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Internal Resistance in battery

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data
    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 there would be if the battery had no internal resistance. Find the ratio r/R


    2. Relevant equations

    R=V/I

    3. The attempt at a solution
    I really have no idea how to even start this problem. Looking at the problem all I can figure out is that R>r. Can anyone point me in the right direction to solve?
     
  2. jcsd
  3. Jun 29, 2009 #2

    jgens

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    Well, for starters, have you drawn a picture (even with simple circuits it can help)? Next, should you consider the battery's internal resistance as a resistor in series with the circuit or a resistor in parallel? Lastly, write out all the equations for each case and solve.
     
  4. Jun 29, 2009 #3

    cepheid

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    1. What is the formula for the power dissipated in a resistor?

    2. If you have two resistors in series, what is the total resistance?

    Can you see that you are comparing two scenarios? In the former scenario, there is only one resistor in the circuit, and the power dissipated is given by the formula in 1. In the second scenario, there are two resistors, and the power dissipated is given by the same formula, but applied to the *total* resistance of the circuit. The two power values from the two scenarios differ by 10%

    3. How does the power dissipated in situation 2 differ from that in situation 1 (i.e. express the 10% result mathematically)
     
  5. Jun 30, 2009 #4
    1. P=I^2 R
    2. 1/Rp = 1/R + 1/r
    3. to find the power dissipated in situation 2 I would use the equation:
    P=I^2 Rp
    because they are parallel resistors. So the difference between the to powers is one is multiplied by R and the other by Rp. If I did do everything right I don't understand how to find the solution when the only number I know is 10%?

    the solution is supposed to be.054
     
  6. Jun 30, 2009 #5

    cepheid

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    Are the resistors in parallel? Think about it. You connect a battery in series with load resistor R. The battery has some internal resistance (which can be modelled as a resistor r in series with an ideal voltage source).

    EDIT: And even if you don't get my hint, I've already given away the answer above. ;)
     
  7. Jun 30, 2009 #6
    so:
    Rs=R+r
    P1=I^2 Rs and P=I^2 R

    I'm pretty sure I've understood everything you have said physics wise regarding the problem, but I don't understand how to come up with the answer when the only number I know is 10%. The only difference I see when comparing the two P's is that in P1 I^2 is multiplied by R+r and in P I^2 is only multiplied by R, but I don't know how to express that mathematically and find the ratio.
     
  8. Jun 30, 2009 #7

    cepheid

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    Ten percent LESS power means only 90% the power compared to the original case.

    Or, ten percent less power means that the difference between the two is 10% of the original.

    These statements are equivalent. Does that help?

    EDIT: Oh yeah, and the total current will not be the same in both cases, so maybe you're better off using the power equation that has V in it.
     
  9. Jun 30, 2009 #8

    vk6kro

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    Sometimes it helps to assign real values to a problem like this if you don't like working in algebraic terms.

    Say it was a 12 volt battery and the external resistor was 15 ohms. Now, what internal resistance would you have to add in series to drop the power in the 15 ohm resistor by 10%?

    You can still work out the ratio required, but with real numbers.
     
  10. Jul 1, 2009 #9
    So using the equation with out the current:
    P=V^2 / R
    P1=V^2/(R=r)

    So according to the example vk6kro suggested:
    V=12 R=15
    P=12^2 / 15
    P=9.6
    (9.6)(.1)=.96
    9.6-.96= 8.64=P1

    P1= V^2/(R+r)
    8.64 = 144/(9.6+r)
    r=7.07
    r/R=7.07/15=.471

    I still don't understand how you can find the ratio algebraically with only knowing that R drops by 10%.
     
  11. Jul 1, 2009 #10

    cepheid

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    Hi,

    EDIT: I changed the equations in this post, because I had originally misread the question, as pointed out by vk6kro. His answer and mine now agree. The second power is the power dissipated in R only, not the total power.

    [tex] 0.9P_1 = P_2 [/tex]

    [tex] 0.9I_1^2 R = I_2^2 R [/tex]

    [tex] I_1 = \frac{V}{R} [/tex]

    [tex] I_2 = \frac{V}{R + r} [/tex]

    Therefore,

    [tex] 0.9 \left(\frac{V}{R}\right)^2 R = \left(\frac{V}{R + r}\right)^2 R [/tex]


    Which gives you a nice ugly quadratic to solve :wink:
     
    Last edited: Jul 1, 2009
  12. Jul 1, 2009 #11

    vk6kro

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    The result you get is not right. In the second power case, the 8.64 watts is only the power in the external 15 ohm resistor, so it does not happen when the resistor has 12 volts across it (because there is an internal resistance in series with it).

    You would have to work out the current when the 8.64 watts occurs in the 15 ohm resistor, (8.64 = i^2 *15) then the total resistance (12 / i), then the internal resistance by subtraction of 15 ohms (R-15).
    Then work out the ratio required.

    If this ratio is independent of the resistor values selected, then you should be able to select any value you like and work it out with real resistor values. Otherwise, you end up trying to solve huge algebraic equations.
     
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