Laplace Circuit Problem: Solving for Capacitor Voltage in the Laplace Domain

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  • #1
kkasper
3
0

Homework Statement



Determine the capacitor voltage in the laplace domain.

Homework Equations



http://picturepush.com/public/8001507
http://picturepush.com/public/8001507

The Attempt at a Solution



http://picturepush.com/public/8001546
http://picturepush.com/public/8001546

Using nodal analysis:
[itex]\frac{Vc-20/s}{3+s}[/itex] + Vc0.5s -4=0
=> Vc= [itex]\frac{8s^2+24s+40}{s(s+1)(s+2)}[/itex]

but the answer should be:
Vc= [itex]\frac{24}{s(s+1)(s+2)}[/itex] which is my answer minus 8/s
After studying the laplace equivalents of capacitors and inductors, i still feel i am doing something wrong in deciding what the polarity of the added sources is, or which voltage in the laplace circuit (the one across the capacitor or the one across the capacitor and the addded source) i should treat as the actual voltage of the capacitor.

I hope someone could help me out. Thank you.
 
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  • #2
kkasper said:

Homework Statement



Determine the capacitor voltage in the laplace domain.

Homework Equations



http://picturepush.com/public/8001507
http://picturepush.com/public/8001507

The Attempt at a Solution



http://picturepush.com/public/8001546
http://picturepush.com/public/8001546

Using nodal analysis:
[itex]\frac{Vc-20/s}{3+s}[/itex] + Vc0.5s -4=0
=> Vc= [itex]\frac{8s^2+24s+40}{s(s+1)(s+2)}[/itex]

but the answer should be:
Vc= [itex]\frac{24}{s(s+1)(s+2)}[/itex] which is my answer minus 8/s
After studying the laplace equivalents of capacitors and inductors, i still feel i am doing something wrong in deciding what the polarity of the added sources is, or which voltage in the laplace circuit (the one across the capacitor or the one across the capacitor and the addded source) i should treat as the actual voltage of the capacitor.

I hope someone could help me out. Thank you.

Hi kkasper, Welcome to Physics Forums.

It looks to me like you've got a pretty good handle on it. The result that you found is the expression for the node voltage where the inductor and capacitor meet, which in the "real world" will be the true voltage across the capacitor. The "answer" that you're comparing it to would be the potential across the capacitor less the initial potential that it started with.

There are two equivalent circuits that you can apply for a capacitor with an initial charge. One has a current source added in parallel as you've done, and the other places a voltage source in series with an uncharged capacitor. In this case that voltage source would be 8/s.
 
  • #3
gneill said:
The "answer" that you're comparing it to would be the potential across the capacitor less the initial potential that it started with.

Hey gneill,

When transforming the correct answer back to the time domain Vc(0)= 8 V (the initial potential) as expected. and Vc(infinity)=20 V as expected, contradicting what you said. Did i misunderstand u ?
 
  • #4
kkasper said:
Hey gneill,

When transforming the correct answer back to the time domain Vc(0)= 8 V (the initial potential) as expected. and Vc(infinity)=20 V as expected, contradicting what you said. Did i misunderstand u ?

Maybe :smile: If I take the inverse Laplace transform of the "answer" I see:
$$ Vc(s) = \frac{24}{s (s + 2)(s + 1)} \Rightarrow Vc(t) = 12V(1 - 2 exp(-t) + exp(-2 t))$$
which, for t = 0, is zero volts; No 8V initial capacitor voltage there!

On the other hand, when I take your expression and find the time domain expression, it gives 8V at time t = 0 as expected.
 
  • #5
gneill said:
Maybe :smile: If I take the inverse Laplace transform of the "answer" I see:
$$ Vc(s) = \frac{24}{s (s + 2)(s + 1)} \Rightarrow Vc(t) = 12V(1 - 2 exp(-t) + exp(-2 t))$$
which, for t = 0, is zero volts; No 8V initial capacitor voltage there!

On the other hand, when I take your expression and find the time domain expression, it gives 8V at time t = 0 as expected.

OK. Now I see : ). In the solution they actually add the initial voltage afterwards. But then wouldn't my answer be correct? The original question was find Vc(s) for t≥0. This is an actual university exam question. Maybe the capacitor voltage in the laplace domain is defined, or is by convention amongst electric engineers defined to be the voltage across the capacitor element in the laplace circuit?

Thanks a lot for your time, I am studying for my exams and this helped me a lot. :)
 
  • #6
kkasper said:
OK. Now I see : ). In the solution they actually add the initial voltage afterwards. But then wouldn't my answer be correct? The original question was find Vc(s) for t≥0. This is an actual university exam question. Maybe the capacitor voltage in the laplace domain is defined, or is by convention amongst electric engineers defined to be the voltage across the capacitor element in the laplace circuit?
I'm not aware of any such definition. I'd say that your answer is correct for the actual time domain voltage across the circuit element.

If the question had asked for an expression for the change in potential across the capacitor after t=0, that would be another story.
Thanks a lot for your time, I am studying for my exams and this helped me a lot. :)
No problem, glad to help.
 

1. What is a Laplace circuit problem?

A Laplace circuit problem is a type of mathematical problem that involves analyzing an electrical circuit using the Laplace transform. The Laplace transform is a mathematical tool that allows for the conversion of time-domain functions (such as voltage or current) into frequency-domain functions, making it useful for analyzing complex circuits and systems.

2. Why is the Laplace transform used in circuit analysis?

The Laplace transform is used in circuit analysis because it simplifies the analysis of complex circuits. By converting the circuit's time-domain functions into frequency-domain functions, the analysis can be done using algebraic equations instead of differential equations, which are often much more complex and time-consuming to solve.

3. How do you solve a Laplace circuit problem?

To solve a Laplace circuit problem, you first need to use the Laplace transform to convert the circuit's time-domain functions into frequency-domain functions. Then, you can use algebraic equations to analyze the circuit and find the desired quantities, such as voltage or current. Finally, you can use the inverse Laplace transform to convert the frequency-domain functions back into time-domain functions to get the final solution.

4. What are some common applications of Laplace circuit problems?

Laplace circuit problems have many practical applications in electrical engineering and physics. Some common applications include the analysis of electronic circuits, control systems, signal processing, and communication systems. They can also be used to study the behavior of mechanical and thermal systems.

5. Are there any limitations or drawbacks to using the Laplace transform in circuit analysis?

While the Laplace transform is a powerful tool for circuit analysis, it does have some limitations. One limitation is that it assumes the circuit is linear, which means that the output is directly proportional to the input. This may not always be the case in real-world circuits. Additionally, the Laplace transform cannot be used for circuits with time-varying elements, such as switches or transistors.

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