Hey! I know this is a simple one. .I just want to reconfirm. .I am getting I0 as Zero. Please confirm guys!!!
OK! So, I tried using the simple Current Loop Method. .however I guess that approach is wrong. I do NOT want to use Nodal Analysis method. .Can this question be solved using a Mesh Current loop method?
It can. There are 3 loops so there will be 3 unknowns and three equations to find those unknowns. Once you know mesh current of the right mesh (I1) and mesh current of the left mesh(I2), the I0 is the difference of the two (I0=I1-I2).
And just to keep it simple, I0 CAN NOT BE ZERO. Why? Because the branch where I0 is shown to be is not an open circuit.
But the current source is in between the Top Mesh & the right mesh. .so, we have to use the Supermesh concept right?
Before you can solve the problem you have to establish, are you using electron flow or current flow conventions, i.e. does the arrow in the 2mA current source represent electron flow or conventional current flow?
Ok. .So, I recalculated. .I am getting I0 as 1.11 A. .Is that correct . . What I did: I considered right mesh as number 3 Left mesh as 2 Top mesh as 1. the 2mA source current is getting split into I1 & I2. .I2 flows through mesh 1 & I1 flows through Mesh 3 So for 3 I got eqn as: 4I1-6I0=0 For Mesh 1 I got: -2I2-3I3=0 (where I3 is the current flowing through the 3k resistor) For Mesh 2 I got: -6 + 3I3 + 6I0 = 0 Please let me know if this is correct
From what i see i believe you do not understand what you are doing because all your equations are in correct. The method you are using works like that : 1)Lets use the same number for the current and the respective mesh so we have I1 for mesh 1 I2 for mesh 2 etc 2) Its the time to choose how is your current flowing ( clock wise or the oposite ) lets assume here that the current flows anticlockwise in every mesh 3)You can now notice that in every resistance two currents "collide" (the one goes up and the other goes down) 4) this case is a special one because as you said earlier you should make a supermesh 5) For example the mesh 2 equation will be -6+6I2-6I3+3I2-3I1=0 => 9I2-6I3-3I1=6 6)The fact you have a supermesh means that you will make one equation for two meshes and then create a second equation from the relation between these two mesh currents p.s i used the mesh names as you did but i changed the current names for obvious reasons Check it out and tell us what you got, this method is a very basic one so i believe you should be able to use it with ease so keep updating this thread with any questions you may have or things you don't understand
Yes, you have to use supermesh. I did a sim in Multisim and I got I0=-1.333 mA. I did solution by hand and I got I0=-1.32 mA (I did round up a few numbers along the way so that is why my answer is slightly smaller and has fewer decimal places.)
Guys I am having a torrid time here.. I am just not grasping the concept at all. . Can you please upload a pic of ONLY the currents labelled in all branches? I1 I2 & I3. .because that is where I am continuously getting stuck. .Please! From there, I can figure it out easier.
My graphine calculator can do simultaneous equations. All I have to do is setup those same simultaneious equations, input them into the calculator and write down the answers. That is how I got I1, I2 and I3, and why I only show how I setup the simultaneous equations.
I am sorry! But the way the problem was explained. .It helped me clear few of my concepts too. .Hence the post. .Sorry!