Circuit with time varying current

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MatinSAR
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Homework Statement
Consider a circuit containing resistors, capacitors, and an inductor. At time t=0, the circuit is closed. Determine the steady current flowing through the inductor and capacitor as a function of time.
Relevant Equations
KVL and KCL
The picture of the question and my try:
1730163583892.jpeg
1730163652232.jpeg
Would it be possible for someone to double-check my answer for accuracy? I would greatly appreciate any feedback or guidance on this matter.
 
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MatinSAR said:
Homework Statement: Consider a circuit containing resistors, capacitors, and an inductor. At time t=0, the circuit is closed. Determine the steady current flowing through the inductor and capacitor as a function of time.
Relevant Equations: KVL and KCL

The picture of the question and my try:
Would it be possible for someone to double-check my answer for accuracy? I would greatly appreciate any feedback or guidance on this matter.
How about using LATEX ?
 
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SammyS said:
How about using LATEX ?
I always try to do that. I apologize for not using it this time. Is the handwriting too difficult to read? I'm aware that people may be hesitant to download PDFs, which is why I shared an image file in the original post.
 
MatinSAR said:
I always try to do that. I apologize for not using it this time. Is the handwriting too difficult to read? I'm aware that people may be hesitant to download PDFs, which is why I shared an image file in the original post.
You could also improve the thread by posting the complete, verbatim problem statement. I strongly suspect 'At time t=0, the circuit is closed. Determine the steady current... ' is a confusing rephrasing.

Could it be that there is nothing that closes at t=0 and the exercise asks for the steady state solution ?

##\ ##
 
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BvU said:
You could also improve the thread by posting the complete, verbatim problem statement. I strongly suspect 'At time t=0, the circuit is closed. Determine the steady current... ' is a confusing rephrasing.

Could it be that there is nothing that closes at t=0 and the exercise asks for the steady state solution ?

##\ ##
Before t=0 the key was open. At t=0 the key is closed. That's what I meant.
 
BvU said:
How come I don't see no key ? Please
I forget to draw that. The problem is not stated in any English book, So I cannot give you a screenshot of that book or sth like this. This problem was stated by my Professor and I translated it in English. Sorry if some information is missing. This is all I remember from the question. I'm not good at writing lecture notes.

BvU said:
You appear to be familiar with complex impedance etc. So why not work out ##\bf I = V/Z## ?
I wanted to use KVL law to solve. I just realize that working with the equation you stated is better here, because we have only one battery. Thanks.
 
The solution provided by the original poster in the post #2 is confusing at the beginning. In the circuit diagram the symbol, which is used for the generator, is a symbol for a current source, but in the equations the expression $$ 2\sin(2t+\frac{17}{9}) $$ is used as an expression for a voltage source.
 
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I just found that why my answer in OP is wrong. Thank you for your help @BvU ...
Gavran said:
The solution provided by the original poster in the post #2 is confusing at the beginning. In the circuit diagram the symbol, which is used for the generator, is a symbol for a current source, but in the equations the expression $$ 2\sin(2t+\frac{17}{9}) $$ is used as an expression for a voltage source.
Your post shows that I should only use LATEX , That's not ##17/9## , it was ##\pi /4##!
And you're right. I shouldn't have changed that symbol.
 
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MatinSAR said:
This is all I remember from the question. I'm not good at writing lecture notes.
Without a clear understanding of the question it's not possible to have a meaningful discussion of the answer.

I suggest you sketch the circuit the best you can and state the problem the best you can. Then we can help you solve that problem, whether or not it's an accurate description of the professor's question. That’s the best we can do.
 
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