Hi,

Need some help regarding this circuit.
Anyone can give me some hint how should i go about starting doing this circuit. as the independent and dependent source confused me.

there is something to do with supermesh, but how do i simplify the 2 current source there

 

Try superposition
1.open 3A source.put i1 current in LH mesh and i2 in RH mesh.Solve for i1,i2.
2.short 50V source.Definitely -3A will be flowing through 10ohm.So this -3+i1 gives Ix.
Vx we can use super mesh or solve for i2 by applying superposition in above 2 cases.Vx is drop over 2ohm,which is 2(i1-i2)

 

I solved it by first transforming the voltage sources to current sources. Then I had all current sources feeding one node at the top.

Second, look at the middle branch. You can combine the current sources into one source as 3+Vx/4. That's not quite as important as being able to see that the combined current source is a function of Vx and the 2Ω resistor.

It doesn't involve a mesh but rather a node. Hopefully that should be enough to get you started.

 

i did a source transformation . but was kinda stuck with all the variable

 

i did a source transformation . but was kinda stuck with all the variableView attachment 46354

Why create a new unknown variable i2? Instead of creating this unknown current variable, why not create a new voltage variable "v" across the node? Then the current through the 5Ω is a function of v, and ix is a function of v and 10Ω.

By the way, you drew your "orange box" way too large. Draw it only around the 2Ω resistor and the current source, i.e. one line in and one line out. You should be able to solve for Vx.

 

As you can see from the responses you got, there are many ways to analyze most circuits. But, I suspect that there are some constraints regarding which ones you are allowed to use on this particular assignment. If so, be sure to say so.

Let's assume that you have to use mesh current analysis (to include using a supermesh, if appropriate). So you have three unknown mesh currents plus two more unknowns to deal with the two controlled sources.

Since mesh current analysis is basically KVL, the problem is that in some of the meshes we go across current sources for which we can't write an expression for the voltage across it very easily. So we generalize our mesh notion to the notion of a 'supermesh' that can go around pretty much any loop we want; we just have to make sure that we use the right mesh currents for each component when writing down the equations. Our goal is to get everything written in terms of the three mesh currents.

So see if you can find a loop that doesn't go through any current sources (current-conntrolled voltage sources are fine, because you can write the controlling currents in terms of the mesh currents).

Then see how many equations you can generate based on the current sources, since they impose relationships between the mesh currents that share them.

Finally, write any control voltages in terms of the resistances and the mesh currents that share them.

 

so, by using supermesh on the 3A current source, i do a KLV on left and right loop. but how should i form the equation for the dependent current source and resisitor?

here what i get:
left mesh:
-50 + 10ix + 2(-vx/4) + 2 (-vx/4)=0
-50+10ix-vx=0

right mesh:
5(ix-vx/4)+4ix + 2(vx/4) + 2 (vx/4)= 0
9ix-1/4vx = 0



after doing some subsitution.
i get ix to be -1.923
vx to be -69.23

seems to be wrong




after some reading again, to work with supermesh i have to disregard all current source right?

so again:

-50+10x+5(ix-(3+vx/4)+4ix=0
19ix-5vx/4=65
76ix-5vx=260

how do i get my 2nd equation?

 

so, by using supermesh on the 3A current source, i do a KLV on left and right loop. but how should i form the equation for the dependent current source and resisitor?

here what i get:
left mesh:
-50 + 10ix + 2(-vx/4) + 2 (-vx/4)=0
-50+10ix-vx=0

right mesh:
5(ix-vx/4)+4ix + 2(vx/4) + 2 (vx/4)= 0
9ix-1/4vx = 0

I don't follow how you are getting either of these and I'm not sure you have units that are working out. But it's hard to tell because you don't track units.

Of course, authors that can't be bothered to provide proper units don't make it easy. You have a dependent current source whose output is a voltage divided by four. That makes no sense. What they mean is that it is a voltage divided by 4Ω. Similarly, a voltage source whose output is 4 times a current is non-sensical. Here, they mean that it is the current multiplied by 4Ω.

If you are going to do mesh analysis, the use mesh currents. You have three meshes, so assign each mesh a current that is circulating in a consistent direction, such as clockwise. Just go from left to right and call them I1, I2 and I3.

Q1) What is ix in terms of the mesh currents?

Q2) What is the output of the dependent voltage source in terms of the mesh currents?

Q3) What is vx in terms of the mesh currents?

Q4) What is the output of the dependent current source in terms of the mesh currents?

Q5) What is the relationship imposed on I1 and I2 as a result of the 3A current source?

Q6) What is the relationship imposed on I2 and I3 as a result of the dependent current source?

Q7) What is KVL around the outer loop, written in terms of the mesh currents?

Q8) How many equations and how many unknowns do the answers to Q5, A6, and Q7 provide?

after some reading again, to work with supermesh i have to disregard all current source right?

I have no idea what might have given you this notion. One of the principle motivations for using a supermesh is to avoid a current source in writing the loop equation but then use the current source to provide the relationship between the multiple mesh currents involved in the supermesh loop. But a blanket statement of just disregarding all current sources doesn't make sense. Remember, there is nothing magical or mysterious here -- this is simply a systematic application of KVL and KCL and Ohm's Law to a circuit.