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Homework Help: Finding 2 resistances in which case the power doesnt change

  1. Sep 13, 2012 #1

    fawk3s

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    1. The problem statement, all variables and given/known data

    http://desmond.imageshack.us/Himg705/scaled.php?server=705&filename=fafadaf.png&res=landing [Broken]

    Find resistances R2 and R3 in such a way that the power on the resistances (shown by the watt-meter) remains constant when the switch SA1 is opened.

    2. Relevant equations

    I=V/R
    1/R=1/R2+1/R3
    Rwhole=R+Rs
    P=I*V

    3. The attempt at a solution

    I've tried many things already but none have got me anywhere really. I understand that P=I*V which means that when "V" decreases n times, then "I" must increase by n times, but I dont really know what do with it. When the switch is opened, V on R2 ought to be bigger than when it is closed and I ought to be smaller than when the switch is closed.

    But thats all I've got. :/

    PS. Dont mind the R3=2,45 in the picture. Both R2=R3=?

    Thanks in advance,
    fawk3s
     
    Last edited by a moderator: May 6, 2017
    fawk3s, Sep 13, 2012
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  3. Sep 13, 2012 #2

    CWatters

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    Yes when S1 is closed R3 will draw extra current through the source resistance Rs. This will cause the voltage on R2 to fall.

    It's a matter of writing enough equations to allow you to solve for the unknowns. Perhaps start with two equations for Power P. One with the switch open and the other closed.
     
    CWatters, Sep 13, 2012
  4. Sep 13, 2012 #3

    fawk3s

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    P1=P2
    V1I1=V2I2
    V1/V2=I2/I1
    I2/I1=(E/Rs+R2) / (E/Rs+Rwhole)

    Rwhole=R2R3/R2+R3

    But dont really know where to go from here.
     
    fawk3s, Sep 13, 2012
  5. Sep 14, 2012 #4

    fawk3s

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    Doesnt anybody have an idea on how to solve it? Or is it too simple for you guys that you dont even bother answering?
    Because I have been thinking about it for quite some time now and cant figure it out.

    Thanks in advance,
    fawk3s
     
    fawk3s, Sep 14, 2012
  6. Sep 14, 2012 #5

    gneill

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    Find expressions for the power for both cases (that is, with and without R3 in the circuit). Equate them. What is the resulting relationship between R2 and R3?
     
    gneill, Sep 14, 2012
  7. Sep 14, 2012 #6

    fawk3s

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    You can use these for the current.
    I2=E/Rs+R2
    I1=E/[Rs+(R2*R3/R2+R3)]

    Yet P=V*I, and since V is different for both cases, which means we have to use V1 and V2 and get 2 more variables. And I cant figure out how to present V to get rid of the extra variables.

    Did you mean I12*R2=I22*(R2*R3/R2+R3), though?
    I wasnt able to get much out of that, even when replacing E and Rs with their given values.

    Did you get an answer though, gneill?

    Thanks in advance,
    fawk3s
     
    fawk3s, Sep 14, 2012
  8. Sep 14, 2012 #7

    gneill

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    You can write separate equations for V1 and V2.

    Hint: Note that the current flowing through Rs will cause the only voltage drop between the battery and the point where the voltage is measured. Thus you are in a position to write expressions for V1 and V2 that don't include new variables.
    No, you can't count on the power in particular resistors being the same, since Rs will be developing power, too, and it will be different for the two cases since the currents will be different.
    Yup. I got an expression relating R2 and R3.
     
    gneill, Sep 14, 2012
  9. Sep 14, 2012 #8

    fawk3s

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    V2=E-Rs*I2 and V1=E-Rs*I1 ?

    Could you elaborate on this, please? Im confused here. Because I cant really understand the difference between writing V*I or I2*R.
    The wattmeter is measuring only the power on resistors, right?

    Thanks in advance
     
    fawk3s, Sep 14, 2012
  10. Sep 14, 2012 #9

    gneill

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    Yup.
    Actually, thinking about it a bit more I see that the wattmeter will indeed be measuring the power drawn by the load resistors. So you should be able to employ I2R here, using the appropriate I's and R's. Sorry about the diversion :smile:
     
    gneill, Sep 14, 2012
  11. Sep 15, 2012 #10

    fawk3s

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    Does the math get all messy and complicated for you too here? Same as with I2*R. Because if you were easily able to solve it then I must be doing something wrong here. Or is it supposed to be messy with all the R's and E's? (Ofcourse you can replace Rs and E with numerical values in the end, but never the less.)

    I would really be interested in seeing how you solved it.

    Thanks in advance
     
    Last edited: Sep 15, 2012
    fawk3s, Sep 15, 2012
  12. Sep 15, 2012 #11

    azizlwl

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    p1=i12r2
    p2=i22rT
    rT=equivalent parallel resistors of r2 and r3.

    i1=[itex]\frac{v}{r1+r2}[/itex]

    i2=[itex]\frac{v}{r1+rT}[/itex]

    p1=[itex](\frac{12}{6.1+r2})^2[/itex]r2

    p2=[itex](\frac{12}{6.1+rT})^2[/itex]rT


    I think no solutions.
    For p1=p2, there are 2 unknowns,only one equation.
     
    azizlwl, Sep 15, 2012
  13. Sep 15, 2012 #12

    gneill

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    Let R stand for whatever the total load resistance is (it'll be R2 in one case and R2||R3 in the other). You've found an expression for the current to be:

    ##I = \frac{E}{Rs + R}##

    If the power is ##I^2 R##, then

    ##P = E^2 \frac{R}{(Rs + R)^2}##

    Clearly, if you equate two powers of the above form then the ##E^2## factors cancel on both sides and you're left with an equation involving only resistances.

    Call the resistances for the two cases Ra and Rb. Equate the expressions and solve for Rb in terms of Ra (shouldn't be too difficult). Then plug in R2 for Ra and R2*R3/(R2 + R3) for Rb into whatever solutions you found. Check for consistency --- one or more of the solutions may end up being impossible once this final constraint is imposed.
     
    gneill, Sep 15, 2012
  14. Sep 15, 2012 #13

    fawk3s

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    Was making a lot of small mistakes along the way, but finally got it to be R2=Rs2/R3. Is this what you got as well?
    Because the task is on the internet, and when choosing a random R2 and calculating R3 from the given equation, the program tells it to be wrong.
    For example, when choosing R3 to be 3,75, we get that R2=22,1 (approx). The program doesnt accept it. But it accepts 25,5 for example. So am I wrong or is the program rounding the I and V values on the meters to much? (Because it does round them quite a lot, and sometimes even wrong when I check it on the calculator.)
     
    fawk3s, Sep 15, 2012
  15. Sep 15, 2012 #14

    gneill

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    Two unknowns with one equation means that there is a relationship between the two unknowns, in other words, a family of solutions. Pick a value for one unknown and the other one is thereby fixed (i.e. a pair that satisfies the conditions).
     
    gneill, Sep 15, 2012
  16. Sep 15, 2012 #15

    gneill

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    No, that's not what I ended up with. Hint: I found the path to solution less complicated by solving for R3 in terms of R2.

    Starting with the desired equality:

    ##\frac{Ra}{(Rs + Ra)^2} = \frac{Rb}{(Rs + Rb)^2}##

    rewrite as:

    ##\frac{Ra}{(Rs + Ra)^2} - \frac{Rb}{(Rs + Rb)^2} = 0 ##

    Combine the fractions. Note that you only need to deal with the numerator, since the expression is set to zero. You may find that factoring the expanded numerator will be enlightening :wink:
     
    gneill, Sep 15, 2012
  17. Sep 15, 2012 #16

    fawk3s

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    Thats odd. I basically did the same thing with P=V*I, only thing is I did it with R2 and R3 right away, and R2=Rs2/R3 popped out.

    Did you get that (Rb-Rs)2/Rs=R2, aka R2=[(R2R3/R2+R3)-Rs]2/Rs ? Because if you didnt, then Im too lazy to start with getting R2 or R3 from there again.
     
    fawk3s, Sep 15, 2012
  18. Sep 15, 2012 #17

    CWatters

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    I've thought about it some more...

    Consider a graph of power vs load resistance..

    If the load resistance is zero the power would be zero (because V=0)
    If the load resistance is infinite the power would also be zero (because I=0)
    Max power occurs when the load resistance is the same as the source resistance.

    In other words the curve has a peak at Rl=Rs and is roughly (very roughly) n shape.

    Any horizontal line on the graph (<Rs) would represent constant power and crosses the curve at two points corresponding to equal power. Lots of possible horizontal lines can be drawn as long as RL<Rs

    There are multiple solutions unless one resistor value is known. So the answer will be an equation for the value of one in terms of Rs and the other.
     
    CWatters, Sep 15, 2012
  19. Sep 15, 2012 #18

    gneill

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    I obtained Rb = Rs2/Ra from the equation as a viable solution. Thus

    ##\frac{R2\;R3}{R2 + R3} = \frac{Rs^2}{R2}##

    Solve for R3 in terms of R2.
     
    gneill, Sep 15, 2012
  20. Sep 15, 2012 #19

    azizlwl

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    From above equation 1 and 2,

    r2=rT

    [itex]r2=\frac{r2r3}{r2+r3}[/itex]

    [itex]\frac{r3}{r2+r3}=1[/itex]

    r2=0, and r3 can be any value.
    So the power equal to zero, irrespective of the value of r3.
     
    Last edited: Sep 15, 2012
    azizlwl, Sep 15, 2012
  21. Sep 15, 2012 #20

    fawk3s

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    I got the same answer as you did now. Took me long enough though. Made so many typos in so many places you could compare the amount to infinity. :shy:

    But I was amazed at how I got R2=Rs2/R3, putting R2 and R2R3/R2+R3 right into the equation instead ob Ra and Rb. So I went back again and found out that I was supposed to get

    R24R3-R23Rs2-R22R3Rs2=0

    but got

    R24R3-R23Rs2=0

    simply because there were so many R's and for some reason I had crossed the last one out of the equation. And from the latter you get that R2=0 or R2=Rs2/R3
    Interesting, isnt it? :bugeye:

    Anyways, thank you for your patience, gneill. I appreciate it. Its weird it took me so long, never been so wrong doing math before.
     
    fawk3s, Sep 15, 2012
  22. Sep 15, 2012 #21

    gneill

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    Glad I could help.

    I think we've all had math problems that flummoxed us due to simple brain farts (recurring repeatedly until the Clue Fairy finally takes pity and smacks one upside the head). Sometimes it just requires putting the problem down for an hour or so and then taking a another look :smile:
     
    gneill, Sep 15, 2012
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