I see that you took the outer route for the Vxn but is there reason why you chose that route? Just chosen this route for the convenience of define Iq in previous step? or always has to take outermost route?

We saw that the current through the 5 ohm resistor is the same as the dependent current source, so I5=0.56*Vx. The current node equation at the node that ties the 3 ohm and 7 ohm resistor is

Vq/7=0.56*Vx+8*Vq/3

hmm... completely lost on this one...

 

It doesn't surprise me that your frustration is rising.

I've no doubt that Steve would encourage you to be very methodical and careful in formulating the solution. The likelihood of making an error which carries through is very high.

I'm attaching an incomplete solution approach which differs from Steve's but which might give some other insights. I strongly suggest you don't simply copy / extend it and then submit it for marking. It's highly likely your professor or your classmates are aware of the AAC homework site and possibly this thread.

I've not included the final correct answer - it should be up to you to find that.

Once you have studied the Thevenin equivalent in class you might find it interesting to find the Thevenin equivalent for this circuit. The result is most surprising.

I understand that Vx, Vq, and Iq as a unknown that I need to find but Ia is not that obvious to me. I guess I don't have skills to tell which unknown I must know to solve the circuit just yet...

And about the Thevenin equivalent, I can't wait to find that out.

 

I see that you took the outer route for the Vxn but is there reason why you chose that route? Just chosen this route for the convenience of define Iq in previous step? or always has to take outermost route? ...

The reason why I thought to suggest this method is because it does not require you to ask "why?" in choosing the next or best equation to use. You are allowed to use any valid relation. The main stumbling block is that it is easy to accidentally use the same relation twice. With experience, you just intuitively know how to proceed without doing this. As you proceed, you will gradually run out of equations. In the end you will automatically use all required equations and all relavent parameters and input sources will be incorporated.

For the relation that you can't see how to derive, use KCL at the node that joins the 3 and 7 ohm resistors. Then look at each term and figure out why each term is what it is. How do you find out? Use the method to define any unknown variables in terms of defined variables. Take as many small steps as you need. I was able to write the equation directly because I have experience. However, this method allows you to break down the steps gradually.

 

I thought I would make a comment about fundamentals. One thing we have not been able to see is how you draw your schematics. For the most part it seems that you understand KVL, KCL and Ohms law, but when solving problems like this it is critical to have conventions and methods solidly in place and to use them religiously. This is just as important for experienced people as for inexperience people. It is just too easy to make an error in signs without a formal method.

I once wrote up a quick sheet for someone else and I was really surprised when they said that this simple thing helped them tremendously. So, I've attached it. Basically, the whole point of it is to rigorously define your current and voltage directions beforehand (as well as unique variable names), and then apply KVL with predefined loop directions. This just minimized the chance of mistake.

This sheet is for KVL, but KCL is simple once the current directions are defined. Simply sum the current arrows going in to the node and set it equal to the sum of current arrows going out of the node.

I know you know this already, but it doesn't hurt to stress fundamentals and conventions. You can set up this "machinery" to minimize the chance of errors in the derivations.

 

Here is how my diagram and KCL and some KVL equations. Maybe I'm making some mistake here in the beginning of problem.

 

I must been very tired yesterday night... The equation that you got at the node (3 ohm and 7 ohm) was very simple to identify.

Finally got the right answer... Thank you guys so much for the help!

I know the other way to solve this problem is to find how many unknown it has and make equations for them(8 unknown -> 8 equations needed) and use calculator to solve for the unknown using matrix function. At least that's how my friend solved it... but I can't get the solution that way for some reason. I may have mistakes on equations I made in the photo attachment above.

 

Why 8 equations?

In reality with 3 unknowns you should be able to form 3 independent equations to solve for any variable.

Here is a possible set

KVL Around the loop ...

\(V_x=-3V_Q+15+5*0.56V_x-6*i_Q-12\)

or

\(6i_Q+3V_Q-1.8V_x=3 \ \ ---(1)\)

KVL around the top left loop ...

\(2i_Q+3V_Q=0 \ \ ---(2)\)

and KCL at the 3Ω, 7Ω, is1 node ....

\(\frac{6i_Q+V_Q}{3}=0.56V_x-\frac{V_Q}{7}\)

or

\(42i_Q+10V_Q-11.76V_x=0 \ \ ---(3)\)

Equations (1), (2) & (3) are sufficient.

-->A
 A  =
 
    6.     3.   - 1.8    
    2.     3.     0.     
    42.    10.  - 11.76  
 
-->B
 B  =
 
    6.     3.     3.  
    2.     3.     0.  
    42.    10.    0.  
 
-->Vx=det(B)/det(A)
 Vx  =
 
  - 6.4009662

 

That's correct.

But my friend found 8 equations with KVL, KCL, and Ohm's law and then find how many unknowns it has, then created a matrix to compute the answer. I've tried that method but I couldn't get the right answer.

 

You can create any number of variables - presumably up to at least 8 given your friend's success.

In an earlier post I gave 5 equations in the pdf attachment which could be used. The five variables would be Ia, Ib, Vx, iQ and VQ

I can formalize the five equations as follows [per the original order in the pdf]

\(0*I_a+0*I_b+1.8V_x-4i_Q+0*V_Q=-3\)
\(0*I_a+0*I_b+0*V_x+2i_Q+3V_Q=0\)
\(-3I_a-7I_b+0*V_x+6i_Q+0*V_Q=0\)
\(I_a-I_b-0.56V_x+0*i_Q+0*V_Q=0\)
\(0*I_a+7I_b+0*V_x+0*i_Q+V_Q=0\)

These can be solved using matrix methods as before ...

-->A
 A  =
 
    0.    0.    1.8   - 4.    0.  
    0.    0.    0.      2.    3.  
  - 3.  - 7.    0.      6.    0.  
    1.  - 1.  - 0.56    0.    0.  
    0.    7.    0.      0.    1.  
 
-->B
 B  =
 
    0.    0.  - 3.  - 4.    0.  
    0.    0.    0.    2.    3.  
  - 3.  - 7.    0.    6.    0.  
    1.  - 1.    0.    0.    0.  
    0.    7.    0.    0.    1.  
 
-->Vx=det(B)/det(A)
 Vx  =
 
  - 6.4009662

So you could presumably use any voltages or currents as N variables of your choice for which you can write N independent algebraic equations each comprising at least two of the selected variables.

 

I'm working on another circuit but I don't know which equations are usable. I know that one thing I should keep in mind is to find independent equations. Im not sure what I'm doing wrong.

What I've done is making equations using kvl, kcl, and ohms law. Then I plug the ohms law information into kvl equations.

I will post my work in a bit.

 

For the first circuit, here are the my friends work.

KCL:
1. I4 + I2 - Iq = 0

2. -I4 + I5 - 0.56Vx = 0

3. -I5 + I7 + Is3 = 0

4. I3 + 0.56Vx - I7 = 0

5. -I3 - Is3 = 0

KVL:
6. 3Vq + Vx + 6Iq + 5I5 - 3 =0

7. 3Vq + 2Iq = 0

8. 3I3 + 6Iq + 7I7 = 0

Ohm's law:
9. V7 - 7I7 = 0

Unknowns are I2, I3, Is3, I4, I5, I7, Iq, Vx, Vq

Creating 9 x 10 matrix and rref gives me the answer.

Here is the new schematics:

My equations are following:
KCL:
-4 - Iq - I3 = 0

-I36 - Is1 + Ix = 0

Is4 - 12 = 0

KVL:
-9Vx - V12 - 7Iq - 5Ix + 36I36 + V4 + V15 + V7 = 0

V4 - V13 =0

V4 - Vq = 0

Vq - V13 = 0

V36 + V11 = 0

V36 - Vs1 = 0

Vs1 + V11 = 0

-V12 - Vs4 = 0

V13 + Vx + V15 - 10Vx - 5Ix - V11 = 0

I know these are not all of the KVL equations but these equations won't solve this problem...

I think the one way of checking the independence of equations is to find the determinant of matrix and should not be equal to 0.

I also attempted the other way (Steveb way)

Vx = Vxp - Vxn
Vxp = -9Iq + 11Ix + 5Ix = -9Iq + 16Ix
Vxn = -10Vx

Vx = -9Iq + 16Ix + 10Vx
-9Vx = -9Iq + 16Ix
Vx = Iq - 16Ix/9

I can't seem to find Iq and Ix for some reason...

 

Since is1 is a constant current source you may immediately find iQ since the 9Ω & 15Ω in parallel form a simple current divider.

Likewise you can easily find Vs1 and ix since Vs1 is dependent only on iQ. Knowing ix gives you Vs3.

Sources is2 and Vs4 are completely bypassed by a short circuit in the main loop and hence they play no part in the the determination of Vx.

It's then a relatively straightforward task to write the KVL equation for the complete loop which has Vx as the only unknown.

 

thanks, it was pretty easy if I knew current divider rule much earlier...

Anyway, I need one more circuit help...

For this circuit, I think I can ignore the left hand corner of the circuit since talking about Vw. And I don't think the Steveb method works here since the can't really put ground point?

No current going through 15 kΩ, Vs3 or Vs6.

The right hand corner has resistors in series which can be rewritten with
Vt = (Vq + V56) = (16.5kΩ)I

Anyway, I wrote some KVL for the circuit3 (attachment)
Outermost:
V17 + V1 + 3Vx - V19 - 22 = 0

Right top: (Vo is voltage across the Is1)
-Vo - 15 - Vx = 0

Left top:
-10 + V17 - V19 = 0

Mid top:
-12 + V1 + Vo + 4Vx = 0

Right bottom:
Vq + V56 + 20 + V62 = 0


KCL
I10 = I17

-I10 = I19

-I17 - .007Vq = 0 -> I17 = -.007Vq -> I10 = -0.007Vq

-I1 - .007Vq = 0 -> I1 = -.007Vq

First thing I did is to computation of I10 using KVL(left)
-10+V17-V19=0
-10+17I10+19I10 = 0
I10 = 5/18 [A]

Then I plugged things into KVL
outermost:
V17 + V1 + 3Vx - V19 - 22 = 0
17I10 + I1 + 3Vx + 19I10 - 22 = 0 (plug I10 = 5/18)
I1 + 3Vx = 11

Also right bottom should have same current going through each resistor since they are in series.
Right bottom:
Vq + V56 + 20 + V62 = 0
4.7I + 5.6I + 6.2I = -20
I = -40/33 [A]

But these KVL equations are not independent thus I can not be solve...

What am I doing wrong?

 

You are correct about the Vw matter - it has no bearing on the result.

You can also ignore the Vs1 loop [which includes the 17k and 19k] since that is shorted within the Vs2 / Vs3 / 1k / is1 loop and has no bearing on the result.

The essence of the problem boils down to finding the voltage drop across is1.

To find is1 you need VQ. VQ may be determined by using the voltage divider rule in association with Vs4 and the three resistor string comprising 6.2k, 4.7k & 5.6k.

Keep in mind the clue that the current in the is1, 1kΩ, Vs2 and Vs3 loop is fixed solely by the is1 source value. Where is1=0.007*VQ

You are incorrect about being able to put a ground or reference point - it simply may not add anything useful to the process.

 

On the matter of good housekeeping.....

I suggest you start a new thread for each new problem - otherwise this will become inordinately cumbersome.

 

Sorry, I will do that next time.

Quick question,
For the 15 kohm resistor, it should have zero voltage correct? since the loop is not closed?

If I calculate everything I get the wrong answer.

Vq = 5.6969 [V]

Is1 = 0.0398787879 [A]

Vx = -8.986707071 [V] as answer

but should be -4.3 [V]

 

If you consider the assigned polarity of VQ and the polarity of Vs4 it should be clear that VQ=-5.6969V