Hi,
I think this problem is easy, but I can't do it.

answer: 2 A

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Node A is on the left, node B is on the right.

So,
A) \( i_2 + i_4 = i_1 + i_3 \)
B) \( i_3 + i_7 = i_6 + i_5 \)

\(i_1=\frac{v_1}{5}\), \(i_2=-3A\), \(i_3=\frac{v_1 - v_2}{20}\), \(i_4 = i_5 = 3 i_1 = \frac{3 v_1}{5}\), \(i_6=\frac{v_2}{10}\), \(i_7= 10 A\)

A) \( -3 + \frac{3 v_1}{5} = \frac{v_1}{5} + \frac{v_1 - v_2}{20}\)

B) \(\frac{v_1 - v_2}{20} + 10 = \frac{v_2}{10} + \frac{3 v_1}{5}\)


I have this system of linear equations:

\(-60 = -7v_1 - v_2\)
\(200 = 13v_1 + 3v_2\)

\(v_1 = -2.5 V
v_2 = 77.5 V \)


\(i_1 =\frac{-2.5}{5} = -.5 A\)

Thanks for the help!

 

The dependent source seems to be a CCVS despite having a symbol typical of a CCCS. However the answer of 2A is correct if it is a CCCS. I note you have I7 as 10A. Do you now see the error?

 

I've tried this so many times that I copied wrong in a moment.
Unbelievable.
Thanks t_n_k for the help, I really appreciated it.

 

The only error I see is that you put 10A instead of 4A for I7.
See the correct result
http://www.wolframalpha.com/input/?i=-3+++(3A)/5+=+A/5+++(A+-+B)/20+,+(A-B)/20+++4+=+B/10+++(3A)/5