Calculating Equivalent Resistance in a Complex Circuit

[discoverer02]

I'm stumped by this problem.

There's no symmetry in the circuit, so I can't ignore the 1 ohm resistor that bisects the triangle. Is there something else I'm missing? I'm supposed to show the Rab = 27/17 ohms.

Please see the attached diagram.

Thanks.

 

[dduardo]

You have to apply a delta-wye transformation. See attached picture to see what I mean.

a = A*B / ( A + B + C )
b = A*C / (A + B + C )
c = B*C / (A + B + C)

Once you get a, b and c, then...

b is in series with the 1ohm resister
c is in series with the 5ohm resister

These two series are parallel with each other and in series with a

 

[discoverer02]

Hmmm... This technique isn't mentioned in my Physics book.

a, b, and c are in ohms? a = 3/5, b = 1/5, c = 3/5.

so b + 1 ohms = 6/5 ohms
c + 5 ohms = 28/5 ohms
The equivalent of these parallel resistors is 85/84 ohms + 3/5 ohm
= 677/420 ohms.

This doesn't agree with the answer in the book which is 27/17 ohms.

The book has been known to be wrong.

Can anyone else verify?

Thanks.

 

[dduardo]

the answer does come out to be 27/17 ohms
.check 85/84 ohms + 3/5 ohm

Here are the steps for people interesting in knowing how to get the answer.:

a = A*B / ( A + B + C )
a = 1*3 / ( 1 + 3 + 1)
a = 3/5 ohms

b = A*C / ( A + B + C )
b = 1*1 / ( 1 + 3 + 1)
b = 1/5 ohms

c = B*C / ( A + B + C )
c = 3*1 / ( 1 + 3 + 1)
c = 3/5 ohms

The circuit should look like this after the transformation
.......b=1/5ohm...1ohm
...a=3/5ohm...---XXXX-----XXXX---
---XXXX----|....... |----
...... ---XXXX-----XXXX---
.......c=3/5ohm...5ohm

please ignore the periods ( . ). They are there for spacing.

The two resistors on top are in series and combined equal 6/5 ohms

The two resistors on the bottom are in series and combined equal 28/5 ohms

so now you have a circuit that looks like this

......6/5ohms
3/5ohms...---XXXX----
--XXXX--|......|-----
......----XXXX----
......28/5 ohms

The top and bottom resistors are in parallel. The equation is R1*R2/(R1+R2)

So the combined parallel resistors equals 84/85 ohms

finally to have a circuit that looks like this


...3/5ohms...84/85ohms
----XXXX-----------XXXX-----

The final two resistors are in series so just add them up

3/5ohms + 84/85ohms = 27/17ohms

 

[discoverer02]

Thanks. You're right. I goofed with the math.

I was able to figure it out the hard way going through the three loops to get three equations using Kirchhoff's rules.

I = I1 + I2 => for the currents entering the the various junctions in the network.
E = electromotive force attached to the network.

5I1 - I2 = 3I
7I2 - I1 = 5I
E - 1.0ohm(I1) - 1.0ohm(I2) = 0

Since E = I(R equivalent) if you do the algebra with the equations above you get E = (27/17ohm)I

Thanks for the help and the shortcut. They're much appreciated.