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**1. Homework Statement**

Using nodal analysis, find [itex]v_0[/itex] and [itex]I_0[/itex] in the circuit below.

http://img248.imageshack.us/img248/7325/chapter3problem301dy.jpg [Broken]

**2. Homework Equations**

KVL, KCL, V = i R, Super-node

**3. The Attempt at a Solution**

So I added 3 current variables, 3 node markers ([itex]V_1\,-\,V_3[/itex]), a super node, a ground node, and marked a KVL loop.

http://img404.imageshack.us/img404/6940/chapter3problem30part26cn.jpg [Broken]

[tex]V_0\,=\,V_3[/tex] <----- Right?

Now I express the currents:

[tex]I_0\,=\,\frac{V_1\,-\,V_2}{40\Omega}[/tex]

[tex]I_1\,=\,\frac{100\,-\,V_1}{10\Omega}[/tex]

[tex]I_2\,=\,\frac{4\,V_0\,-\,V_1}{20\Omega}[/tex]

[tex]I_3\,=\,\frac{V_0}{80\Omega}[/tex]

KCL at [itex]V_1[/itex]:

[tex]I_0\,=\,I_1\,+\,I_2[/tex]

[tex]\left(\frac{V_1\,-\,V_2}{40}\right)\,=\,\left(\frac{100\,-\,V_1}{10}\right)\,+\,\left(\frac{4\,V_3\,-\,V_2}{20}\right)[/tex]

[tex]7\,V_1\,-\,V_2\,-\,8\,V_3\,=\,400[/tex]

KCL at super-node:

[tex]I_0\,+\,2\,I_0\,=\,I_3[/tex]

[tex]3\,\left(\frac{V_1\,-\,V_2}{40}\right)\,-\,\left(\frac{V_0}{80}\right)\,=\,0[/tex]

[tex]6\,V_1\,-\,6\,V_2\,-\,V_3\,=\,0[/tex]

KVL inside super-node:

[tex]V_3\,-\,V_2\,=\,120[/tex]

Now I put those 3 equations into a matrix and rref to get [itex]V_1\,-\,V_3[/itex].

[tex]\left[\begin{array}{cccc}0&-1&1&120\\7&-1&-8&400\\6&-6&-1&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{cccc}1&0&0&-1688\\0&1&0& -1464\\0&0&1&-1344\end{array}\right][/tex]

[tex]V_3\,=\,V_0\,=\,-1344\,V[/tex]

But -1344 Volts seems too high (or low) doesn't it? Should I have also expressed the currents that I did not mark at the short wires between ground and [itex]V_1[/itex] and [itex]V_3[/itex]? Maybe I should use KVL 1 loop instead of the super-node KVL expression?

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