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Power dissipated in circuit combination

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    three resistors R1=R2 and R2=2R3, are connected to a battery, as show in the figure, if the power dissipated for each one is p1,p2 and p3 respectirvely;p3 must be equal to:

    2. Relevant equations

    P=VI
    P=V^2/R
    P=I^2*R

    3. The attempt at a solution

    I have tried to find the total resistance in the circuit by knowing that R2 and R3 are in parallalel and this equivalent resistance lets call it R23 will be in series with R1 so the equivalent resistance would be Req=R1+R23 this will allow me to find the total current in the circuit which is I=V/Req while doing so I should be able to find the power in R1 with this P=I^2*R1 since the total current in the circuit is the same going through R1, but I cant find the relationship with the power dissipated with P2 and as you can see P1 is in terms of P2
     

    Attached Files:

  2. jcsd
  3. Dec 9, 2013 #2

    gneill

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    Staff: Mentor

    Your third Relevant Equation is important. The question wants you to express the power dissipated by R1 in terms of the power dissipated by R2. So aim towards expressing R1 in terms of R2 and the current through R1 in terms of the current through R2. KCL and the current division rule will come in handy :wink:
     
  4. Dec 9, 2013 #3

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Hint -- a simpler approach would be to assign values to the resistors (like 1 Ohm and 2 Ohms), and then look at how current divides between R2 and R3. Label the current through R2 = I. Then what is the current through R3? And what does that mean the current is through R1?

    And solve...Please show us your work toward the solution! :smile:
     
  5. Dec 9, 2013 #4
    lets call i the total current in the circuit therefore the current that goes through R1. applying kirchhoff's law to the junction where R2 and R3 splits i=i1+i2 the total current is the sum of the current that goes through R2 and R3
    so we have:
    i=i1+i2
    i=V/R2 + V/R3
    i=(R3+R2)V/R2*R3
    but R2=2R3
    i=(R3+2R3)V/R2*R3
    i=3V/R2

    I will assume V is the voltage drop on R2 so
    V=iR2/3
    And the power dissipated P2=V^2/R2 and P1=I^2*R1 so we have
    P2=(iR2/3)^2/R2
    P2=i^2R2^2/9R2
    P2=i^2R2/9
    but R1=R2 and P1=I^2*R1
    P2=1P1/9
    P1=9P2
     
  6. Dec 9, 2013 #5

    berkeman

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    Correct! :smile:
     
  7. Dec 9, 2013 #6
    it would make any sense to ask P3?
     
  8. Dec 9, 2013 #7

    berkeman

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    Staff: Mentor

    For extra credit? Sure!

    What is P1/P3?
     
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