Metropoulos Forum

Hi ,
I want to use a Vishay Draloric AC05 as a dropping resistors (8.2k/10k)
But the datasheet does not speak about the maximum working voltage, and gives a formula : V=√(PxR)

The datasheet : http://www.mouser.com/ds/2/427/acseries-222859.pdf

The formula give me something around 202V (for 8.2k), but I don't understand how it correlate to the circuit.... does it mean that it can dissipate 200v at 5W intensity ? for example, if I put 460v trough the resistor, will the resistance's "Power handling" lower, and will it act as a 2W or 3W resistor? I'm lost, I don't understand

Can I use them as dropping resistors in a super lead build ?
I need help, thank you

The working voltage is the voltage drop across the resistor not the B+ voltage with respect to ground. For instance, if you measure less than 202V across the resistor worst case condition then you comply with specifications. But you have to also watch power handling capacity as well. In this example the 8.2K resistor with 202V across it barely makes the 5W power rating. V=√(PxR) is just a reworking of the basic power formula P = V x I and substituting V/R for current I and solving for V. The provided formula is just an easy way to calculate maximum working voltage for a given power rating and resistance.

THANK YOU !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

So I can only drop the voltage by 202V with a 8.2k resistor ? AM i right
Thank you I was feeling bad it's been a week that I try to find an answer on net net

So I can only drop the voltage by 202V with a 8.2k resistor ? AM i right

That is true for a 5W resistor. A higher wattage resistor can withstand a greater voltage drop. Lower wattage less voltage drop. Again this goes back to basic ohm's law.