I have this circuit problem

  • #1

Homework Statement



I have this circuit problem, The Resistance is not given yet the Current is 10A and Voltage is 220V
my professor told me that A is the only series, A's value could be = to 3. D to Y is 1 over ? and B to C is also 1 over ? it's really hard but here's the picture link of the circuit.

He even told me that if you add D to Y and B to C it should be = to 1 over 19
can someone give me the value of Ra to Rz ? with a little bit explanation on how you come up
to the answers.

https://fbcdn-sphotos-e-a.akamaihd.n...72073306_n.jpg [Broken]

thank you~

Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
6,054
391


The image is not accessible. The link to it is messed up.
 
  • #3


sorry~

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash4/s480x480/488389_518468254845450_1272073306_n.jpg
is it accessible already?
 
  • #4
gneill
Mentor
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So, you want to solve for 26 unknowns given only two known values (voltage and current)? That ain't gonna happen!

I think you'll have to make some "creative" assumptions (i.e., cheat :smile: ).
 
  • #5


My other classmates gave up too. Some of them says that it's impossible / hard to answer this circuit if the R = Resistance is not given. Since Resistance could be any value in this situatuion.

I'm hopeless. :)
 
  • #6
6,054
391
Since no other conditions a given, you can approach this pragmatically: do whatever it takes to solve it with the least effort. So you could just say: let them be all equal. What is the value then?
 
  • #7
gneill
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Since no other conditions a given, you can approach this pragmatically: do whatever it takes to solve it with the least effort. So you could just say: let them be all equal. What is the value then?

Still too much work in my opinion. I've got a devious mind; There's another possible approach that would make for a trivial solution :wink:
 
  • #8
6,054
391
Well, you could make some R's zero. But that could be considered, like you said, cheating. But that is a valid approach if R's can really be ANYTHING.
 
  • #9
gneill
Mentor
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Well, you could make some R's zero. But that could be considered, like you said, cheating. But that is a valid approach if R's can really be ANYTHING.

I'd say use whatever latitude the problem statement does not specifically forbid or constrain. It's the problem author's responsibility to clearly define the parameters and limits of the problem, and if he leaves the door wide open to trivial solutions based on convenient assumptions for unspecified values, then take advantage of it.

If there is nothing in the problem statement that forbids making some or all of the values zero, go for it and save the brain sweat.
 
  • #10
I guess its a really hard problem isn't? Well then, thank you guys so much for dropping by here. At least your presence made me believe that even me, I can't answer it too.

And thanks for your time again.. :)
 
  • #11
CWatters
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I don't think it's hard. Just missing some information.

When the prof said A "could be 3" perhaps he was just suggesting you just pick non zero values that give a workable solution. eg to show you know how to solve series and parallel circuits.

He may even be hoping that you manage to do it with standard/preferred values :-)
 
  • #12
SammyS
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What if you assume that Ra - Rz all have the same value?
 
  • #13
2,010
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Obviously, your teacher wants the full set of solutions to
a+(b+c)(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j)/(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)+((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j))+d+g)/(b+c+(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j)/(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)+((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j))+d+g)) = 22

can't be too hard.
 
  • #14
6,054
391
  • #15
Obviously, your teacher wants the full set of solutions to
a+(b+c)(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j)/(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)+((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j))+d+g)/(b+c+(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j)/(((q+r+s+tu/(t+u)+z)(m+n+op/(o+p)+vw/(v+w))/((q+r+s+tu/(t+u)+z)+(m+n+op/(o+p)+vw/(v+w)))+x+y)+((e+f)(h+i)(k+l)/(e+f+h+i+k+l)+j))+d+g)) = 22

can't be too hard.

Yes~

that's what he said to us. B to C and D to Y value should be equal to 1 over 19 since it's the reciprocal of 19. So, if A's value is 3. 19 + 3 you'll come up with 22.

1/19 = 1/? + 1/?
or
1/19 = BtoC + DtoY
 
Last edited:
  • #16
NascentOxygen
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Setting all resistances equal would make this a good quick exercise in series/parallel analysis. ❪It's almost obvious that that's what should be intended. :wink:
 

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