Recent content by Xiao Xiao

No he did not, unfortunately. So I asked him about the solution in hope he respond, if he uploads the solution for the question later (because this was in our assignment) or answers my question, I'll update with the righ question or method to use here.

Yeah it's completely messed up, and I couldn't figure on what basis they changed the question. So I'm sticking with the solution that I ended up with in my previous reply. I did end up asking my professor about if there's an error in the Question or not but he hasn't responded yet.

Nothing else

The third one should be ##(v_b-v_c)/2+0.2v_3=(v_c-v_a)/2## ##(v_c-v_a) /2= i_2## and ##0.2v_3=i_2## That's why I wrote it that way. I solved it by hand and got an answer for it, but I'm not sure if it's correct. I couldn't find out if it's 10V or 10A but I decided to count is a current source...

Yes, these are my equations so far. 1. ##0.02v_1+i_2=(v_a-v_b)/3## 2. ##(v_a-v_b) /3+10=(v_b-v_c)/2## 3. ##(v_b-v_c)/2+i_2=i_2## I do use this ##0.2v_3=i_2## and through equation 3 I find that ##v_b-v_c=v_1=0## and solve the rest of the equation using that.

Ooooh, yes you're right, I completely missed that, thank you.

Yeah, I guess I was tired when I wrote them and didn't copy some stuff correctly. I got my hands on the solution manual and their equations are the same as me, except they didn't include 0.2*V3. But they considered 10V as 10A. I'll just send it to my professor and ask.

Oh yes, starting from left, first node is Va, second is Vb, third is Vc.

I see, thanks a lot.

Yes, I get it now, thank you.

Okay, so I took a long time thinking about it. So, in the original Kirchhoff's equation, we assigned a negative because of the passive sign convention and direction of the loop, but in Ohm's law when we did V=iR we gave it a negative sign because the original current (not just the direction of...

I thought that when applying the law, and I start in the direction of clockwise loop in my example, the current enters the nagetive terminal for the 12V and I started with -, so now any source with current entering the negative terminal, the current will be given a negative sign, and sources...

So I know I have to use kirchhoff's Voltage Law so when I apply it's: -12+4i+2Vo-4-Vo=0 and Vo=6i so --> -16+4i+2(6i)-(6i)=0 but apparently that's wrong and Vo should be =-6i and so when I substitute it in the equation it should be -16+4i+2(-6i)-(-6i)=0 and I don't understand why.

Okay, my professor's lecture was very confusing and didn't properly explain initial conditions, but now I used outside sources to study and my confusion about the question is cleared so I'll delete this in a bit. Edit: I guess deleting questions doesn't work.