Circuit analysis, problems with Laplace?

In summary, the homework statement states that a switch is turned on at t=0. I_k(t) is the current flowing through the resistor in parallel with the inductor when t starts from 0. The equation for i_k(t) is given by 4*sqrt{2}*e^{-500t}*sin(500*sqrt{3}t).
  • #1
Kruum
220
0

Homework Statement



The switch k is turned at t=0. Calculate [tex]i_k(t)[/tex], when t starts from 0. You can find the circuit in my link.

Homework Equations



All of them, basically. :tongue:

The Attempt at a Solution



http://www.filefactory.com/file/af19a7h/n/index_mht You can find my attempt there. It's in .mhtml format, but any basic browser should open it. I'm not quite sure, if my Laplace to time transformation is correct. The factor before sin is so small. I'd appreciate it, if someone could take a look.

My symbols might differ from your's. But j is imaginary unit (i.e. 1+j), small letters are in time plane, captiol letters are in complex plane an captiol letters with (s) are in Laplace plane.
 
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  • #2
When I click on your link all I get is...

index​.mht
Size: 251.71 KB
Description: No description

...and a login prompt. Do I have to join FileFactory to read this?
 
  • #3
There should be free to download link a bit farther down, if not I'll have to upload somewhere else.

Yep, the second option is free and without registering.
 
  • #4
I just realized, I might have messed up even more than I thought. I'm missing [tex]e^{-t}[/tex] from my transformation. I'd greatly appreciate it, if someone can locate my mistake. I've been banging my head to the wall and I really wouldn't like to start from scratch.
 
  • #5
For t>0 you are using only the capacitor and the inductor. What about the resistor, that is in parallel with the inductor?
It is the resistor thet will provide the attenuation (the [tex]e^{-\alpha t}[/tex] that is missing).
 
  • #6
CEL said:
For t>0 you are using only the capacitor and the inductor. What about the resistor, that is in parallel with the inductor?
It is the resistor thet will provide the attenuation (the [tex]e^{-\alpha t}[/tex] that is missing).

Oh yeah, you're right! So [tex]I_L(s)= \frac {U_{C0}}{L(s^2+ \frac{s}{RC} + \frac {1}{LC}}= \frac {2000 \sqrt{2}}{5}*\frac {1}{s^2+1000s+1000}=\frac {2000 \sqrt{2}}{5*500 \sqrt {3}}* \frac {500 \sqrt {3}}{(s+500)^2+(500 \sqrt {3})^2}[/tex]
Then [tex]i_k(t)=\frac {4 \sqrt{2}}{5 \sqrt {3}}e^{-500t}sin(500 \sqrt{3}t)[/tex].

Now it should make sense! Thank you!
 
Last edited:

What is circuit analysis?

Circuit analysis is the process of studying and understanding how electrical circuits work. It involves analyzing the components and connections in a circuit to determine the voltage, current, and power at different points.

What is Laplace transform in circuit analysis?

The Laplace transform is a mathematical tool used in circuit analysis to simplify complex differential equations. It allows us to convert a time-domain function into a frequency-domain function, making it easier to solve circuit problems.

What are some common problems encountered in Laplace circuit analysis?

Some common problems encountered in Laplace circuit analysis include improper choice of initial conditions, incorrect application of the Laplace transform, and difficulty in converting back to the time-domain.

How can I improve my Laplace circuit analysis skills?

To improve your Laplace circuit analysis skills, practice solving a variety of circuit problems using the Laplace transform. You can also refer to textbooks, online resources, and seek help from experienced engineers or professors.

What are some real-world applications of Laplace circuit analysis?

Laplace circuit analysis has many real-world applications, including in the design of electronic circuits, control systems, and signal processing. It is also used in fields such as telecommunications, power systems, and biomedical engineering.

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