, for that market. The equipment was used, and was all 220/50Hz and destined for a plant in the USA.
Though I could buy a Variable Frequency Drive for the 500 AMP service required, nobody makes a less-expensive "freak drive" latched for one frequency conversion. I would be taking the 50 Hz and expressing only 60 Hz, using a machine that can output on a huge range of frequencies. When you consider the number of systems that swap sides of the Atlantic, this seems most unusual, but there it is. The client bought the old standard motor/generator system.
Most switchmode power supplies that are configured for 10,000VDC are in the fractional or single digit amp power range at 200VDC.
Note: Don't step off the wooden stool!

Answer 3

A resistor alone is not going to do it for you.

Although Sad has shown how to get 10 volts (across the 1k resistor) in order to draw 200 amps through the circuit, power =I^2R for the 1 meg ohm => 1,000,000*40,000 Watts !!!

This is one reason no one uses high voltage DC. You need AC and a transformer OR a special switching power supply.

Answer 4

I think you wanted a dropping resistor in series to your load. Assuming that your load current is 200 A in 12 volts, then R load is 12/200 which is equal to 0.06 ohm.

Used the voltage divider method,

V2 = (R/(R+Rs))xV1

where V2 = 12 V
R = 0.06 ohm
V1 = 10,000 V
So, Rs = (RxV1/V2) - R
Rs = 0.06(10,000/12) - 0.06
Rs = 49.92 ohm

Power rating is, P = (I^2)R
P = 200^2 x 49.94
P = 1997600 Watts

You need resistor having 49.94 ohm with 1.9976 MWatts rating. This is very impossible.

Answer 5

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Answer 6

12 Volts / 200 Amps = 0.06 Ohms (60 milliOhms)

Your load is very small, and nominally would be part of the voltage divider, itself, but.....
That means that the main portion of the voltage divider (the one you speak of in your question) will need to drop approximately 10,000 volts at 200 amps, and that means it will have to dissipate 2 MegaWatts.

This is totally impractical -- not only would your generator have to supply that kind of power (which I seriously doubt it can), but it would be a total waste of that power (which I seriously doubt you can afford).

No... the practical solution is to get a switch-mode inverter, and it will probably have to be custom-made for the 10 kV input voltage.


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Answer 7

A resistor? Are you goofy? Let's see, 12V@200A = 2400W. You don't regulate a 2.4KW load with a resistor, you get a really big honking convertor of some sort, and believe me, it won't have just a couple of resistors inside.

Answer 8

While everyone (correctly) pointed out why a resistor would not work in this situation, no one proposed a solution. Probably the best approach here would be a motor-generator set.

Answer 9

you need 2 resistors in series with a ratio of 1000:1

eg 1k ohm and 1Meg ohm

Answer 10

You need two resister R1+R2/R2=10012/12