CIRCUIT ANALYSIS: Use superposition - 2 Current source, 1 Voltage source, 4 resistors

Your equations are written in terms of i's, but you're calculating the voltage v_{01}. Here's how to do it in terms of i's:1. Calculate i1, i2, i3, i4 from the six equations. You can use the inverse matrix to get the values of i1, i2, i3, i4. Here's what I got: i1 = 0.8 A, i2 = -0.2 A, i3 = 0.2 A, i4 = 0.2 A2. Calculate v01 from the equation v01=6i4. Therefore, v01 =
  • #1
VinnyCee
489
0
Using superposition, find [itex]v_0[/itex] in the following circuit.

ch4prob14.jpg



My work so far:

ch4prob14_Part1.jpg


[tex]v_1\,=\,2\,i_2\,+\,v_{0\,1}[/tex]

[tex]v_1\,=\,4\,i_1[/tex]

[tex]v_{0\,1}\,=\,3\,i_3[/tex]

[tex]v_{0\,1}\,=\,6\,i_4[/tex]

KCL @ v1:
[tex]1\,A\,=\,i_1\,+\,i_2[/tex]

[tex]i_2\,=\,i_3\,+\,i_4[/tex]


Using these six equations, with 6 variables:

[tex]v_{0\,1}\,=\,\frac{12}{7}\,V[/tex]

Is that correct?

My real question is how to put the six equations above into matrix form to enable solving using RREF.

Thanks:tongue:

EDIT: I have fixed this part with Paallikko's guidance (thanks!)

[tex]i_1\,=\,\frac{1}{4}\,v_1[/tex]

[tex]i_2\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}[/tex]

[tex]i_3\,=\,\frac{1}{3}\,v_{01}[/tex]

[tex]i_4\,=\,\frac{1}{6}\,v_{01}[/tex]

[tex]i_1\,+\,i_2\,=\,1\,A\,\,\,\Rightarrow\,\,\,\frac{3}{4}\,v_1\,-\,\frac{1}{2}\,v_{01}\,=\,1[/tex]

[tex]i_3\,+\,i_4\,=\,i_2\,\,\,\Rightarrow\,\,\,\frac{1}{3}\,v_{01}\,+\,\frac{1}{6}\,v_{01}\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}[/tex]

[tex]v_{01}\,=\,1\,V[/tex]
 
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  • #2
Apply superposition two:

ch4prob14_Part2.jpg


[tex]i_3\,=\,i_2\,-\,2\,A[/tex]

[tex]i_1\,=\,i_0\,+\,i_2[/tex]

[tex]v_{0\,2}\,=\,6\,i_1\,=\,6\,i_3\,=\,3\,i_o[/tex]

From these five equations, I put into a matrix and solve.

[tex]v_{0\,2}\,=\,-6\,V[/tex]


Next, I do the last superposition:

ch4prob14_Part3.jpg


The circuit above can be simplified:

ch4prob14_Part4.jpg


[tex]v_{0\,3}\,=\,3\,I[/tex]

[tex]v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right)\,=\,20\,V[/tex]

Total them up:

[tex]v_0\,=\,v_{0\,1}\,+\,v_{0\,2}\,+\,v_{0\,3}[/tex]

[tex]v_0\,=\,\left(\frac{12}{7}\,V\right)\,+\,(-6\,V)\,+\,(20\,V)[/tex]

[tex]v_0\,=\,15.7\,V[/tex]

Does this look right? If not, please explain the errors and how to correct them. Thanks:tongue2:
 
  • #3
Well, you've understood how to use superposition (setting the sources to 0), you even seem to have most of the equations right*. However, all your calculations are wrong.

The first one in matrix form would be (solve for ix from the first equations and substitute into the KCL equations):

[tex]
\left[ {\begin{array}{*{20}c}
{3/4} & { - 1/2} \\
{ - 1/2} & 1 \\

\end{array} } \right]\left[ {\begin{array}{*{20}c}
{V_1} \\
{V_2} \\

\end{array} } \right] = \left[ {\begin{array}{*{20}c}
1 \\
0 \\

\end{array} } \right]
[/tex]

* the two equations below are wrong
[tex]i_1=i_0 + i_2[/tex]
[tex]v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right) \,=\,20\,V[/tex]PS. If you're not familiar with PSpice, there's a free version available for download eg. here: http://www.electronics-lab.com/downloads/schematic/013/ you can quickly check if your answers are correct with it.
PPS. the topic's more engineering or introductory physics than advanced physics.
 
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  • #4
For the first incorrect equation (in the 2nd part of problem), I see that all the currents are going out of the node. Does this mean that the actual equation is [tex]i_0\,+\,i_1\,+\,i_2\,=\,0[/tex]?

In the third part of the problem, is the error becuase I didn't include both resistors in the I = equation?

[tex]v_{03}\,=\,3\,I[/tex]

[tex]v_{03}\,=\,3\,\left(\frac{20\,V}{6\,\ohm}\right)\,=\,10\,V[/tex]

Is that right now?
 
  • #5
Yep, they're right now.
 
  • #6
How do I solve the second part? I keep getting 0V for [tex]v_{02}[/tex].

I also used Pspice and got an 8V differential at the resistor for [tex]v_0[/tex].

[tex]v_0\,=\,v_{01}\,+\,v_{02}\,+\,v_{03}[/tex]

[tex]v_0\,=\,1V\,+\,0V\,+\,10V\,=\,11V[/tex]

This is 3V higher than Pspice reports though.
 
  • #7
You've written:
v02 = 3i0 = 6i1 = 6i3 (1)
i2 - i3 = 2 (2)
i0 + i1 + i2 = 0 (3)

Substituting (2) into (3) (Eliminating i2):
i0 + i1 + i3 = -2

Substituting (1) into the above:
v02(1/3 + 1/6 + 1/6) = -2

=> v02 = -3 (V)
 
Last edited:

1. What is superposition in circuit analysis?

Superposition is a method used in circuit analysis to simplify complex circuits by breaking them down into smaller, simpler parts. It involves analyzing the effects of individual sources in a circuit separately, and then combining the results to determine the overall behavior of the circuit.

2. How do you apply superposition in circuit analysis?

To apply superposition in circuit analysis, you first need to identify all the sources (voltage and current) in the circuit. Then, for each source, you need to calculate the voltage or current at the point of interest while holding all other sources at 0. Finally, you add all the individual results to get the total voltage or current at the point of interest.

3. Can superposition be used for any circuit?

Superposition can only be applied to circuits that are linear and have only independent sources. This means that the output of the circuit is directly proportional to the input and there are no dependent sources, such as voltage or current controlled sources.

4. What are the limitations of using superposition in circuit analysis?

The main limitation of superposition in circuit analysis is that it can only be used for circuits with a small number of sources. As the number of sources increases, the number of calculations required also increases, making it impractical to use. Additionally, superposition cannot be used for circuits with dependent sources.

5. What are the benefits of using superposition in circuit analysis?

Superposition allows for the simplification of complex circuits, making it easier to analyze and understand their behavior. It also helps in identifying the contributions of individual sources to the overall behavior of the circuit, which can be useful in troubleshooting and designing more efficient circuits.

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