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Homework Help: Power dissipation in resistive circuit

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A single resistor is wired to a battery as shown in the diagram below. (Figure 1) Define the total power dissipated by this circuit as P(subscript 0). Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to the left (figure 2). What is the power, in terms of P(subscript 0) dissipated by this circuit?

    2. Relevant equations

    I am not sure! This problem seems so easy -- I don't think I'm really understanding what they want from me? This is the feedback I got.
    The correct answer does not depend on the variables: I, R
    So I tried it again...
    The correct answer does not depend on the variables: U, t

    3. The attempt at a solution

    I uploaded the pictures from the question. Can anyone help?

    Attached Files:

  2. jcsd
  3. Feb 23, 2010 #2


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    [EDIT] Deleted what I had previously said: "Will your program allow you to input the answer in terms of the variables V and R?"

    I think I understand now. The problem statement is asking you for a power ratio. A power divided by a power (not an actual power itself). So once you find the two powers, divide one by the other.
    Last edited: Feb 23, 2010
  4. Feb 23, 2010 #3
    Huh?? Well, my first guess was IsquaredR and then it was U/t...because both those formulas give power. What do you mean a power ratio?
  5. Feb 23, 2010 #4
    The answer not a formula. Don't understand...
  6. Feb 23, 2010 #5


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    That's fine to start with equations such as

    [tex] P _0 = (I _0)^2 R [/tex].

    For the first circuit. But you'll find it easier in this case, if you stick with the given variables (Both circuits are given only in terms of V and R, so you might try to stick with those). Since you already know

    [tex] V = I _0 R [/tex]

    you'll find it easier to use

    P _0 = \frac{V^2}{R}

    Now do the same thing for the second circuit, except call it [tex] P _1 [/tex].

    It think your physics program is looking for [tex] P_1 / P_0 [/tex]. Give this a try and I think you'll see what I mean.
  7. Mar 5, 2010 #6
    You should also try to put 0.5P0 since by adding one resistor of the same magnitude in series the power should go down by half, since P = V^2/(Sum of R)
  8. Mar 5, 2010 #7


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    Don't give people full solutions to problems in the homework help section, no matter how simple the problem may be.

    EDIT: It's against forum rules, and it is also counterproductive for the original poster, who learns nothing and gains nothing from having other people do his/her homework for him/her.
    Last edited: Mar 5, 2010
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