# Help with intro circuit analysis problem! Thevenin and max power prob!

1. Mar 16, 2013

### nchin

Problem #1:

What i did so far was I source transformed the left side with the Io and 4Ω into a voltage of 4Io. Then i did mesh analysis, io and i2.

the eq'ns i got are
4vx - 2I2 - 2Io = 0
10i2 - vx - 2io = 0

IM not sure if this is right but does i2 = vx/4?? because of I = V/R?

I worked everything out and I got confused.
7vx/4 = io
Vx(10/4 - 14/4) = vx

i need help please!!!!! (see attached pic)

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Last edited: Mar 16, 2013
2. Mar 16, 2013

### Staff: Mentor

Your loop equations do not appear to be correct; why is there an Io term in both? Can you show your work detailing how you arrived at them?

3. Mar 16, 2013

### nchin

i used two mesh loops. instead of i1, i used io. so the two loops are io and i2.

so after that the mesh equations would be,

For the io loop:
-4Io + 4Vx + 2(Io-I2) = 0

For the i2 loop:
-Vx + 2(I2-Io) + 2I2 + 6I2 = 0

after simplifying I get:
4Vx - 2I2 - 2Io = 0
10i2 - Vx - 2io = 0

and on the i2 loop there is a voltage drop of Vx across the 4Ω resistor so i2 = Vx/4 ??

Am i correct?

4. Mar 16, 2013

### Staff: Mentor

Io is a fixed source, and after the Thevenin equivalent conversion, it becomes a fixed voltage source of magnitude 4Io. So it is NOT a mesh current. You should choose another variable name for the mesh current in the first loop.

Yes, Vx is related to the current in the second loop. Be sure to take note of the direction of the potential drop caused by that mesh current when you write the expression for Vx; Vx's potential is 'measured' in a particular way as designated by the "+ -" designation.

5. Mar 16, 2013

### nchin

if i use another variable name then i would have 4 unknowns and and only 2 equations, io, i1, i2 and Vx? How would i solve that?

6. Mar 16, 2013

### Staff: Mentor

In this context io is a given value (granted, it's not numerical value, but it is a constant in this context). i1 and i2 are the unknown mesh currents, and Vx is strictly dependent on i2. So only the two mesh currents are unknowns.

I'm not sure where you're going with this particular analysis. Sure, you can determine the Thevenin voltage (as a function of io) by multiplying the mesh current i2 by the 6Ω resistor in the second loop, but you won't have the Thevenin resistance. How were you planning to find that?

7. Mar 16, 2013

### nchin

so would this be correct now using the mesh variables as i1 & i2?

For the i1 loop:
-4Io + 4Vx + 2(I1-I2) = 0

for i2:
-Vx + 2(I2-I1) + 2I2 + 6I2 = 0

so how would we find a numerical value of i1 and i2 if io and Vx are not a numerical value?

I assume i simplify the resistors? like for example, the 2Ω parallel with 6Ω + 4Ω....? some thing like that?

8. Mar 16, 2013

### Staff: Mentor

They look okay. Note that your Vx really is a voltage drop across a real resistor in loop two.
It can be treated as just another resistor in loop two. And you can substitute the appropriate voltage drop for Vx in loop one. Thus the "variable" Vx disappears.

9. Mar 16, 2013

### nchin

I see. so i can just substitute everywhere with Vx with I2/4?

Hmm looking over my notes I see a way to solve the Rth is Dividing the open circuit voltage with the short circuit current?

RTH = VOC/ISC?

10. Mar 16, 2013

### Staff: Mentor

Yes, sort of, if you meant I2*4, but BE SURE TO PAY ATTENTION TO THE SIGNS. Vx is NOT +4I2. Note the direction of i2 and the resulting polarity of the potential drop across the 4Ω resistor. Note also the indicated polarity for measuring Vx. Are they the same?
Yes, that's one way. It's quite a bit of work though. If I may make a suggestion, why not leave RL in the circuit, adding another (but very simple) mesh. If you solve for the voltage across RL by finding the current in its mesh, you'll have an expression from which you can easily pick out both the Thevenin voltage and the Thevenin resistance. You'll also have an expression for the current through RL which you can use to determine the power dissipated... Three birds with one stone, to paraphrase an old saying.

11. Mar 16, 2013

### nchin

i see. so then it would be, for the i2 loop:

-(-4I2) + 2(I2 - I1) + 2I2 + 6I2 = 0 ?

since the mesh loop goes through the neg side first

12. Mar 16, 2013

### Staff: Mentor

Yes. Of course, the 4Ω resistor is just a 4Ω resistor in its loop. Vx is just a measurement taken there; Vx has no effect at all on that loop, since it is only a measurement.

13. Mar 16, 2013

### nchin

i really don't know what to do with these equations. So i leave RL in there so now i have three mesh eq'ns.

After simplifying the mesh eq'ns i have:

I1:
-4Io - 18I2 + 2I1 = 0

I2:
14I2 - 2I1 = 0

I3:
6I3 - 6I2 + RLI3 = 0

what can i do now?

14. Mar 16, 2013

### Staff: Mentor

Check your equation for mesh 1; in particular the coefficient for the i1 term. In the second equation there should be a term for [/SUB]I3. The third mesh equation looks fine.

Now you want to solve for an expression for i3. Hint: if you set up the equations in matrix form and use Cramer's Rule, you can solve for i3 without having to deal with the other two...

15. Mar 16, 2013

### nchin

im not sure whats wrong with mesh 1?

i checked it and i got:
-4Io + 4Vx + 2(I1-I2) = 0
-4Io + 4(-4I2) + 2(I1-I2) = 0
-4Io - 16I2 + 2(I1-I2) = 0
-4Io - 18I2 + 2I1 = 0
??

my mistake i did forget the I3.
i now have
14I2 - 2I1 - 6I3 = 0

Last edited: Mar 16, 2013
16. Mar 16, 2013

### Staff: Mentor

What happened to the 4Ω resistor in loop 1? Here's the circuit we're dealing with:

No worries, these things happen.

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17. Mar 17, 2013

### nchin

Ah you're right! I forgot about that as well.

so for mesh 1 i have
-4Io + 6I1 - 18I2 = 0

18. Mar 17, 2013

### nchin

So i can solve i3 just from using mesh 3 equation alone?

6I3 - 6I2 + RLI3 = 0

i understand how to do 2x2 or 3x3 cramer rule but how would i do 1x1 with cramer?

19. Mar 17, 2013

### Staff: Mentor

No, set up the 3x3 impedance matrix and the voltage vector as usual, then solve for i3.

20. Mar 17, 2013

### nchin

Will I get a numerical value for I3?