Power Dissipated by Resistors

Discussion in 'Homework Help' started by Elizabeth Paredez, Mar 22, 2011.

  1. Elizabeth Paredez

    Thread Starter New Member

    Mar 22, 2011
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    0
    The only thing I know about resistors is that V = IR. I think I can simplify the circuit but I really have no idea how to answer the following questions.

    What is the power dissipated by each of the resistors in the following circuit and what is the electric potential difference between points (a) and (b)?

    Please any help would be greatly appreciated because I don't even know where to start.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Take the two parallel resistors, make them one, then you will have two series resistors. Make them one, then you have it.
     
  3. Elizabeth Paredez

    Thread Starter New Member

    Mar 22, 2011
    3
    0
    Here's what I did.

    V = IR

    I combined the 2.0 Ω resistor with the 24.0 Ω resistor to give me "26.0 Ω" resistor. Since I made it a parallel circuit, volts are constant.

    For R(equivalent) I got (26^-1) + (8^-1) = .163461....then ^-1 it equals approx 6.0

    Then I did, I26 = V26/ R26 (12/26) and got approximately .5 Amps
    Then I did, I8 = V8/R8 (12/8) and got approximately 1.5 Amps

    Then I did, I(total) = V/Req (12/6) and got approximately 2.0 Amps

    Does anybody know if what I did is correct???
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    You can't do it that way. You must combine the parallel, since both of them are in series with the 2Ω resistors.

    Combine the two parallel resistors first.
     
  5. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    The reason you can't combine the 24Ω and the 2Ω resistor is that they are neither parallel nor series.
    They would be parallel if they had common endpoints.
    They would be series if they had only one common endpoint, which would be unique to the two of them.
    In your case, that endpoint is shared with the 8Ω resistor which spoils the "serial-ity".
     
  6. saikat36

    Member

    Jan 28, 2009
    16
    0
    U have 2 parallel resistor 24? & 8?.So

    Req=(R1*R2)/(R1+R2)=6?

    Now 6? & 2? r in series, so
    Req=6+2=8?
    then current I=(12/8)=1.5A
    now branch current of 24? & 8?

    I24=(8/32)*1.5=.375A

    I8=(24/32)*1.5=1.125A

    Power dissipation

    P24=24*0.375*0.375=3.375W
    P8 =8*1.125*1.125 =10.125W
    P2 =2*1.5*1.5 =4.5W

    Vab=-(2*1.5)=-3V

    I think now all is cleared.
    Thank u.
     
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