# Solve RC Integrator Circuit Problem | Romania Student

• Engineering
• hmzi123
In summary, the conversation was about a physics problem involving C=150 nanofarads, R=1000 ohms, and a frequency of more than 10kHz. The student was unsure about the formula for V1 and requested help understanding how to choose V1 and its formula to calculate and find V2. The expert explained that V1 can be taken as coswt where w=2*pi*10000 and it satisfies R >> 1/(omega C). The expert also suggested finding out why the solution uses the approximation symbol instead of an equal sign and if solving in the frequency domain has been covered in class.
hmzi123
Homework Statement
RC integrator circuit
Relevant Equations
RC integrator circuits

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 don't 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.

Last edited:
Hello hm, !

hmzi123 said:
told us to solve this problem
Did you forget to mention the 'problem'? I don't see what is asked !

BvU said:
Hello hm, !

Did you forget to mention the 'problem'? I don't see what is asked !
The value of V2. Sorry!

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.

BvU said:
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?

It satisfies ##R >> {1\over \omega C}##, so yes.

 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 ##\ =\ ## ?

BvU said:
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?

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?

Joshy said:
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.
hmzi123 said:
After solving

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

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## What is an RC integrator circuit?

An RC integrator circuit is an electronic circuit that uses a resistor (R) and a capacitor (C) to integrate a signal over time. It is commonly used in electronic filters, signal processing, and audio applications.

## How does an RC integrator circuit work?

An RC integrator circuit works by charging and discharging the capacitor to integrate the input signal. When the input signal changes, the capacitor voltage also changes, and this change is integrated over time by the resistor. The output voltage of the circuit is proportional to the integral of the input signal.

## What is the time constant of an RC integrator circuit?

The time constant of an RC integrator circuit is equal to the product of the resistance (R) and the capacitance (C). It represents the time it takes for the capacitor to charge or discharge to 63.2% of its final value.

## How can I solve an RC integrator circuit problem?

To solve an RC integrator circuit problem, you can use Kirchhoff's laws and the equations for the charging and discharging of a capacitor. You can also use circuit analysis techniques such as nodal analysis or mesh analysis. Additionally, you can use simulation software to model and analyze the circuit.

## What are some applications of RC integrator circuits?

RC integrator circuits have various applications, including low-pass filters, audio amplifiers, signal processing, and analog-to-digital converters. They are also used in electronic synthesizers, oscilloscopes, and other electronic instruments.

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