# Find potential between A and B

1. Dec 4, 2008

### Altairs

1. The problem statement, all variables and given/known data
Find potential between A and B

2. Relevant equations

3. The attempt at a solution
The R1 is confusing me. If it hadnt been there simple voltage divider would have taken care of things quite easily...I cant make logic of how to solve it now...How do I deal with R1 ?

File size:
5.7 KB
Views:
74
2. Dec 4, 2008

### The Electrician

What methods of circuit analysis have you studied so far?

Do you know how to do loop, mesh or node analysis?

Do you know about Thevenin equivalents?

3. Dec 4, 2008

### Altairs

Everything except the Thevenin Equivalents. The circuit I have attached is the result of applying superposition principle. Actually there was another Voltage source (V') connected between R3 and the common joint of R3,R4 and R7, with positive towards the right. I had to find the voltage across AB (One Port). SO by using the superposition principle and suppressing the V' this is the circuit I got. I applied loop analysis one by one i.e. once on this circuit and then by suppressing the V source but I don't get the right answer. And to make matters worse when I apply the loop analysis without suppressing any of the sources I get a completely different answer.

The values given are :-
V = 18V
V' = 13V
R1=R2=R3=3ohm
R4=R5=6ohm
R6=20ohm
R7=10ohm

1. Suppressing the V' then current through R7 = 0.09A
2. Suppressing the V then current through R7 = 0.571A
3. Solving without superposition then current through R7 = 0.125A

4. Dec 4, 2008

### The Electrician

I would be better able to help you if you will post the original circuit, including the V' source.

5. Dec 4, 2008

### Altairs

here is the original circuit

File size:
6.7 KB
Views:
61
6. Dec 4, 2008

### The Electrician

I won't be able to see the jpg until it gets approved, which could take hours.

and post the URL here, I will be able to get it sooner.

7. Dec 4, 2008

8. Dec 5, 2008

### The Electrician

Ok, I got it. This problem is a natural for the loop method. You don't need to suppress any voltage sources. I labeled the loop currents as I1 for the bottom left loop, I2 for the top loop, and I3, I4 for the next two loops to the right of the I1 loop.

Can you write the loop equations for those loops and show your work?

Just to inspire you, I got a current of 1.5 amps in R7, and that makes 15 volts at the A-B port. I have been known to make mistakes though. :-(

9. Dec 5, 2008

### Altairs

For loop 1:-

$$3(i_1 + i_4) + 6(i_1 - i_2) = 18$$

For loop 2:-

$$6(i_2 - i_1) + 3(i_2 + i_4) + 20(i_2 + i_3) = 13$$

For loop 3:-

$$6(i_3 - i_4) + 20(i_3 + i_2) +10i_3 = 0$$

For loop 4:-

$$3i_4 +3(i_4 + i_1) + 3(i_4 + i_2) + 6(i_4 - i_3) = 13$$

10. Dec 5, 2008

### Altairs

11. Dec 5, 2008

### The Electrician

You have it exactly right. Your equations give the same currents I got, even though I selected my loops slightly differently.

What currents do you get when you solve the system?

Last edited: Dec 5, 2008
12. Dec 5, 2008

### Altairs

I get i3 =0.125

13. Dec 5, 2008

### The Electrician

How are you solving it? I get:

I1 = 3.9333
I2 = 2.4
I3 = -1.5
I4 = -1.0

14. Dec 5, 2008

### Altairs

Simlifying the equations then making i1 as the subject from last equation and substituting it in two of above and then solving simultaneously to get i2 , i3 and i4....must be some algebraic mistake

15. Dec 5, 2008

### The Electrician

I used a matrix solver after setting up the matrix and excitation vector like this; just collect the coefficients for each current:

Code (Text):
[  9  -6   0  3 ]  [ I1 ]     [ 18 ]
[ -6  29  20  3 ]* [ I2 ]  =  [ 13 ]
[  0  20  36 -6 ]  [ I3 ]     [  0 ]
[  3  3   -6 15 ]  [ I4 ]     [ 13 ]
then use Matlab, or a calculator with matrix capabilities to solve it.

If you have the matrix set up properly, this method guarantees no algebra mistakes! :-)

Given the magnitude of the voltage sources and the resistor values, I think .125 amps is definitely too small.

16. Dec 5, 2008

### Altairs

which loops did you make?

17. Dec 5, 2008

### The Electrician

I used yours.

18. Dec 5, 2008

### The Electrician

Look at your 2nd equation, for example. You can collect terms and rewrite it as:

-6*I1 + 29*I2 + 20*I3 + 3*I4 = 13

and similarly for the others. Thats how you get the entries in the matrix.