# CIRCUIT ANALYSIS: 4 resistor, 2 Voltage Source, V.C.V.S., C.C.C.S. - Find Io & Vo

1. Jan 23, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

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

3. 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 [Broken]

$$V_0\,=\,V_3$$ <----- 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 $V_1$:

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

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

$$7\,V_1\,-\,V_2\,-\,8\,V_3\,=\,400$$

KCL at super-node:

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

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

$$6\,V_1\,-\,6\,V_2\,-\,V_3\,=\,0$$

KVL inside super-node:

$$V_3\,-\,V_2\,=\,120$$

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

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

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

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?

Last edited by a moderator: May 2, 2017
2. Jan 23, 2007

### 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?

Last edited: Jan 23, 2007
3. Jan 23, 2007

### VinnyCee

Nope, they are there.

4. Jan 23, 2007

### 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

Last edited: Jan 23, 2007
5. Jan 23, 2007

### VinnyCee

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

6. Jan 23, 2007

### mjsd

you wanna 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

7. Jan 23, 2007

### 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?

8. Jan 23, 2007

### 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).

9. Jan 23, 2007

### 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?

10. Jan 23, 2007

### 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