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I have a 1k ohm rheostat that has a 25W power rating, and a current rating of 0.15 amps.

I got it because I was running a 12v/24v circuit at 1amp draw. I just looked at the power rating of 25W and thought it would be OK in my circuit. However, that's a low current rating (which I did not notice till just now). I have not used the rheostat yet, but I am guessing it will not handle the 1amp current even though it would be within the power rating.

Please confirm. Thank you.

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The rheostat current rating is 0.15 amps, period. You need to draw 1 amp and that's far above its rating. So you bought the wrong rheostat.

The rheostat wattage is not its most important spec, but the current rating. The current rating is what limits how much power the rheostat will be able to dissipate, for any arbitrarily low resistance setting. Note that this is precisely what you would expect from a high-power load: the ability to cope with a high current draw.

Guidelines for selecting a rheostat:

  1. Select those rheostats with a current rating equal to or higher than the current that you need to draw. I would recommend a derating of at least 20-30%.

  2. Once the previous requirement has been verified, then you can proceed to select a rheostat that can dissipate the power you need. This is the same as selecting a rheostat that can provide the resistance needed by your load to draw the required current at your working voltage.

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The power rating for a rheostat applies only when the entire resistance is in circuit.

The heating only occurs in the bit of wire carrying the current. If only half, or 10% of the wire is in circuit, then cooling only takes place on that half or 10%, and the power rating is proportionally lower.

This variable area for heat loss is taken account of by giving the rheostat a current rating. At 150mA, \$I^2R\$ gives 22.5 watts (more or less 25!) for the entire 1k\$\Omega\$, but only 2.25 watts for 10% of the resistance and 10% of the heat loss area.

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Your Rheostat will die young if connected .The power rating of 25W is an overall statement about total heating .Yes it can do 25W without getting too hot when it is at or near its max resistance .But the current rating is really a function of wire diameter .Your rheostat is 1 K so the wire will be thin .A lower resistance rheostat will have a higher current rating even if the power rating is still 25Watts.

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All of the answers above are over simplified and provide incorrect guidance. The power rating of a rheostat is actually more important than the current rating. The power rating is the amount of power that can be dissipated without producing destructive temperature. It is determined not only by the conductive windings but also the enclosure size and material. Heat can be dissipated by those windings not carrying a current and those that are. Of course the non-current-carrying windings will be at a lower temperature than those are current carrying. The manufacturer of a rheostat should specify a power rating that is suitable for all settings of the rheostat.

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  • \$\begingroup\$ No, Bob, the current rating determines the power per unit length of the resistance element and that's what's important. Yes, the construction matters, of course. I would expect relatively little heat to be conducted along the non current-carrying section and the temperature would rapidly fall off as distance from the wiper increased. "The manufacturer of a rheostat should specify a power rating that is suitable for all settings of the rheostat." There isn't one. It will vary with the setting, hence we calculate the maximum current. Welcome to EE.SE. \$\endgroup\$ – Transistor Aug 9 '18 at 6:36
  • \$\begingroup\$ I guess looking at it your way we both are right. The maximum suitable power Pmax can be calculated from the maximum suitable current Imax by Pmax = Imax^2*Rmax where Rmax is that of the full winding. That is, the manufacturer could equivalently specify either Imax or Pmax. (you need to consider Rmax because it is a possible setting of the rheostat). \$\endgroup\$ – Bob Roesser Aug 10 '18 at 17:45
  • \$\begingroup\$ It's not clear what you mean by "suitable" current or power. In any case, another way of looking at it is, "the maximum power you can dissipate is the rated power by % of full resistance." Note that saying "All of the answers above ..." doesn't work on this site as answers are sorted by votes or user preferences. \$\endgroup\$ – Transistor Aug 10 '18 at 18:03

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