Solve V_0 & I_0 in Circuit w/ Nodal Analysis

[VinnyCee]

Homework Statement

Using nodal analysis, find v_0 and I_0 in the circuit below.

http://img248.imageshack.us/img248/7325/chapter3problem301dy.jpg

Homework Equations

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

The Attempt at a Solution

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

http://img404.imageshack.us/img404/6940/chapter3problem30part26cn.jpg

V_0\,=\,V_3 <----- Right?

Now I express the currents:

I_0\,=\,\frac{V_1\,-\,V_2}{40\Omega}

I_1\,=\,\frac{100\,-\,V_1}{10\Omega}

I_2\,=\,\frac{4\,V_0\,-\,V_1}{20\Omega}

I_3\,=\,\frac{V_0}{80\Omega}

KCL at V_1:

I_0\,=\,I_1\,+\,I_2

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

7\,V_1\,-\,V_2\,-\,8\,V_3\,=\,400KCL at super-node:

I_0\,+\,2\,I_0\,=\,I_3

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

6\,V_1\,-\,6\,V_2\,-\,V_3\,=\,0KVL inside super-node:

V_3\,-\,V_2\,=\,120Now I put those 3 equations into a matrix and rref to get V_1\,-\,V_3.

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

V_3\,=\,V_0\,=\,-1344\,V

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 V_1 and V_3? Maybe I should use KVL 1 loop instead of the super-node KVL expression?

 

[mjsd]

your diagram as it stand contains two wires that short out V1 and V3 making them trivial with respect to ground chosen..should they not be there?

 

[VinnyCee]

Nope, they are there.

 

[mjsd]

in that case your (effective) diagram changes dramatically, i m not even sure whether it is consistent?...anyway, if it is consistent: your circuit theory then tells you: I3=0, V_0=0, I2=0, V1=V3=0 so can join them up and form a loop containing the 120V source and 40 ohm resistor..and I_0=3A

it is actually ok i think... consistency wise speaking

 

[VinnyCee]

So how would I go about getting a system of equations for V_1 thorugh V_3?

 

[mjsd]

you want to find V_0 and I_0, why worry about v1, v2 and v3? they are introduced for node voltage analysis purposes only... but if you really want a "system" of equations: you have
V1=0, V3=0, V2 = -120

all w.r.t. the ground chosen

 

[VinnyCee]

OIC, V_1 and V_3 are 0 becasue they are both connected to ground through the wires you were asking about.

So for I_0 we have

I_0\,=\,\frac{V_1\,-\,V_2}{40\Omega}\,=\,\frac{(0)\,-\,(-120\,V)}{40\Omega}\,=\,3\,A

Right?

 

[mjsd]

yes. NB: but for this question, introducing V1-V3 are not necessary, just observe that those wires short out several components leading to a simpler circuit and solve by inspection or doing a simple KVL loop (in this case).

 

[VinnyCee]

So, to get V_0, I do a KVL loop around the 80\Omega and 40\Omega and the 120 V source?

V_0\,+\,40\,I_0\,-\,120\,V\,=\,0\,\,\longrightarrow\,\,V_0\,=\,120\,-\,40(3)\,=\,0\,V

So V_0 is really zero?

 

[mjsd]

by the way, I got V0=0 by inspection. but what you have done is also correct. i guess showing that the circuit is consistent after all