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Resistance???

Discussion in 'General Electronics Discussion' started by Young, May 18, 2013.

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  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Yeah, I tried. It doesn't work.
     
  2. Young

    Young

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    Feb 6, 2013
    The resistance of the load is 5ohm,R1 is 3.7ohm,R2 is 5ohm,try it an you will still get 5ohm,this is the link "hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html" thanks
     
  3. Young

    Young

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    Feb 6, 2013
    Type this on google,hyperphics voltage divider
     
  4. komalbarun

    komalbarun

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    Nov 25, 2011
    xxxxx
     
    Last edited: May 19, 2013
  5. komalbarun

    komalbarun

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    Nov 25, 2011
    I try to help where I can .. and dude..am also here to learn...and I also edited my post where I have used the ratio approach, pls check it before insulting people..it's because of people like you that people like me don't get discouraged at learning electronics !
     
    Last edited: May 19, 2013
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, so you changed your post to use the same method that I suggested.

    Unfortunately, nobody gets notified of your changes. I certainly don't re-read every post before I reply.

    Far better to post another post with corrections if there are posts following the one you want to retract.

    edit: Actually, I'll retract that. Your edited post is still wrong if the bulb draws any appreciable power. Either that or you need to use very low values resistors which will eat up an unnecessary amount of power.
     
    Last edited: May 19, 2013
  7. Young

    Young

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    Feb 6, 2013
    Last edited by a moderator: May 19, 2013
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,501
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    Jan 21, 2010
    Let me explain again, for the benefit of Young and komalbarun...

    If you have a voltage divider let's say the 7 ohms and 5 ohms suggested by komakbarun.

    If you put 12 volts across it, you will get 5 volts across the 5 ohm resistor.

    Now let's put a load on it. Let's assume our load is 5 ohms (that would be similar to a 5V 5W lamp).

    The voltage the load sees is now 12* ((5||5)/((5||5) + 7)

    5||5 is 2.5 ohms so

    V = 12*(2.5/(2.5+7)) = 12* (2.5/9.5) = 12* (0.2632) = 3.158 volts.

    Clearly it's not 5V.

    My solution would be to put a 7 ohm resistor in series with the bulb, and the voltage across the bulb would be...

    5V

    But apparently it's far too complex to get this right...

    Now, my solution depends on Young answering the question of the lamp rating in Watts, or the current in Amps, which he has not, despite being asked more than once.

    Now, it's also possible to allow for the resistor in parallel with the bulb.

    You could have a resistor of 3.5 ohms and a resistor of 5 ohms.

    Unloaded, this would produce a voltage of 7.06V, but with the bulb in place you would again get 5V.

    However this is simply wasting the power flowing through the 5 ohm resistor.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    And try it yourself with

    Vin = 12V
    Z1 = 7 ohms
    Z2 = 9999999 ohms
    ZL = 5 ohms

    What is VOut?

    Z2 = 999999 ohms is a large number because using that calculator we can't say "I don't actually want a Z2"
     
  10. komalbarun

    komalbarun

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    Nov 25, 2011
    Steve is right.

    A resistor in series to the bulb is much better than a V-divider.
     
    Last edited: May 19, 2013
  11. Young

    Young

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    Feb 6, 2013
    Steave calm down when ever you input a number into the calculator it loads number into other blank boxes dont care about that all i want you to do is make sure z1,z2,zl,and v input is typed before you conclude.the only box you should leave alone are the v unounted and vout and you will get 4.8v i think thats enought note z1=3.7ohm(not 7ohm),z2=5ohm,zl(load resistance)=5ohm the vout(voltage ouput even after the load resistance)=4.8v,v unmounted is vout put before appling a load resistance
     
  12. Young

    Young

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    Feb 6, 2013
    i didnt say its oot a better way but his calculation was wrong i used 39ohm as he calculate with a voltage of 12v i got 8v this may even damage my load thats why i went with your cos i got 4.8-5v even with the load resistance,all i did was asume the load resistance was another resistance and i inputed it to d voltage divider calculator and found out i needed r1=3.7,r2=5,load resistance=5 and i got 4.8-5v try it and see
     
  13. komalbarun

    komalbarun

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    Nov 25, 2011
    ok got it so its like this :

    Rp = total R of R2 and RL in parallel...
    Rp = r2xRL / r2+rL = 25 / 10 = 2.5

    Rtotal= R1+Rp = 3.7 + 2.5 = 6.2

    VRp = (2.5/6.2) * 12
    = (approximately) 0.4 x 12
    =(approximately) 4.8V

    But Since VRp = I x Rp;

    I = VRp / Rp
    I = 4.8 / 2.5 = 1.92 A

    Power Dissip. in load will be(I think) :

    PL = IV = 1.92 X 4.8 = 9.2W !

    Does your bulb withstand a power dissip. of more than 9.2Watts?
     
    Last edited: May 19, 2013
  14. Young

    Young

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    Feb 6, 2013
    Power rating 20watts
     
  15. komalbarun

    komalbarun

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    Nov 25, 2011
    and voltage rating is 5V?
     
  16. Young

    Young

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    Feb 6, 2013
    Mistake its 12
     
  17. komalbarun

    komalbarun

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    Nov 25, 2011
    Are you using a bulb or an LED? btw the power rating should be mentionned on the bulb...
     
  18. komalbarun

    komalbarun

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    Nov 25, 2011
    So its 5V @ 12W?
     
  19. komalbarun

    komalbarun

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    Nov 25, 2011
    I would still prefer steve to confirm this ;)
     
  20. BobK

    BobK

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    1,688
    Jan 5, 2010
    A resistor in series with the bulb IS a voltage divider. As Steve said, adding another resistor does nothing but waste power.

    Bob
     
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