Finding 2 resistances in which case the power doesnt change

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
R_whole=R+R_s
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

 

P₁=P₂
V₁I₁=V₂I₂
V₁/V₂=I₂/I₁
I₂/I₁=(E/R_s+R₂) / (E/R_s+R_whole)

R_whole=R₂R₃/R₂+R₃

But dont really know where to go from here.

 

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

 

You can use these for the current.
I₂=E/R_s+R₂
I₁=E/[R_s+(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 R_s with their given values.

Did you get an answer though, gneill?

Thanks in advance,
fawk3s

 

V₂=E-R_s*I₂ and V₁=E-R_s*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

 

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 R_s 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

 

Was making a lot of small mistakes along the way, but finally got it to be R₂=R_s²/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.)

 

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=R_s²/R₃ popped out.

Did you get that (R_b-R_s)²/R_s=R₂, aka R₂=[(R₂R₃/R₂+R₃)-R_s]²/R_s ? Because if you didnt, then Im too lazy to start with getting R₂ or R₃ from there again.

 

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=R_s²/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-R₂³R_s²-R₂²R₃R_s²=0

but got

R₂⁴R₃-R₂³R_s²=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=R_s²/R₃
Interesting, isnt it?

Anyways, thank you for your patience, gneill. I appreciate it. Its weird it took me so long, never been so wrong doing math before.