Power resistor 15 Ohm/50W

minhtam

Joined Aug 28, 2023
7
Hi everyone,

I have a question about power resistor and I'd like to hear your opinion.
I have a pre-magnetize circuit for a power transformer (60kVA) and they use 2 resistors 15 Ohm 50W. The voltage here is 440Vac.
As far as I understand, the current flow those power resistors should not exceed 1.82 A. But if you apply 440Vac to them, the current is now 50A flowing to resistors. The process lasts 0.7 s and after that the contact K1 will be opened and the pre-magnetize is finished.
You can see the figure below. My question is that my calculations is correct and how to understand correctly the power resistors rated 15 Ohm/ 50W?

AnalogKid

Joined Aug 1, 2013
11,022
Not all 15 ohm resistors are created equal.

The 50 W rating is for continuous dissipation, not a short transient current surge. In your situation, a more important parameter is the resistor's fusing capability, the peak current it can handle before it blows like a fuse. This has to do with the resistor's internal construction.

ak

BobTPH

Joined Jun 5, 2013
8,922
That is the current that would have to flow in a 15 Ohm resistor to make it dissipate 50W. The resistor does not ensure that, the circuit must do so. If you simply put 440V across a 15 Ohm resistor, Ohm’s law says 29A flows and the resistor burns up.

crutschow

Joined Mar 14, 2008
34,386
You need to know the internal thermal time-constant of the resistors you are using to see how long they can tolerate the maximum current before the internal resistor temperature reaches its maximum rated value.

Last edited:

AnalogKid

Joined Aug 1, 2013
11,022
If you simply put 440V across a 15 Ohm resistor, Ohm’s law says 29A flows and the resistor burns up.
Not in 0.7 s. Not great design, but I've seen it before. And I've done a less extreme version of the same thing in a 2 kW active power factor corrector.

ak

BobTPH

Joined Jun 5, 2013
8,922
Not in 0.7 s. Not great design, but I've seen it before. And I've done a less extreme version of the same thing in a 2 kW active power factor corrector.

ak
But my point was that you cannot calculate the current from the power rating of the resistor. He expected the current to be limited to that which would result in 50W dissipation. That is where the 1.82A came from.

Edited to add: OK, maybe I misinterpreted the question, I thought he was asking why 50A flowed when he calculated 1.82A from the resistance and power rating. You interpreted it as asking why the 50A current did not blow up the resistor. Reading and re-reading the OP, it is still not clear to me what he is asking.

Last edited:

minhtam

Joined Aug 28, 2023
7
But my point was that you cannot calculate the current from the power rating of the resistor. He expected the current to be limited to that which would result in 50W dissipation. That is where the 1.82A came from.

Edited to add: OK, maybe I misinterpreted the question, I thought he was asking why 50A flowed when he calculated 1.82A from the resistance and power rating. You interpreted it as asking why the 50A current did not blow up the resistor. Reading and re-reading the OP, it is still not clear to me what he is asking.
Hi ak,

I calculated 1.82 A from the resistor ratings. I meant that it is the current flowing the resistor should not greater than 1.82 A. (I used the formula I = sqrt of (P/R).
In the circuit you can see the voltage applied to the resistors is 440Vac. I used the formula for 3 phase for the current I= (V * sqrt of 3)/R and the current is 50A.

Janis59

Joined Aug 21, 2017
1,846
I would put there the incandescent lamps. 440/15=29 Amps. Exist 220 V lamps. What demands two in series 220*29=6450 W each. Hmmm, indeed problematic. Okay, then shift to TEH, two 220 V, 6 kW heaters in series. From teapot 2.2 kW three pieces parallel.

No, thats all not right. Firstly, let explore isnt it possible to feed the contactor via diodes or via third source atc let the grid consumption have no need to be sustained. Secondly, if it isnt possible, then calculate the voltage between those lines and max time, for what resistors will sustain the second grid. Then calculate the Watt*seconds (Joules) and approximate that any resistor before burning off "waits" about 3 seconds. Thus the max Joules are Wattage divide to 3. Or maybe 10, but sure not 100.

Last edited:

jiggermole

Joined Jul 29, 2016
161
duty cycle probably. As stated earlier not all resistors are created equal. And in applications where you are operating close to the physical limits of the device to reduce cost you'd probably consult the manufacturer of the resistors. They can give you the engineering limits of the device. It takes time for things to heat up and cool down so my guess would be that if you calculate the power dumped into the resistor over that time period its under the thermal limit of the device. Run it for more than 0.7s and you will definitely burn it up, but as long as there are safeguards to ensure that doesn't happen, it stays under that thermal limit. So the limit is no longer the 50W its the heat buildup in the resistor.
And assuming I am reading the question as you intended to write it.