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

1. Sep 27, 2006

### VinnyCee

Here is the circuit:

We are supposed to find v1, v2 and v3.

My work so far:

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

(KVL 1): $$10\,i_3\,+\,4\,i_0\,=\,5\,i_1\,+\,20\,i_4\,+\,15V$$

(KVL 2): $$15\,i_2\,-\,10\,V\,-\,4\,i_0\,-\,5\,i_3\,-\,5\,i_1\,=\,0$$

(KVL 3): $$5\,i_1\,+\,5\,(i_1\,-\,i_3)\,+\,10\,i_o\,=\,0$$

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

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.

Last edited: Sep 27, 2006
2. Sep 27, 2006

### VinnyCee

Just checking, but can everyone see the picture with the circuit schematic?

3. Sep 27, 2006

### jpr0

I can't see the schematic. You're trying calculate what the $i_n$ are in terms of the constants 10V, 15V etc?

--I can see it now

You need one more equation to solve this set, you have 5 unknowns and only 4 equations (unless $i_0$ is given?)

Last edited: Sep 27, 2006
4. Sep 27, 2006

### VinnyCee

$i_0$ is not given. What would you suggest as the 5th equation? A supernode, but where?

Do the first four equations look right?

5. Sep 27, 2006

### Corneo

Do you see that $$i_0 = \frac {v_1-v_3}{10}$$

BTW I think solving this using KCL is much easier than what you are trying. You only need 3 equations.

Last edited: Sep 27, 2006
6. Sep 27, 2006

### doodle

I agree with Corneo. Since you wanted to solve for the nodal voltages, then nodal analysis (which is KCL) should be more accomodating than mesh analysis (or KVL, which is what you are using now). And like Corneo says, nodal analysis would only give you three equations which appears more reasonable for a exercise of this kind.

But if you should insist on using KVLs, then please check that I3 = (V2 - 4*I0)/20 is incorrect.

7. Sep 27, 2006

### VinnyCee

Where would the three KCL's be applied at? Can you show please?

8. Sep 28, 2006

### doodle

I'll show you one and you try to obtain the rest.

The KCL on node 1 (which has the nodal voltage V1) gives
(V1 - 15)/20 + (V1 - V2)/5 + (V1 - V3)/10 + 3 = 0
or
(7/20) V1 - (1/5) V2 - (1/10) V3 = -9/4
and this becomes Equation #1.

Try to generate the other two KCLs with respect to nodes 2 and 3.

Last edited: Sep 28, 2006
9. Sep 28, 2006

### VinnyCee

Yes, I see that when KCL is applied to a $v_n$ node, that the terms that are not constant in the resulting equation all begin with $v_n$.

KCL @ $v_2$:

$$\frac{v_2\,-\,v_1}{5}\,+\,\frac{v_2\,-\,4\,i_o}{5}\,+\frac{v_2\,-\,v_3}{5}\,=\,0$$

KCL @ $v_3$:

$$\frac{v_3\,-\,v_1}{10}\,+\,\frac{v_3\,-\,v_2}{5}\,+\,\frac{v_3\,+\,10}{15}\,+\,3\,=\,0$$

Is that correct?

10. Sep 28, 2006

### doodle

$$\frac{v_3\,-\,v_1}{10}\,+\,\frac{v_3\,-\,v_2}{5}\,+\,\frac{v_3\,+\,10}{15}\,-\,3\,=\,0$$
since we are summing up the currents leaving the node.

Noting next that I0 = (V1 - V3)/10, the KCL for node 2 can then be written in terms of the nodal voltages only. You will then have 3 unknowns and 3 equations which then becomes a mathematical exercise.

11. Sep 28, 2006

### VinnyCee

$$RREF\,\left( \begin{array}{cccc} \frac{7}{20} & -\,\frac{1}{5} & -\,\frac{1}{10} & -\,\frac{9}{4} \\ -\,\frac{11}{20} & \frac{3}{5} & -\,\frac{9}{50} & 0 \\ -\,\frac{1}{10} & -\,\frac{1}{5} & \frac{11}{30} & \frac{7}{3} \end{array} \right)\,=\,\left( \begin{array}{cccc} 1 & 0 & 0 & -5.544 \\ 0 & 1 & 0 & -0.691 \\ 0 & 0 & 1 & 4.475 \end{array} \right)$$

$$v_1\,=\,-5.544\,V$$
$$v_2\,=\,-0.691\,V$$
$$v_3\,=\,4.475$$

Does it look right?

12. Sep 28, 2006

### doodle

I am not sure if the second row is correct. You might want to check.

13. Sep 28, 2006

### VinnyCee

Double checking KCL @ $v_2$:

$$\frac{3}{5}\,v_2\,-\,\frac{1}{5}\,v_1\,-\,\frac{4}{5}\,\left(\frac{v_1\,-\,v_3}{10}\right)\,-\,\frac{1}{5}\,v_3\,=\,0$$

$$RREF\, \left( \begin{array}{cccc} 0.35 & -0.2 & -0.1 & -2.25 \\ -0.28 & 0.6 & -0.12 & 0 \\ -0.1 & -0.2 & 0.367 & 2.34 \end{array} \right)\,=\,\left( \begin{array}{cccc} 1 & 0 & 0 & -7.178 \\0 & 1 & 0 & -2.767 \\0 & 0 & 1 & 2.912\end{array} \right)$$

Does this mean that $v_1\,=\,-7.18\,V,\,v_2\,=\,-2.77\,V,\,v_3\,=\,2.91\,V$?

14. Sep 28, 2006

### doodle

Yup, that looks good to me.