Solving Circuits: Finding V_1-V_3

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In summary, the conversation discussed determining the voltages in a circuit using KCL and the equation V=iR. The solution involved adding variables for currents and a super-node, and using KCL and matrix operations to solve for V_1 through V_3. The final values for V_1, V_2, and V_3 were given as 18.86V, 6.286V, and 13V, respectively.
  • #1
VinnyCee
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Homework Statement



Determine voltages [itex]V_1[/itex] through [itex]V_3[/itex] in the circuit below.

http://img264.imageshack.us/img264/9311/chapter3problem161oz.jpg [Broken]

Homework Equations



KCL, V = iR

The Attempt at a Solution



So I added some variables to represent currents and a super-node. The variables are in red and the super-node in light-blue.

http://img264.imageshack.us/img264/1875/chapter3problem16part21ki.jpg [Broken]

[tex]V_A\,=\,V_2[/tex]

[tex]V_3\,=\,13\,V[/tex]

[tex]I_1\,=\,\frac{V_1\,-\,V_3}{\frac{1}{2}\Omega}\,=\,2\,V_1\,-\,2\,V_3[/tex]

[tex]I_2\,=\,\frac{V_1\,-\,0}{1\Omega}\,=\,V_1[/tex]

[tex]I_3\,=\,\frac{V_2\,-\,0}{\frac{1}{4}\Omega}\,=\,4\,V_2[/tex]

[tex]I_4\,=\,\frac{V_2\,-\,V_3}{\frac{1}{8}\Omega}\,=\,8\,V_2\,-\,8\,V_3[/tex]

Now I use KCL at the super-node:

[tex]I_1\,+\,I_2\,+\,I_3\,+I_4\,=\,2\,A[/tex]

[tex](2\,V_1\,-\,2\,V_3)\,+\,(V_1)\,+\,(4\,V_2)\,+\,(8\,V_2\,-\,8\,V_3)\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,-\,10\,V_3\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,-\,10(13\,V)\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,=\,132[/tex]

And get the voltage equation from inside the super-node:

[tex]V_1\,-\,V_2\,=\,2\,V_A[/tex]

[tex]V_1\,-\,V_2\,-\,2\,V_2\,= \,0[/tex]

[tex]V_1\,-\,3\,V_2\,=\,0[/tex]

Now put into a matrix and rref to get [itex]V_1[/itex] and [itex]V_2[/itex]:

[tex]\left[\begin{array}{ccc}3&12&132\\1&-3&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{ccc}1&0&\frac{132}{7}\\0&1&\frac{44}{7}\end{array}\right][/tex]

So I get these for [itex]V_1[/itex] through [itex]V_3[/itex]:

[tex]V_1\,=\,\frac{132}{7}\,V\,\approx\,18.86\,V[/tex]

[tex]V_2\,=\,\frac{44}{7}\,V\,\approx\,6.286\,V[/tex]

[tex]V_3\,=\,13\,V[/tex]

Does this look right?
 
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  • #2
Yup, looks right to me.
 
  • #3


I would say that your solution looks correct. You have used the appropriate equations and methods to solve for the unknown voltages in the circuit. Your answer also makes sense intuitively, as V_1 and V_2 are both positive and smaller than V_3, which is the voltage source. Good job on solving the circuit!
 

1. How do I find V1 in a circuit?

To find V1 in a circuit, you can use Kirchoff's Voltage Law (KVL) which states that the sum of voltages around a closed loop in a circuit must equal 0. You can also use Ohm's Law (V = IR) if the circuit only contains resistors.

2. What is the importance of finding V1 in a circuit?

Finding V1 is important because it allows us to understand the voltage drop across a specific component in a circuit. This can help us determine the power dissipated by that component and troubleshoot any issues in the circuit.

3. Can I use V1 to find V2 and V3 in a circuit?

Yes, V1 can be used to find V2 and V3 in a circuit if you know the circuit's total resistance and the values of the other components in the circuit. You can use a combination of KVL and Ohm's Law to calculate these voltages.

4. What happens if V1 is equal to 0 in a circuit?

If V1 is equal to 0 in a circuit, it means that there is no voltage drop across that specific component. This could be due to a short circuit or a malfunctioning component.

5. Are there any tools or software that can help with solving circuits and finding V1-V3?

Yes, there are many online circuit simulators and software programs that can assist with solving circuits and finding V1-V3. These tools use mathematical equations and circuit analysis techniques to provide accurate results.

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