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Wirewound Resistor - Can it be used with 9 V battery ??

Discussion in 'Power Electronics' started by rajat, Jun 3, 2012.

  1. rajat

    rajat

    16
    0
    Jun 3, 2012
    Hey All,

    I have a wirewound resistor that says on it 5W 24 KJ.

    I want to find out what is the maximum voltage it can work with ??

    Any help will be appreciated.

    Kind Regards
    Rajat
     
  2. CocaCola

    CocaCola

    3,635
    4
    Apr 7, 2012
    Watts = Amps x Volts

    Don't exceed the 5 Watt rating of the resistor...

    @ 9 volts that would be no more than 0.55 Amps
     
  3. rajat

    rajat

    16
    0
    Jun 3, 2012
    Thanks

    Thanks Cocacola for your reply :)

    Taking a step further calculations.

    For 9 V

    W = VI
    I = 0.55 amps

    For 240 V (normal power circuit)

    W = VI
    I = 0.020833 amps

    So does this mean my resistor will get more 'Heated Up' as it is resisting the higher voltage.

    Or will it burst ?

    If I miss something please do let me know.


    Rajat
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
    2,689
    Jan 21, 2010
    OK, several things...

    Firstly, resistors DO have a voltage rating. Often you can assume it is around 200V if not specified. his voltage limit applies even if the resistor would not go over its power rating if the voltage was applied.

    Applying voltages in excess of the limit can result in the resistor exhibiting unusual changes in resistance (often lower) with the risk of this change becoming permanent.

    Having said all of that, there is a relationship between resistance, voltage, and current. Knowing either of these allows you to work out the third. Any two of these also allows you to determine the power dissipation too.

    Thus since your resistance is fixed (I'm not sure what value -- but let's say it is 24000 ohms) you can determine the power dissipation at a particular voltage.

    The basic relationships I will use are I = V/R and P = V*I. Substituting the former in the latter, we get P = V^2/R, and that's what I'll use.

    Let's start off with 9V. The power dissipated is (9*9)/24000 = 0.003375W (or 3.4mW approximately)

    And lets try at 240V (understand that this may be beyond the voltage rating, you need to check the specs to be sure). The power dissipated is (240*240)/24000 = 2.4W

    Here is an example of a datasheet for a wirewound resistor.

    Note that on page 2 it gives details about the voltage ratings. For these resistors, a 5W resistor has a working voltage of 350 volts and an allowable maximum of 700V.

    Also note that the smallest have a working voltage of 250V, which is close to my suggested default assumption of 200V.

    Assuming that the resistor you have is similar to these ones, they could withstand about 345 volts while dissipating a shade under 5W.

    All of my calculations are based on the assumed value of your resistor. If you don't know the resistance you can't do the calculations. However you can measure the current through it when a certain voltage is applied, and that will allow you to calculate the resistance. This is essentially what a multimeter does when measuring resistance.

    edit: if you allow the resistor to dissipate too much power for too long it may start to smoke, catch fire, or simply become open circuit like a bulb or a fuse.
     
    Last edited: Jun 3, 2012
  5. BobK

    BobK

    7,592
    1,636
    Jan 5, 2010
    24K is an unlikely value for a wirewound resistor.

    Bob
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
    2,689
    Jan 21, 2010
    I agree. But the math is the same, just with different (i.e. correct) figures
     
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