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Junction Rule and Loop Rule

  • Thread starter 4real4sure
  • Start date
26
0
For the network shown, show that the resistance Rab = (27/17)ohms .

http://www.webassign.net/pse/p28-28.gif
Figure P28.28
My attempt to the problem:
I got the first following equations:
I1-I3-I2=0
I4+I6-I1=0
I2+I5-I6=0
I3-I4-I5=0

Note that I3 and I4 are currents flowing at the top
I2 and I6 are flowing through 3 and 5 ohm resistances
I1 is connected to a battery at tyhe left side of I2
Any help will be highly highly appreciated!!!
 

tiny-tim

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welcome to pf!

hi 4real4sure! welcome to pf! :smile:

(hae an omega: Ω and try using the X2 icon just above the Reply box :wink:)

what are your equations involving the amounts of the resistances? :smile:
 
26
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they are :
10-I3-I4=0
-I3-I5+3I2=0
-I4+5I6+I5=0
10-3I2-5I6=0

I am just not sure about these equations... Please can you check them out and give your precious precious advice.
 

tiny-tim

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hi 4real4sure! :wink:
I1-I3-I2=0
I4+I6-I1=0
I2+I5-I6=0
I3-I4-I5=0
yes, these are fine (except you don't need I1)

but will be easier to use them if we summarise them as:

I2 + I3 = I4 + I6

I6 - I2 = I5 = I3 - I4
10-I3-I4=0
-I3-I5+3I2=0
-I4+5I6+I5=0
10-3I2-5I6=0
no, you can't have loop equations involving I1, because you don't know the voltage supplied

you can only have loop equations for the two loops in the diagram

(actually there are three loops, the left one, the right one, and the joint one, but there's only two independent loop equations)

so the correct ones are the second and third one …

I3 + I5 = 3I2

I4 = I5 + 5I6

carry on from there :smile:
 
26
0
Thank you so much!! But whne I am trying to find the values of the current the first three are giving me a wierd value which equals infinity..
 

tiny-tim

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show us :smile:
 

tiny-tim

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4real4sure said:
Actually when I am finding the values of the current using MATLAB I am getting
NaN
NaN
NaN
0
0
then don't use MATLAB :wink:

do it yourself! :smile:
 
26
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can you kindly check your inbox please
 
26
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I don't think its possible without using any tool... I have tried
 

tiny-tim

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substitute from one of the loop equations into one of the junction equations …

what do you get?​

(and please reply on the forum, not by pm :frown:)
 
26
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Okay but I ma confused about one thing.. I am not using emf here when I am actually supposed to use it in loop rule
 

tiny-tim

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ah, you only have to use emf if there's an emf in the loop :smile:

lots of loops have no emf …

(in fact, we often prefer loops with no emf, since they're simpler to calculate!)

it's nothing to worry about! :wink:
 
26
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That doesn't make sense because if I have no voltage then my current should be zero.
How can you have current with no emf
 

tiny-tim

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take the left loop …

from the left end to the bottom right, there are two ways to go …

diagonally down, or right and then down …

there's no emf in that loop, but current is still flowing along both paths because of the potential difference between the two points …

what https://www.physicsforums.com/library.php?do=view_item&itemid=93" say for that circuit is that the sum of the potential differences is zero if you go round the same way, ie if you go out along one path and back along the other path …

(and that's the same as saying that if you go out along one path, the potential difference is the same as if you go out along the other path! :wink:)
 
Last edited by a moderator:
26
0
Okay honestly, I am completely stumped by the algebra of those five equations.. I am trying since yesterday and I am getting more and more confused while substituting
 

tiny-tim

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Okay honestly, I am completely stumped by the algebra of those five equations..
which five equations?

can you please set them out again, so we can be sure they're the right equations :wink:
 
26
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Here they are:
I3 + I5 = 3I2
I4 = I5 + 5I6
I2 + I3 = I4 + I6
I6 - I2 = I5
I3 - I4 = 15
 

tiny-tim

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ok

first, you can ignore the fifth equation because it equals the difference between the third and fourth equations

now take I5 from the fourth equation, and substitute it into the first two equations :wink:
 
26
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Okay now I have a new equation... 3I2 = I4+I3... don't we have any new software for calculating this type of simultaneous equation
 
26
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my fault.. its 23I2 = I4+6I3
 

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