equations for figuring resistor power rating

[texwest]

I'm wondering what are the correct equations for figuring the power rating given a certain voltage, resistance and current.

I put a 500r 5 wt from the rectifier to the first filter cap in a silverface champ to lower the voltage.

The plate voltage is now 355,  but the original voltage coming off the rectifier was 415v.

The ma on the power tube is 42ma now.

I'd love to see your equations!

[sluckey]

All you need...

E = I x R
P = E x I

... well you do need to be able to shuffle and substitute.

[texwest]

OK, how do you use those to determine if the resistor you are using is a high enough rating?

Lets say I use 415v .  I will use .05 amps to round up from the 42 ma and include the preamp tube current.

415 x .05 = 20.75 wts

I don't think that is the final solution  as people often use 5 or 10 watt resistors in this position,  there must be something else to it.

[d95err]

You use the voltage across the resistor, not the voltage from B+ to ground.

(415-355)*0.042 = 2.52W

[Williamblake]

P=I*U being a substitution for the formulas already mentioned.

[sluckey]

You use the voltage across the resistor, not the voltage from B+ to ground.

(415-355)*0.042 = 2.52W

Yes. That gives you the actual wattage dissipated by the resistor. Then as a safety factor, double the actual wattage and buy the next larger wattage that is commonly available. In this example, the actual wattage is 2.5W. Double that to 5W. Buy a 5W resistor.

[texwest]

OK that makes sense to me now.  The voltage drop multiplied by the current gives you the amount of power being dissipated by the resistor.

And some of the voltage drop at the plate is being caused by the OT resistance.  I can't remember the voltage at the first node but I think it was close to 370.

So we are really looking at 415 -370 = 45v drop  X .05ma to include some preamp tube current.  2.25 watts dissipated.  Looks like a 5 watt resistor is plenty for this application.


Awesome!  Slowly learning more.  Thanks a bunch. 

[sluckey]

So we are really looking at 415 -370 = 45v drop.

There's an easier way to measure voltage drop across a resistor. Simply put one meter probe on one side of the resistor and put the other meter probe on the other side of the resistor. Wala! No math needed.

[AmberB]

Of course, if you want to drop the voltage a certain amount and you want to know what resistance you need to do that, so you can buy the proper resistor, or closest to it, then you need the math.
Some days, the math tends to make my head hurt...

[shooter]

Knot doing math can be an expensive hobby 

[ckilgore]

Something is not jiving.  You stated the voltage before the 500R is 415V, and after the 500R is 355V.  This is a 415-355=60 volt drop across the 500 ohm resistor.  Since Ohms Law states I=E/R when we solve for I we get 60/500=.12A or 120mA, not the 42mA you stated the power tubes are drawing.  If we solve for the power dissipated by the dropping resistor P=E^2/R= 60*60/500=7.2W.

[sluckey]

Haha. We all jumped on the red herring! 42mA had nothing to do with the example. Thanks for clearing that up.