Circuit analysis - Laplace

In summary, the user provided a detailed solution for a circuit problem, using KCL to write equations at two nodes and solving for V1 and V2. However, they should consider the direction of current flow and the initial conditions, and use the superposition principle to incorporate the 5 ohm resistor and 5A current source into their solution.
  • #1
ineedmunchies
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Homework Statement



The question is as shown in the first picture. Question1.jpg

Homework Equations


The Attempt at a Solution


It asks that initial condition current generators are used, what I believe models the circuit for t>0 is shown in the second picture with the switch open, and a current generator added. (I am not sure if this is correct.)

I then used KCL to write equations at node 1 and node 2.

Node 1:
[tex]\frac{5}{s}[/tex] - [tex]\frac{2}{s}[/tex]=[tex]\frac{V_{1}}{1}[/tex]+[tex]\frac{V_{1}-V_{2}}{2s}[/tex]Node 2:
[tex]\frac{2}{s}[/tex]+[tex]\frac{2}{s}[/tex]=[tex]\frac{V_{2}}{1}[/tex]-([tex]\frac{V_{1}-V_{2}}{2s}[/tex])These can then be rearraged to give

[tex]\frac{3}{s}[/tex]=[tex]V_{1}[/tex](1+[tex]\frac{1}{2s}[/tex])-[tex]V_{2}[/tex]([tex]\frac{1}{2s}[/tex])

and

[tex]\frac{4}{s}[/tex]=[tex]V_{1}[/tex]([tex]\frac{-1}{2s}[/tex])+[tex]V_{2}[/tex](1+[tex]\frac{1}{2s}[/tex])

Which I then put into a matrix and solved for [tex]V_{1}[/tex] and [tex]V_{2}[/tex]

Giving
[tex]V_{1}[/tex] = [tex]\frac{3}{s}[/tex]-[tex]\frac{2}{1+\frac{1}{2s}}[/tex]
which can be simplified to
[tex]V_{1}[/tex] = [tex]\frac{3}{s}[/tex]-[tex]\frac{4}{s+2}[/tex]

and
[tex]V_{2}[/tex] = [tex]\frac{\frac{-3}{2}}{1+\frac{1}{2s}}[/tex]+[tex]\frac{4}{s}[/tex]
Which can be simplified to
[tex]V_{2}[/tex] = [tex]\frac{4}{s}[/tex]-[tex]\frac{3}{s+2}[/tex]

Then convert these back to the time domain to give:
[tex]V_{1}[/tex](t)=3-4[tex]e^{-2t}[/tex]
and [tex]V_{2}[/tex](t)=4-3[tex]e^{-2t}[/tex]

Can anyone tell if there is a mistake here or not?
I don't feel confident this is the correct answer. I think the 5 ohm resistor and 5A current should effect the circuit somehow but do not know how to encorporate it into my equations.
 

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  • #2
Any help would be appreciated.



Thank you for your detailed solution. Your approach seems to be correct, but there are a few things that you may need to consider in order to ensure the accuracy of your answer.

Firstly, when using KCL to write equations at node 1 and node 2, you have to take into account the direction of current flow. In this case, the current flowing through the 5 ohm resistor and the 5A current source should be in opposite directions, which would result in a negative sign in your equations.

Secondly, when solving for V1 and V2, you have assumed that the initial condition for the circuit is zero. However, this may not be the case, as the circuit has been running for some time before the switch is opened. In this case, you would need to incorporate the initial conditions into your equations by adding a term representing the initial voltage at each node.

Lastly, in order to incorporate the 5 ohm resistor and 5A current source into your equations, you can use the superposition principle. This means that you can solve for V1 and V2 separately when the 5 ohm resistor is present and when the 5A current source is present, and then add the two solutions together to get the final answer.

I hope this helps. Happy problem-solving!
 
  • #3


Your approach to solving the circuit using Laplace transforms is correct. However, there are a few mistakes in your equations and final solutions.

Firstly, in your initial equations at node 1 and node 2, you have not included the effect of the 5 ohm resistor and 5A current generator. These should be included in the equations as follows:

Node 1:
\frac{5}{s} - \frac{2}{s}=\frac{V_{1}}{1}+\frac{V_{1}-V_{2}}{2s}+\frac{5}{s}

Node 2:
\frac{2}{s}+\frac{2}{s}=\frac{V_{2}}{1}-(\frac{V_{1}-V_{2}}{2s})+5

Note that the 5 ohm resistor is in parallel with the 2 ohm resistor, so its effect is added to the current source.

Next, in your final solutions for V1(t) and V2(t), you have not included the initial conditions. The initial conditions for V1 and V2 are V1(0) = 0 and V2(0) = 0, since the switch is open for t<0. These initial conditions should be added to your final solutions, giving:

V_{1}(t)=3-4e^{-2t}-\frac{2}{s}
and V_{2}(t)=4-3e^{-2t}-\frac{4}{s}

Finally, note that your final solutions for V1(t) and V2(t) should also have a unit of voltage, not just a number. So the correct final solutions are:

V_{1}(t)=3V-4e^{-2t}V-\frac{2}{s}V
and V_{2}(t)=4V-3e^{-2t}V-\frac{4}{s}V

Overall, your approach to solving the circuit using Laplace transforms is correct, but make sure to include all elements in the circuit and to include the initial conditions in your final solutions.
 

1. What is Laplace analysis and how is it used in circuit analysis?

Laplace analysis is a mathematical technique used to analyze and solve differential equations, which are commonly used to model circuits. It is particularly useful for solving circuits with time-varying components and inputs. Laplace transforms are used to convert the time-domain equations into complex frequency-domain equations, making it easier to analyze and understand the behavior of the circuit.

2. What are the advantages of using Laplace analysis in circuit analysis?

Laplace analysis allows for the use of complex numbers and transforms, which simplifies the analysis of circuits with multiple components and inputs. It also provides a more intuitive representation of the circuit behavior in the frequency domain, making it easier to understand and design circuits. Additionally, Laplace analysis can be used to solve differential equations that are difficult to solve using traditional methods.

3. What are some common applications of Laplace analysis in circuit design?

Laplace analysis is commonly used in the design and analysis of filters, amplifiers, and control systems. It is also used in the design of power systems and communication systems. Additionally, Laplace analysis is used in the study of transient response and stability of circuits.

4. What are the limitations of Laplace analysis in circuit analysis?

Laplace analysis assumes linearity and time-invariance in the circuit, which may not always be the case in real-world circuits. It also requires the use of complex numbers and transforms, which may be challenging for some users. Additionally, Laplace analysis may not be suitable for analyzing circuits with nonlinear or time-varying components.

5. How can Laplace analysis be applied to practical circuit problems?

Laplace analysis can be applied to practical circuit problems by using software tools such as MATLAB or SPICE. These programs allow for the easy conversion of time-domain equations into Laplace domain equations and provide a graphical representation of the circuit behavior. Additionally, understanding the basics of Laplace analysis can help in the design and troubleshooting of circuits in various applications.

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