dc series circuit problem

Discussion in 'Homework Help' started by pidge, Apr 26, 2010.

  1. pidge

    Thread Starter New Member

    Apr 26, 2010
    i have 3 resistors in a 329VDC circuit series R1=9V R2=25ohms R3=15ohms how can i find out the ohms value for R1 or what i'm thinking is it's a trick qu. and i need more values
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    Ohm's law is all you need.
  3. TheWeasel

    New Member

    Aug 28, 2009
    beenthere's right, there's no trick, you have all you need, and it can be worked out as follows;

    The volts drop given across R1 is 9v which leaves the remaining 320v to be dropped across R2 + R3 (40 ohms).
    I=V/R so 320/40 gives 8 Amps.

    Knowing the current now permits calculation of the value of R1 needed to drop 9v at 8A ;
    R= V/I or 9/8 which makes R1 = 1.125 ohm

    As a double check in the complete circuit;
    V/R = I or 329V/41.125R = 8A......so it all fits together.
    Last edited: Apr 26, 2010
    kingdano likes this.
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    To TheWeasel - it might be more instructive to get the OP to make some of the calculations and see some of the relationships himself. More learning goes on that way.
    kingdano likes this.
  5. kingdano

    Senior Member

    Apr 14, 2010
    i agree with beenthere - but for me it was always helpful to see an example worked through once in a while.

    so kudos to both of you for the helpful posts to the EIT.