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Internal resistance ratio problem

  1. Mar 9, 2007 #1
    1. The problem: A resistor has a resistance R, and a battery has an internal resistance r. When the resistor is connected across the battery, ten percent less power is dissipated in R than would be dissipated if the battery had no internal resistance. Find the ratio r/R.


    2. Relevant equations:
    P=IV, P=I^2R, P=V^2/R;
    V(IR)=emf - Ir
    Obviously a series circuit when the battery's internal resistance and the resistor are both considered... so those relationships apply also



    3. The attempt at a solution
    I have been struggling with this one for awhile. I don't really know how to approach it. What I am having trouble understanding is how to choose which of the relationships describing power I need to use to solve the problem. There are three which could be relevant: P=IV, P=I^2R, P=V^2/R ... I am not sure which ones to use to set up my initial ratio and thereby solve the problem.

    So far I have these (where o is "naught..."), describing power in the no-current and current scenarios, respectively:

    P(o) = IV(o)
    P(1) = IV(o) - (I^2)r

    But I am not sure to proceed...
     
  2. jcsd
  3. Mar 9, 2007 #2

    berkeman

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    Staff: Mentor

    When there is no internal resistance r, then the power dissipated in the external resistor is P=V^2/R. Do you see why?

    When the internal resistance r is included in the circuit, the total power dissipated in both of the resistors is P=V^2/ (what?)

    Does that help?
     
  4. Mar 9, 2007 #3
    Okay... the power is V^2/R because the circuit is a series circuit? And likewise, the total power dissipated in both resistors has to be P=V^2/(R+r) because resistances just add in a series situation...

    To generalize... would it be reasonable to say that you always want to try to analyze things like power in terms of the other physical quantities that are VARIABLE in the types of circuit you're dealing with?

    Is that analysis correct? Let me see if I can figure out how to solve the problem now.
     
  5. Mar 9, 2007 #4
    Also, I am not at all certain how to extrapolate from this in order to set up a ratio between r and R.

    Oh how I wish my professor spoke English...
     
  6. Mar 9, 2007 #5

    berkeman

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    You're doing fine, just keep going. You wrote the correct equation for the 2nd situation. Now the problem says that in the 2nd situation the R resistor dissipates 10% less power. Just set up the powers for the two cases (call them P1 and P2 or something), show the ratio relationship, and then I'll bet you can solve the problem.

    As to how to think about this kind of problem in general, just visualize the voltages and currents as the fundamental quantities. Then you'll be able to figure out which form of the power equations you want to use.
     
  7. Mar 9, 2007 #6
    After substituting EMF - Ir based on the definition of internal resistance, I arrive at a ratio that looks like this:

    P(1)/P(2) = ((V[o]^2)(R) + (V[o]^2)(r)) / ((r)(V[o]-Ir)^2)

    Which expands out into nastiness if I do the exponential expansion... the problem is that I've got current in the expression, and I don't know how to get rid of it...
     
  8. Mar 9, 2007 #7

    berkeman

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    Maybe try it a different way. Draw the 2nd situation, where you have the voltage source on the left, r across the top, and R on the right. Now, you know that the power disspiated in R in this setup is P2=0.9P1, right? And you can write an expression for the voltage across R in this situation using a voltage divider, right?

    So write the expression for the voltage across R, and use the fact that you know P1=V^/R, and that should be an easier equation to solve. You still will have to be careful as you square out terms and then gather stuff back again.
     
  9. Mar 9, 2007 #8
    A voltage divider? I'm not sure I follow what you mean by that.

    Many thanks for your help so far.
     
  10. Mar 9, 2007 #9

    berkeman

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    Staff: Mentor

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