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Circuit problem- resistance, power, voltage

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data

    We connect two resistors to a battery, once with the resistor of the resistance R1 = 2 Ω, and the second time the resistor of the resistance R2 = 0.5 Ω. In both cases the resistor uses up the power of 2W. What is the internal resistance of the battery? What is the driving voltage of the battery?

    2. Relevant equations

    P= V*I
    V= R*I
    P= R*I²
    P= V²/ R

    3. The attempt at a solution

    - Now I derivate three general equations:

    1) V(0)= R(1)I(1) + R(0)I(1)
    2) I(1)= V(0)/ (R(1) + R(0))
    3) I(2)= V(0)/ (R(2) + R(0))

    - Now I plugged these three equations in the power formula and got:

    P= R(1)I(1)²= R(1)*[V(0)²/ (R(1) + R(0))²]
    V(0)²= [P*(R(1) + R(0))²]/ R(1)

    P= R(2)I(2)²= R(2)*[V(0)²/ (R(2) + R(0))²]

    P= R(2)* [(P*(R(1) + R(0))²)/ (R(1)* (R(2) + R(0))²)]
    1= [R(2)/R(1)]*[ (R(1) + R(0))²)/ (R(2) + R(0))²)]
    R(2)/R(1)= (R(1) + R(0))²)/ (R(2) + R(0))²)
    sqrt(R(2)/R(1))= (R(1) + R(0))/ (R(2) + R(0))
    sqrt(R(2)/R(1))* (R(2) + R(0))= R(1) + R(0)
    R(0)*(sqrt(R(2)/R(1))) – 1)= R(1) – ((sqrt(R(2)/R(1)))*R(2))
    R(0)= [R(1) – ((sqrt(R(2)/R(1)))*R(2))]/ [sqrt(R(2)/R(1))]
    R(0)= 1/3 Ω

    V(0)²= [P*(R(1) + R(0))²]/ R(1)
    V(0)²= 2.33 V

    Are my calculations correct?
    Thank you for helping!:smile:
  2. jcsd
  3. Jan 8, 2010 #2


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    Homework Helper

    I am getting different answer.
    P= R(1)*[V(0)²/ (R(1) + R(0))²] = R(2)*[V(0)²/ (R(2) + R(0))²]
    Cancel V(0)^2 from both and rearrange the terms. You get
    R(1)/R(2) = [R(1) + R(0)]^2/[R(2) + R(0)]^2
    Substitute the values of R(1) and R(2) and solve for R(0).
  4. Jan 9, 2010 #3
    OK, I rewrote my equations and recalculated and this are my new results:

    R(0)= 1 Ω
    V(0)= 3 V

    Are my calculations correct?

    Thank you for helping!:smile:
  5. Jan 9, 2010 #4


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    Homework Helper

    Yes. Your calculations are correct.
  6. Jan 10, 2010 #5
    Thank you for reviewing and all the help!:smile:
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