Here is the circuit:(adsbygoogle = window.adsbygoogle || []).push({});

http://img208.imageshack.us/img208/6185/ch3prob26mn0.jpg [Broken]

We are supposed to find v1, v2 and v3.

My work so far:

[tex]i_1\,=\,\frac{v_3\,-\,v_2}{5\,\Omega}\,,\,i_2\,=\,\frac{v_3\,+\,10\,V}{15\,\Omega}\,,\,i_3\,=\,\frac{v_2\,-\,4\,i_0}{20\,\Omega}\,,\,i_4\,=\,\frac{v_1\,-\,15\,V}{20\,\Omega}[/tex]

(KVL 1): [tex]10\,i_3\,+\,4\,i_0\,=\,5\,i_1\,+\,20\,i_4\,+\,15V[/tex]

(KVL 2): [tex]15\,i_2\,-\,10\,V\,-\,4\,i_0\,-\,5\,i_3\,-\,5\,i_1\,=\,0[/tex]

(KVL 3): [tex]5\,i_1\,+\,5\,(i_1\,-\,i_3)\,+\,10\,i_o\,=\,0[/tex]

(KVL 4 - Loops 1 and 2): [tex]15\,i_2\,-\,10\,V\,-\,15\,V\,-\,20\,i_4\,-\,5\,(i_1\,-\,i_3)\,-\,5\,i_1\,=\,0[/tex]

Using these equations, I get infinte answers. I did not list KVL 5, which is the whole outer loop, but I think that it is incorrect anyways becuase I am getting 0 = 0 for the last two rows in the matrix. Can someone please help?

EDIT:The 3V source at the top of the schematic should actually be a 3A independent current source.

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# Simple (supposedly) circuit analysis involving a C.C.V.S. - 2 I.V.S. - I.C.S - 6 Res.

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