In this Q use Mesh analysis to find Current through each mesh in the given?

Question

In uploaded diagram Use Mesh analysis to find Current through each mesh in the given network also find the current Io. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks. Draw the complete circuit diagram and also mention the units of each derived value. Download Cercuit Diagram from http://siel.edu.pk/MeshAnalysis.doc

Answer 1

I set up that the 4 ohm resistor to the upper left has I1 downward
The 6 ohm resistor below has I2 downward
The 2 ohm resistor has I4 downward
And the voltage across the 4A current source is VI with positive downward.

Start by summing voltages around the upper left mesh using V=I*R

6+4I1+6Io+VI=0
Next the bottom mesh voltages sum
6I2-2I4-6Io=0

Sum the voltages around the right loop
12-2I4+VI=0

Sum the currents at the left center node
-I1+Io+I2=0
Sum the currents at the center node
4+I4-Io=0
Then sum the currents at the node at the top of the current source
-I1+I2+I4+4=0

This results in the following matrix of equations
(note that each line sums to zero according to the equations above)
i1 i2 Io I4 VI Constant
4 0 6 0 1 6
0 6 -6 -2 0 0
0 0 0 -2 1 12
-1 1 1 0 0 0
0 0 1 -1 0 -4
-1 1 0 1 0 4



This is tricky to solve the way I did it. I remember from my BSEE circuit analysis that there is a simple matrix inversion method, but I don't remember how to do it, so I solved it by finding factors that reduce the quations:

Step one, subtract the third equation from the first and you have the following left
i1 i2 IoI 4 VI Constant
4 0 6 2 0 -6
0 6 -6 -2 0 0
-1 1 1 0 0 0
0 0 1 -1 0 -4
-1 1 0 1 0 4


See how I eliminated VI?
The next step was harder since there are no simple additions and subtractions. I was able to isolate Io with the following coefficients
i1 i2 Io I4 VI Constant
12 0 18 6 0 -18
0 -12 12 4 0 0
-16 16 16 0 0 0
0 0 6 -6 0 -24
4 -4 0 -4 0 -16


when you add all of the equations with these coefficients together simultaneously, you get
52*Io-58=0
Io=58/52

Then carefully substituting back, you get the following

i1=1.26923076
i2=0.153846154
Io=1.115384615
I4=-2.88
VI=-17.76923077

I checked all of the equations using Excel and they all check correctly.

j

Answer 2

you won't be able to apply the supermesh. nicely possibly you do not favor to at any fee. they seem to paintings nicely even as a node has a "bare" modern source. And even as that would seem to be the case, the depending source is paired with both ohm resistor. and on condition that the source relies upon, i imagine in straightforward words one loop equation outcomes: V - 3*i1 - 2*(i1+i2) = 0 (V in the course of the a million A source) yet obviously i1 = a million and i2 = 0.25*V, so that's common to scrub up for V and then both voltages. yet another mind-set must be to rework the depending source to Thevenin Equivelent. that ought to provide you a the voltage source that you element out. It replaced into in basic terms hidden as modern source. contained in the precise, that's all a similar. As for the full loop no longer appearing to sum to 0 volts, the present sources do have a voltage in the course of them. that's common to ignore that because theoretically they have a limiteless resistance. in this circuit, it truly is why the currents from both source do no longer bypass into the different source's leg, and ultimately, reason the the loop voltages to sum to 0.