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

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The discussion revolves around determining the capacitor voltage in the Laplace domain using nodal analysis. The initial attempt yielded a voltage expression that included an additional term, leading to confusion about the correct representation of the capacitor voltage. Clarifications were provided regarding the interpretation of initial conditions and the distinction between the voltage across the capacitor and the total voltage including initial charge. The consensus is that the original expression accurately reflects the voltage across the capacitor, while the standard answer represents a modified version accounting for initial conditions. The conversation highlights the importance of understanding conventions in electrical engineering when solving such problems.
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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:
\frac{Vc-20/s}{3+s} + Vc0.5s -4=0
=> Vc= \frac{8s^2+24s+40}{s(s+1)(s+2)}

but the answer should be:
Vc= \frac{24}{s(s+1)(s+2)} 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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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:
\frac{Vc-20/s}{3+s} + Vc0.5s -4=0
=> Vc= \frac{8s^2+24s+40}{s(s+1)(s+2)}

but the answer should be:
Vc= \frac{24}{s(s+1)(s+2)} 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.
 
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 ?
 
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.
 
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. :)
 
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.
 

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