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Resistors in parallel

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure, the rate at which R1 is dissipating electrical energy is 20.0 W.
    [​IMG]
    A) Find R1
    B) Find R2
    C) Find emf of battery
    D) Find current through R2
    E) Find current through 10 ohm resistor.
    F) Find total electrical power consumption of the resistors
    G) Find total electircal power delivered by battery

    2. Relevant equations
    Ohm's Law: V = IR
    Resistors in parallel: I/V_ab = 1/R_eq = 1/R1 + 1/R2 + 1/R3
    Current through resistors in parallel: I1 = V_ab/R1
    Power delivered by battery: P = Ei
    Energy dissipated by resistors: (i^2)R

    3. The attempt at a solution
    A) I set the power dissipation 20 = (i^2)R, having i = I1 = 2 A, and solved for R1.

    B) I'm stuck here. Only thing I can think of using is the resistors in parallel equation but I don't know V, and I can't find V because I don't know R_eq yet.

    C) Find R_eq and use V = IR_eq, which equals E.

    D) I2 = V/R2

    E) Same as part D

    F) Apply the power consumption equation to each resistor and add them up.

    G) P = Ei
     
  2. jcsd
  3. Oct 13, 2007 #2
    Let us deal one step at a time.
    A) good
    B) There are three equations for power. P=VI, P=V^2/R, and P=I^2R. Think and execute.

    Please show us some more of your work so that we can help if you need it.


    <<post edited slightly by berkeman>>
     
    Last edited by a moderator: Oct 13, 2007
  4. Oct 13, 2007 #3
    Many thanks for the quick reply.

    B) 20 = V*I1, so V = 20/I1 = 10

    I/V = 1/10 + 1/R1 + 1/R2
    1/R2 = I/V - 1/10 - 1/5
    R2 = (above)^-1 = 20

    C) V = E in this case, right? E = 10

    D) I2 = V/R2 = .5

    E) (I'll call this one I3) I3 = V/10 = 1

    F) (I3^2)10 + (I2^2)R2 +20 = 40

    G) P = EI = 10*3.5 = 35? Can the power consumed by the resistors be greater than the power given by the battery? I think I made a mistake.
     
  5. Oct 13, 2007 #4
    It should equal 35 Watts.
    P3 = V*I =10V*1A=10W
    P2 = V^2/R= 10V^2/20ohms=5W
    P1 = 20W
     
  6. Oct 14, 2007 #5
    Ahh, thanks. I see what I did wrong. I forgot to square the I2 in (I2^2)R2.

    Thank you very much.
     
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