# Vout=-1/RCVindt

Discussion in 'Homework Help' started by ronybhai, Sep 18, 2012.

1. ### ronybhai Thread Starter Member

Sep 18, 2012
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Vout=-(1/RC)(∫Vin dt) how to solve these equation when Vin=5v, R is 10kΩ and C is 0.01µF?

Last edited: Sep 18, 2012
2. ### WBahn Moderator

Mar 31, 2012
23,408
7,115
By, perhaps, doing what the equation says? Integrate Vin when it is a constant 5V and divide by the product of R and C.

Q1) What is the integral of 5V as a function of time?

Q2) What are your initial conditions (i.e., what is Vout at t=0)?

3. ### ronybhai Thread Starter Member

Sep 18, 2012
97
0
question ask to calculate the output voltage of the integrator. Assume the 100k ohm resistor behave as an open circuit.
this is the diagram
http://i45.tinypic.com/2zgevb6.jpg

Sep 18, 2012
97
0
5. ### WBahn Moderator

Mar 31, 2012
23,408
7,115
Okay, so assume that for t<0 that the input voltage was 0V. What is the output voltage at t=0s? Now assume that, at t=0s, the input voltage steps up to 5V. What is the current into the capacitor? What is the output voltage in terms of the capacitance and the current into the capacitance?

Yes, you will come up with the same equation you started with, which brings us back to my first question - what is the integral of 5V as a function of t?

Given that Vout as a function of t starting at t=0s, how long is that function good for (i.e., until t=what)?. What does the function become then?