# Circuit Analysis

#### gicode0823

Joined Feb 8, 2012
31
I just learned Kirchoff's law and Ohm's law and as a homework problem I got this circuit diagram.

I'm just confused on how to approach this circuit. Maybe confused on dependent/independent sources... ?

since I didn't learn the nadal analysis method or mech method just yet, I can't use them on this problem.

What I did was write all the KVL, KCL, and ohm's law and trying to substitute into equations that I got from KVL, KCL, and Ohm's law.

Any suggestion?

Circuit: Find Vx #### t_n_k

Joined Mar 6, 2009
5,447
This is rather a complex problem. Perhaps too much for a newcomer to circuit analysis. Unless I'm missing something completely obvious.

Effectively you are being asked to find the Thevenin equivalent of the circuit.

The Thevenin resistance [Rth] is relatively easy to find - by inspection. [See EDIT below]

Rather than trying to solve directly for the unknown open circuit value of Vx - I would then suggest you make an assumption that a pure voltage source V'x is placed across the open terminals and that this source value sets the current iQ to 1A.

You can deduce what that source voltage V'x will be, since the unknown sources and control parameters then have solvable values based on an assumed value of iQ=1A. Once you deduce the controlled current source value in particular, you then deduce the applied V'x. From there you may also find the current drawn from V'x.

At that stage you assume a Thevenin model for the entire circuit with the deduced source V'x across the output terminals. Given you know V'x, Rth and the current draw it remains only to then solve for Vth - which will be the previously unknown open circuit voltage Vx.

Another approach may be to short the terminals Vx [making Vx=0] and then finding the short circuit current [Isc] draw from the Thevenin equivalent. Vx=Vth=Isc*Rth

EDIT: Re the Thevenin resistance - on reflection that's not a trivial matter to find by simple inspection. So one might have to 'solve' the problem twice by proposing another case with iQ=2A say.

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• gicode0823

#### gicode0823

Joined Feb 8, 2012
31
Thanks, I will try that. Only concern is that I won't get full credit for this problem if I use the Thevanin method since I haven't learn them yet.

#### t_n_k

Joined Mar 6, 2009
5,447
Thanks, I will try that. Only concern is that I won't get full credit for this problem if I use the Thevanin method since I haven't learn them yet.

That's probably the least of your worries.

Your prof is a tough problem setter!

Unless you've a lot of experience in circuit solutions it's asking a lot of you to do this problem.

In any case you will need to show some working to get any serious help for your questions.

#### gicode0823

Joined Feb 8, 2012
31
Quick question:

1. When the independent voltage source (Vs4 = 15) is connected with resistor (4 ohm) in series as in diagram, then can I say that voltage across resistor should be 15 volt?

2. By using KCL on closed surface right next to the independent voltage source, Vs2, the current going through the voltage source should be zero (left side of the voltage source)? If so, does the right side of the voltage source still should have zero current?

3. When the circuit like above is not closed, only current provider such as independent/dependent current source can provide the current to circuit?

4. Does the current change when going though the independent/dependent voltage source? (I don't think it does but I wanted to make sure)

#### t_n_k

Joined Mar 6, 2009
5,447
1. No

2. Yes

3. Current can only flow where there are closed loops.

4. No

#### gicode0823

Joined Feb 8, 2012
31
Forgot to mention, prof. said something about making simultaneous equations and solve them.

#### steveb

Joined Jul 3, 2008
2,436
Forgot to mention, prof. said something about making simultaneous equations and solve them.
I don't know if my comment will be helpful or hurtful, so if you find it confusing, then just ignore it. I agree this is a bit of a difficult problem. Sometimes on the tough ones you need to just grind through it. So, writing the equations out and then solving them is a good way. Along the way, you should see some simplifying relations since current can't flow across the gap of Vx. As a hint, the current through the 4 ohm resistor on the left is zero.

Anyway, my comment is that there are two ways to set up simultaneous equations. The usual way is to use traditional nodal or loop analysis in a formal way. You identify the number of nodes or loops and write out the relevant equations. However, there is another heuristic method. Figuratively, you can think of it as weaving the equations as you need them. You start by writing an expression for the variable you want (in this case Vx) and then work backwards until you have equations for all unknowns and have included all of the input independent sources.

I can start you as follows.

Vx=Vxp-Vxm, where Vxp is the voltage on the left/positive side and Vxm is the voltage on the right/minus side.

In the above equations you have two unknown variables, so you try to write an expression for each of these. I'll do the Vxp for you and I'll leave the more difficult Vxm for you.

First, I'm going to define a ground point to be at the negative terminal of Vs4. This node then has a potential of 0V. You could pick any other node, but this looks convenient to me.

Vxp=15-3Vq

This now leaves Vq as an unknown, but it is on the other side and the relation for it will become apparent when you solve that side.

Now, why is it Vxp=15-3Vq? Well, since no current flow across the gap of Vx, no current can flow through the 4 ohm resistor. Hence, there is no voltage drop across the 4 ohm resistor.

On the left, you can also establish the relation Iq=-3Vq/2, which will be useful on the right hand side for Vs3

Now, you can figure out a relation for Vxm. You can note that the current through Vs2 is zero and you can note that the current through the 5 ohm resistor is 0.56 Vx.

Once you work backwards until all unknown variables are defined and you have included all independent sources, you will have a set of linear equations that can be solved by a number of known methods. In my old age, I now just plug them into a symbolic math processor like Maxima, MuPad, Maple or Mathematica (hmmm, why do they all begin with the letter "m"?), but you should only do this once you know how to do it yourself; otherwise, you'll fail all of your exams.

Like I said, this is one method, and perhaps not the one usually taught, but it is good in these unusual problems with a mix of dependent sources. If this is too confusing a method, then ignore it and work with the usual node and loop equations and substitute in the dependent source equations as needed.

#### gicode0823

Joined Feb 8, 2012
31
Vxp=15-3Vq

This now leaves Vq as an unknown, but it is on the other side and the relation for it will become apparent when you solve that side.

Now, why is it Vxp=15-3Vq? Well, since no current flow across the gap of Vx, no current can flow through the 4 ohm resistor. Hence, there is no voltage drop across the 4 ohm resistor.
Thank you for the reply. I have couple questions.

To obtain equation Vxp = 15 - 3Vq,
did you use kvl on the left top loop? I understand that current though 4 ohm resistor is zero thus no voltage across the resistor.

And how do you know that the across 5 ohm resistor has same current as is1 ?

#### gicode0823

Joined Feb 8, 2012
31
How do I find out what voltage is across 3 ohm resistor?

Do I use kvl over Vs3, Vq and 3 ohm resistor? or Voltage divider rule apply on 3 ohm resistor and 7 ohm resistor?

Then I will get V3 = -6iq - Vq ?

Since the grounded node is selected at negative terminal of Vs4, then 5 ohm resistor and is1 has zero voltage as well?

Sorry for the many questions

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#### steveb

Joined Jul 3, 2008
2,436
To obtain equation Vxp = 15 - 3Vq,
did you use kvl on the left top loop? I understand that current though 4 ohm resistor is zero thus no voltage across the resistor.

And how do you know that the across 5 ohm resistor has same current as is1 ?
Yes, KVL is used for that equation. I defined the ground point, and then defined Vxp relative to that ground. Hence, KVL can be applied around that loop.

The current in the 5 ohm resistor must be the same as the current source since no current flows through Vs4=15 V. This voltage source is in series with the 4 ohm resistor, hence the current is zero.

• gicode0823

#### steveb

Joined Jul 3, 2008
2,436
How do I find out what voltage is across 3 ohm resistor?

Do I use kvl over Vs3, Vq and 3 ohm resistor? or Voltage divider rule apply on 3 ohm resistor and 7 ohm resistor?

Then I will get V3 = -6iq - Vq ?
I wouldn't use the voltage divider formula because current is being drawn by the current source. This will invalidate that relation.

Your other approach seems fundamentally sound.

Since the grounded node is selected at negative terminal of Vs4, then 5 ohm resistor and is1 has zero voltage as well?
I'm not following you here. If you are saying that the 5 ohm resistor has no voltage drop across it, then no that is not correct.

• gicode0823

#### gicode0823

Joined Feb 8, 2012
31
I'm not following your method completely. Is there name for that approach? But here is my trial.

Vxn = 12 + 6Iq +V3 - Vq +Vq

According to the KVL withing the loop (Is1, 7ohm, 5ohm), there should not have any voltage drop across the dependent current source, thus voltage across 5 ohm should be +Vq with respect to Vxn.

By using the KVL over the right loop I get:
6Iq + Vq - V3 = 0
V3 = -6Iq - Vq

Now substitute into the Vxn equation
Vxn = 12 + 6Iq + (-6Iq - Vq)
Vxn = 12 - Vq

Since Vxp = 15 - 3Vq and Vx = Vxp - Vxn

Vx = (15 - 3Vq) - (12 - Vq)
Vx = (15 - 3Vq) - 12 + Vq
Vx = 3 - 2Vq

Now I'm stuck here...
Vq = -7 * I7

Using KCL at the right hand corner node

I7 - I5 = 0

I5 = I7

I5 is .56Vx then

.56Vx = I7

Now substitute into Vq

Vq = -7 * .56Vx

And substitute into the Vx
Vx = 3 - 2(-7 * .56Vx)
Vx = 3 + 14 - 1.12Vx
2.12Vx = 17
Vx = 8.018867925 [V]

#### steveb

Joined Jul 3, 2008
2,436
I'm not following your method completely. Is there name for that approach?
I'm not aware of a name for this approach, but it is a key part of the traditional signal flow graph approach. Basically, the way I'm suggesting deriving the equations is similar to the way you make a signal flow graph. However, I think it is too big a jump to go into the details of this.

The method itself is basically a "define as you go" approach. Personally, I derive all circuit equations this way, but many people don't like it. I wouldn't have mentioned it for someone just starting on circuit analysis, but for the fact that your teacher has given you a challenging problem that is particularly well suited for this approach. Like I said above, if you find it too confusing, then don't struggle with it. However, you seemed to be stuck, and this seemed like a way to get you unstuck.

But here is my trial.

Vxn = 12 + 6Iq +V3 - Vq +Vq
Which loop are you trying to apply KVL too with this equation? I don't follow this at all.

My suggestion is to use the following formula, derived by going from ground, through the 5 ohm resistor, through Vs3 and then through Vs2 and back down to ground through Vxn.

Vxn = 12 + 6Iq - 5*0.56*Vx

The last term is a result of the fact that the current source forces it's current through the 5 ohm resistor.

Then remember that Iq=-3Vq/2 which gives you the following.

Vxn = 12 - 9Vq - 5*0.56*Vq

I suggest retrying with this approach. However, remember that I am also human and can still make a mistake. So, double check my suggestions and make sure you agree with what I'm suggesting.

I didn't follow why you say there is no voltage on Is1. I don't think that is correct.

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#### gicode0823

Joined Feb 8, 2012
31
To obtain the Vxn, I use KVL to find them. I thought I had to go though each components.

If I choose to use KVL over outer loop (excluding 7 ohms and Is1), what happens to the components that are excluded?

And why the 12 [V] is included in the KVL? Is this means that 12V is powering up the circuit and terminates at the ground point?

I'm confused about how each components behave. To find the behavior, do I use KVl and KCL to find which way current is going and what's powering which components and such?

For the Vxp, if KVL is used, why does the 15V included? I thought KVL states that sum of all the voltage across each component within a loop is equal to zero?

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#### gicode0823

Joined Feb 8, 2012
31
By using Vxn as Vxn = 12 + 6Iq + V3 - 5*0.56Vx

Which I can simplified into
Vxn = 12 - 9Vq + V3 - 2.8Vx

Since V3 = -6Iq - Vq and Iq = -3Vq/2

V3 = -6(-3Vq/2) - Vq
= 9Vq - Vq
= 8Vq

Vxn = 12 - 9Vq + 8Vq - 2.8Vx
= 12 - Vq - 2.8Vx

Vx then

Vx = Vxp - Vxn
= (12 - 2Vq) - (12 - Vq - 2.8Vx)
= -Vq + 2.8Vx

-1.8Vx = Vq
Vx = -Vq/1.8

Using KVL,
-Vq + V5 - Vo = 0
-Vq = -V5 + Vo

since V5 = 5(.56Vx) = 2.8Vx

-Vq = -2.8Vx + Vo
Vq = 2.8Vx - Vo

Now define Vo using KVL
-Vo + V3 + 6Iq + V5 = 0
-Vo + (8Vq) + 6(-3Vq/2) + 2.8Vx
-Vo = -Vq + 2.8Vx
Vo = Vq - 2.8Vx

Now substitute into Vq equation above
Vq = 2.8Vx - Vq + 2.8Vx
2Vq = 5.6Vx
Vq = 2.8Vx

Now substitute into Vx equation which was
Vx = -Vq/1.8
Vx = -2.8Vx/1.8

which both Vx cancels...

What am I doing wrong!? This problem is very frustrating...

#### steveb

Joined Jul 3, 2008
2,436
I'm sorry. I had a typo in my KVL formulas. There should not be a V3 in the relation. I corrected it above. I'm in a remote location with an iPhone, so it is hard to type and check. I'll have to wait till I get to my computer to give better answers.

• gicode0823

#### gicode0823

Joined Feb 8, 2012
31
Thank you so much for taking time to answering my questions.

#### t_n_k

Joined Mar 6, 2009
5,447
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.

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#### steveb

Joined Jul 3, 2008
2,436
t_n_k has a nice approach and if you find that one more intuitive then by all means go with it.

I'll go a step further in the way I was suggesting just in case you also want to consider alternatives. In the end you should basically get the same equations by any approach.

First, I'll summarize what I've mentioned above, trying to eliminate my unfortunate typo which no doubt just added confusion.

We have the following.

Vx=Vxp-Vxn (note that Vxp and Vxn are unknown so far)
Vxp=15-3Vq (note that Vxn and Vq are unknown so far)
Vxn=12-9Vq-5*0.56*Vx (note that only Vq is unknown so far)

This is where we left off and you should see that if you can get one more relation to define Vq, you will have defined Vx completely.

There are a few ways, but this is your challenge to find any way, and then you only need to solve a linear set of equations to get the value of Vx.

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

Notice how this equation has Vq and Vx which provides a way to define Vq in terms of Vx, and Vx has been defined already. Hence you are done.

Now, there are a few steps to derive the above equation using KCL at the node joining the 3 ohm and 7 ohm resistor. Can you figure that part out?

I expect this looks complex because you are still developing intuition on how to approach a problem like this, but the main point here is to understand the method. You start at your output variable and use valid relations to work back to your known input sources. Also, always keep in mind that there are two steps to solving these circuit problems. The first step is to identify the correct linear equations for the system. The second step is to solve those linear equations. These steps are completely independent and can be isolated from each other. The idea of "divide and conquer" can sometimes help you psychologically.

Both t_n_k and I have probably given too much of the answer away, but this is such a tricky problem and you appear to be making an effort, so I think it is ok. In the end you need to fully understand your chosen method and all the steps involved. Neither one of us has shown all steps, so you need to fill in the gaps. I worked the equations through and get the answer that you claim is correct. Vx=-6.4V