0
\$\begingroup\$

I need to find Ix with mesh analysis method. I will really appreciate your help.enter image description here So, I turned Is2 to a voltage source and tried writing equations for a super mesh and KCL. anyway, I miss something in this problem or I got something totally wrong with my approach towards the problem enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Hi, welcome to EE.SE. You have to show some effort to get help. Start solving the problem and come back to here when you get stuck somewhere. You can explain your attempt by pressing on edit button. \$\endgroup\$ – Hazem Jan 5 at 11:48
  • \$\begingroup\$ Ok so I tried to solved it and I got stuck, will really appreciate any assistance. \$\endgroup\$ – BaRaKL Jan 6 at 19:10
0
\$\begingroup\$

It's not complicated. Try it without bothering to do the Norton to Thevenin conversion:

schematic

simulate this circuit – Schematic created using CircuitLab

$$\begin{align*} 0\:\text{V} + V_\text{IS1} - R_1\cdot\left(I_1-I_2\right) &=0\:\text{V}\\ 0\:\text{V} - R_1\cdot\left(I_2-I_1\right)-R_2\,I_2-V_{1.5\,I_x}&=0\:\text{V}\\ 0\:\text{V}+V_{1.5\,I_x} - R_3\,I_3-R_4\cdot\left(I_3-I_4\right)&=0\:\text{V}\\ 0\:\text{V}- R_4\cdot\left(I_4-I_3\right)-V_\text{IS2}&=0\:\text{V}\\\\ I_X&=I_1-I_2\\ 1.5\,I_X &=I_3-I_2\\ I_1&=2\:\text{A}\\ I_4&=-5\:\text{A} \end{align*}$$

I think you should be able to follow how I developed each of those, above. (I assumed that the voltage across each current source has its positive terminal on the top side and its negative terminal on the bottom side -- which doesn't have to be true for the above equations to work correctly. It just has to be consistently applied, is all.)

You certainly can, if you want, do the conversion first and reduce the number of loops by one. But there's no real need to do that.

If you use Sage (freely available to anyone), then you might set up the above like this:

var('i1 i2 i3 i4 r1 r2 r3 r4 ix vis1 vis2 vix')
eq1=Eq(vis1-r1*(i1-i2),0)
eq2=Eq(-r1*(i2-i1)-r2*i2-vix,0)
eq3=Eq(vix-r3*i3-r4*(i3-i4),0)
eq4=Eq(-r4*(i4-i3)-vis2,0)
eq5=Eq(ix,i1-i2)
eq6=Eq(1.5*ix,i3-i2)
ans=solve([eq1,eq2,eq3,eq4,eq5,eq6],[i2,i3,ix,vis1,vis2,vix])

It solves just fine.

\$\endgroup\$
  • \$\begingroup\$ yeah, it helped a lot. thank you. I think I shouldn't convert the right side current source to a voltage source, and I used for not making KVL's for meshes with current sources in them but on "super meshes" only. still, it helped a lot. thank you again! \$\endgroup\$ – BaRaKL Jan 9 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.