Resistor in paralell...

Discussion in 'Homework Help' started by ESPguitar, Sep 25, 2005.

  1. ESPguitar

    Thread Starter Member

    May 15, 2005
    16
    0
    Hi..

    I have a relay the coil is 960 Ω and i need to couple a resistor in paralell with this relay, the total resistance should be 1547,32 Ω..

    When i use the formula R2=(Rtot * R1) / (R1 - Rtot) i get this answer:

    R2= (1547,32 * 960) / (960 - 1547,32) = 1455427,2 / -587,32 = (-2529,161615)

    This ain't the right answer :S

    Can somebody help me with this (i guess it is a easy question for you but, you know:S)

    Thanks,

    Robin
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    You have a coil with 960 Ω resistance.

    You have to couple a resistor to that coil to make the a total resistance of 1547.32

    Am I correct?
     
  3. Z2148

    New Member

    Sep 21, 2005
    1
    0
    Are you sure that Rtot=1547.32 Ω and R1=960 Ω ?

    From what I know, Rtot should be smaller than the smallest resistance in the circuit when connected in parallel..
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    Correct, you can't have the above mentioned situation for two resistive devices in parallel. Either the numbers are wrong or there is an additional resistive element in series with the parallel network.
     
  5. aibelectronics

    Member

    Aug 26, 2005
    24
    0
    unless you're using a faulty calculator that answer is impossible.

    Rtotal = Rcoil*R / (Rcoil+R)

    you can make R subject of formula. remember abs(Rcoil) = 2*3.142*freq*L

    what's your inductance?
     
  6. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    You must add a 580-Ohm resistor in series with the 960-Ohm coil. Assuming that 960-Ohms is the DC resistance.
     
  7. sanjeev203

    New Member

    Sep 19, 2005
    7
    0
    Robin,

    In relay, resistor always goes in series with the inductor consituting (RL+R) as total resistance.

    As other member suggested you, the equivalent resistance in parallel network is always lesser then the smallest one hence that has to be in series.

    I found a nice article for you.

    http://www.leachintl2.com/english/english2...erties/how8.htm

    Hope this will help you.

    Best regards,
    Sanjeev
     
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