Solving Circuit for v1, v2, and v3

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In summary: Great job!In summary, the problem involved finding the values of v1, v2 and v3 in a circuit with multiple voltage and current sources. Using KVL and KCL equations, a system of equations was created and solved using row reduction to obtain the values for the nodal voltages. The final values obtained were v1 = -7.18V, v2 = -2.77V, and v3 = 2.91V.
  • #1
VinnyCee
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Here is the circuit:

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

ch3prob26.jpg
 
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  • #2
Just checking, but can everyone see the picture with the circuit schematic?
 
  • #3
I can't see the schematic. You're trying calculate what the [itex]i_n[/itex] are in terms of the constants 10V, 15V etc?

--I can see it nowYou need one more equation to solve this set, you have 5 unknowns and only 4 equations (unless [itex]i_0[/itex] is given?)
 
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  • #4
[itex]i_0[/itex] is not given. What would you suggest as the 5th equation? A supernode, but where?

Do the first four equations look right?
 
  • #5
Do you see that [tex]i_0 = \frac {v_1-v_3}{10}[/tex]

BTW I think solving this using KCL is much easier than what you are trying. You only need 3 equations.
 
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  • #6
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
Where would the three KCL's be applied at? Can you show please?
 
  • #8
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.
 
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  • #9
Yes, I see that when KCL is applied to a [itex]v_n[/itex] node, that the terms that are not constant in the resulting equation all begin with [itex]v_n[/itex].

KCL @ [itex]v_2[/itex]:

[tex]\frac{v_2\,-\,v_1}{5}\,+\,\frac{v_2\,-\,4\,i_o}{5}\,+\frac{v_2\,-\,v_3}{5}\,=\,0[/tex]

KCL @ [itex]v_3[/itex]:

[tex]\frac{v_3\,-\,v_1}{10}\,+\,\frac{v_3\,-\,v_2}{5}\,+\,\frac{v_3\,+\,10}{15}\,+\,3\,=\,0[/tex]

Is that correct?
 
  • #10
VinnyCee said:
KCL @ [itex]v_3[/itex]:

[tex]\frac{v_3\,-\,v_1}{10}\,+\,\frac{v_3\,-\,v_2}{5}\,+\,\frac{v_3\,+\,10}{15}\,+\,3\,=\,0[/tex]

The last equation should read...
[tex]\frac{v_3\,-\,v_1}{10}\,+\,\frac{v_3\,-\,v_2}{5}\,+\,\frac{v_3\,+\,10}{15}\,-\,3\,=\,0[/tex]
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
[tex]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)[/tex]

[tex]v_1\,=\,-5.544\,V[/tex]
[tex]v_2\,=\,-0.691\,V[/tex]
[tex]v_3\,=\,4.475[/tex]

Does it look right?
 
  • #12
VinnyCee said:
[tex]\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)\end{array} \right)[/tex]
I am not sure if the second row is correct. You might want to check.
 
  • #13
Double checking KCL @ [itex]v_2[/itex]:

[tex]\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[/tex]

[tex]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)[/tex]

Does this mean that [itex]v_1\,=\,-7.18\,V,\,v_2\,=\,-2.77\,V,\,v_3\,=\,2.91\,V[/itex]?
 
  • #14
Yup, that looks good to me.
 

What is a circuit?

A circuit is a closed loop or path through which an electrical current can flow. It is made up of various components, such as resistors, capacitors, and power sources, connected by conductive wires.

How do you solve a circuit for v1, v2, and v3?

To solve a circuit for v1, v2, and v3, you need to use Kirchhoff's circuit laws and Ohm's law. First, use Kirchhoff's current law to determine the currents at each node in the circuit. Then, use Kirchhoff's voltage law to calculate the voltage drops across each component. Finally, use Ohm's law to find the values of v1, v2, and v3.

What are Kirchhoff's circuit laws?

Kirchhoff's circuit laws are two principles that govern the behavior of electric circuits. Kirchhoff's current law states that the sum of the currents entering a node in a circuit must equal the sum of the currents leaving that node. Kirchhoff's voltage law states that the sum of the voltage drops around a closed loop in a circuit must equal the sum of the voltage sources in that loop.

What is Ohm's law?

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. It can be expressed as I = V/R, where I is current in amperes, V is voltage in volts, and R is resistance in ohms.

Why is it important to solve circuits for v1, v2, and v3?

Solving circuits for v1, v2, and v3 allows us to understand the behavior of electric circuits and predict the values of voltages and currents at different points in the circuit. This is crucial in designing and troubleshooting electrical systems, as well as in many scientific and engineering applications.

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