Here v has a different potential than Vs relative to the ground. V1 can be expressed in terms of V, that is V1=V/3. I would like to solve this using KVL, if possible. I tried, and got this (for three loops): i*2R+v1+i3*3R-Vs=0 -4V1+i2*3R+Vs-i3*3R=0 i*2R+V1-4V1+i2*3R=0 (i being the left side current, i3 middle and i2 right). Is this KVL loop analaysis correct?
Currents and voltages are both SIGNED quantities. This means that saying that i is the left side current is incomplete since it doesn't tell us if i is the current flowing clockwise or counterclockwise through that branch. Same with the other two. Either indicate the currents, with their directions, on the diagram or give an unambiguous description in words. Don't make people have to figure out things that you should be making explicit. This particularly applies to homework and the proper care and feeding of homework graders.
I forgot to write the current direction. Here is an illustration how I assigned them. Which means that i = i2+i3. Is the KVL I did earlier correct for this setup?
Can I use it to solve the task? Or maybe KCL would be simpler. I wonder though how KCL would be defined for the current i3? If the Vs wasn't there it would obviously be V/3R. How does the Vs being in between affect it?
The convention of doing KVL is to use the first polarity sign that one "meets". Going counter-clockwise, we "meet" V1 from the + side, hence +v1. Vs on the other hand is met from the - polarity side. Could someone please help how to get to the answer? Just write how you would solve it.
Since you have: i*2R+v1+i3*3R-Vs=0 -4V1+i2*3R+Vs-i3*3R=0 i*2R+V1-4V1+i2*3R=0 i1 = i2 + i3 We also know that: V1 = R*i1 You can then say: i*2R+v1+i3*3R-Vs=-4V1+i2*3R+Vs-i3*3R Collecting terms gives: 5V1 - 2Vs = -2R*i1 + 3R*i2 - 6R*i3 Also rearranging i*2R+V1-4V1+i2*3R=0 gives 3V1 = 2R*i1 + 3R*i2 or 3V1 = 2(V1) + 3R*i2 ----> V1 = 3R*i2 You should be able to finish it.
This circuit only has 2 independent loops. When you add a third redundant equation, you only cause confusion for yourself. You also have extraneous unknowns. Since i=i2+i3, get rid of i right away--substitute (i2+i3) for i. Also, V1=(i2+i3)*R so eliminate V1. You only need two equations in two unknowns to solve this circuit using the loop method. If you use the nodal method, only one equation is required.
Your equations are correct, as far as they go, but your approach is going to probably leave you scratching your head because there are some sutble points you are missing. Count up how many unknowns you have. You might think at first glance that you have three, namely {i,i2,i3}. But you also have v1, so your unknowns are {i,i2,i3,v1}. How many equations do you have. At first glance you might think that you have four (the three loop equations plus the KCL equation i=i2+i3. But this is not true because your equations have to be independent and your loop equations aren't. To see that this is the case, add the first two equations together and you will see that you get the third: i*2R+v1+i3*3R-Vs=0 -4V1+i2*3R+Vs-i3*3R=0 ------------------------------- i*2R+V1-4V1+i2*3R=0 So you really only have two loop equations and one KCL equation, but you still have four unknowns. Another problem you are going to have is that none of your equations incorporate R, the resistor across which v1 is taken. Now, either this is because it doesn't matter what R is, or because you are missing another equation. A casual inspection should convince you that you can't change that resistor to just any value without affecting the result. But since you already know you are missing an equation, you can see what equation you can come up with that will relate the value of this resistor to the unknowns. Once you've done this, you should be able to solve your four equations to get your four unknowns. Now, if you are thinking that there must be a way to approach this that doesn't leave you at the mercy of having to spot subtle points like whether you truly have independent equations, there is. Several of them, actually. The two basic ones are called Mesh Current Analysis (MCA) and Node Voltage Analysis (NVA). If you haven't been exposed to them, go check them out. If you use Node Voltage Analysis, you can write down a single equation and solve for v in terms of Vs directly. Once you know v, finding the three circuits and the control voltage for the dependent source is trivial. So hopefully that provides some motivation to go and learn MCA and NVA.
Thanks. I prefer to usually use the NVA method, because it seems to be the method that "always works" (correct me if I'm wrong). For some reason, MCA isn't included in my course. Would you still say it's necessary to learn it anyway, or is NVA enough? More specifically: are there any instances when NVA can't be used and you would HAVE to use MCA?
the more methods you know, the greater yout "toolbox" is. NVA is simple and it always works but some things are much easier (faster) solved using particular method (using thevenin for example). this is why one learns variety of methods. in your courses you may be forced to things certain way. this is good way to get consistent results but once you get comfortable there is nothing to say you can't feel rebellious (as long as you play by the rules, you can do anything you like). for example sometimes i prefer to use shorter equations (node to node using all circuit branches). one can always combine them into longer equations but reverse is not so easy. also when solving problems it is a good idea to number each equation so one can refer to them... for example: (1) v1 = i*R (2) v = i*2R +v1 = i*3R (3) v = vs - i3*3R (4) v = 4v1 - i2*3R (5) i = i2 + i3 ---------------------- using (1) and (2) (1) v1 = i*R (2) v = i*3R = 3*v1 (2') v1= v/3 ---------------------- using (2) and (4) you can solve for i3 in terms of i .... ---------------------- then using (5) you can do the same for i2 (also in terms of i) .... ---------------------- using (2) and (3) you can solve for i in terms of vs and R ....
You don't have to have NVA, either. You could always use KVL/KCL directly. Keep in mind that NVA is nothing more than a systematic application of KCL that incorporates the contraints imposed by KVL in its very setup. Similarly, MCA is nothing more than a systematic application of KVL that incorporates the constraints imposed by KCL in its very setup. In a similar vein, you don't HAVE to use phasors to do AC analysis; instead, you could use differential equations. The diffy-Qs will always work (are at least always apply, though you may not always be able to solve them), while the phasors only work under very restrictive conditions, namely linear circuits in sinusoidal steady state. Think of all of these analysis techniques as tools in a toolbox. Like a good mechanic that has a wide selection of tools available and who has developed the skills to use each and to choose the proper tool for the job, they can do most jobs faster, better, and with few mistakes than someone that has a a pair of pliers, a screwdriver, and a hammer. Or, for that matter, someone that has the first person's tools but who uses a Cresent wrench instead of a box-end wrench because the Crescent wrench "always works". There are lots of times when NVA will work, but is a pain, while MCA will allow you to write the answer down by inspection. The same with all of the other techniques.