- #1

- 138

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter discoverer02
- Start date

- #1

- 138

- 1

- #2

dduardo

Staff Emeritus

- 1,891

- 3

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

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

- #3

- 138

- 1

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.

- #4

dduardo

Staff Emeritus

- 1,891

- 3

.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

- #5

- 138

- 1

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.

Share: