RC integrator circuit

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  • Thread starter hmzi123
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Homework Statement:

RC integrator circuit

Relevant Equations:

RC integrator circuits
1.png

Hello, i am student from Romania ( first year) and my physics teacher told us to solve this problem. We know that C=150 nanofarad, R= 1000 ohms, and the frequency we should take is more then 1061 ( 10kHz, to be more exactly). My problem is that i dont know what to take as V1 ( i took it coswt, where w=2*pi*10000), can you help me understand how to choose the V1, and what its formula is so that I can calculate and find V2.
 
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Answers and Replies

  • #2
BvU
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Hello hm, :welcome: !

told us to solve this problem
Did you forget to mention the 'problem'? I don't see what is asked !
 
  • #3
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Hello hm, :welcome: !

Did you forget to mention the 'problem'? I don't see what is asked !
The value of V2. Sorry!
 
  • #4
BvU
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No need to apologize. Can you render the complete problem statement ? Asking for V2 when nothing is given for V1 is weird. The picture does not help much either. The only thing I can suppose is that you are asked to derive ##V_2\; {\bf \approx}\; \displaystyle {1\over RC}\int V_1 dt\quad ## i.e. show what is small enough to be ignored wrt what else.
 
  • #5
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No need to apologize. Can you render the complete problem statement ? Asking for V2 when nothing is given for V1 is weird. The picture does not help much either. The only thing I can suppose is that you are asked to derive ##V_2\; {\bf \approx}\; \displaystyle {1\over RC}\int V_1 dt\quad ## i.e. show what is small enough to be ignored wrt what else.
I took v1 as coswt where w is the 2*pi*frequency ( in this case the teacher said to pick the frequency 10000). Is this correct as a formula for v1 or not?
 
  • #6
BvU
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It satisfies ##R >> {1\over \omega C}##, so yes.

[edit] actually more ##R > {1\over \omega C}## but you can calculate that for yourself, can't you.

Can you find out why it says ##\ \approx\ ## and not ##\ =\ ## ?
 
  • #7
Joshy
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Can you find out why it says ##\ \approx\ ## and not ##\ =\ ## ?
It should be ##\approx## there and the line above it (if I'm seeing that tiny font correct) because ##i \neq V_1/R##. They're omitting the small impedance from the capacitor, which is being dominated by a much larger ##R##.

Another question is why not solve this in frequency domain? Has that been covered in the class?
 
  • #8
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1.jpg

After solving this is what i got. I know i should calculate v2 max value so where i wrote sin(2pi*10000t) i thought that the max value of any sin is 1 so that means the fraction will be 1/(2*pi*10000). Did i solve this any good?
 
  • #9
BvU
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It should be ##\approx## there and the line above it (if I'm seeing that tiny font correct) because ##i \neq V_1/R##
Correct, but you are giving away the "why" -- and that is what I think the exercise wants Hm to find out.
After solving

1590743731166.png
No, not exactly. Work out the correct expression and then apply the approximation.
 

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