Electronics - Operational Amplifiers - Discussion

My answer coming 1.8khz

Can you explain the process?

I'm getting 4.8kHz i.e fc=(1/2*pi*C*R)

Sub values v get 4.822 khz.

Please any one tel me is it correct or not ?

I too getting 4.8 khz please explain.

Please tell me cutoff frequency formula for low pass filter.

fc=1/(2*pi*R*C)
= 1/(2*3.14*33*10^3*0.001*10^-6)
= 4.8KHZ

At cut off freq output of opamp should be 1/2^.5.
At frequency 0hz output is=vin

So at cutoff freq vout= vin/1.414

Because vout = vin/(r*s*C+1)
Or |v|=vin/((r*s*c)^2+1)^.5

So (r*s*c)^2=1
So r*s*C=1
So 1.8k ans.

Could you please elaborate your answers @D Sati and @Vikas. If we calculate with the formula you gave it again gives the answer as 4.8KHz. And the other thing I would like to put forth before you is that the advantage of this configuration is that the op amps high input impedance prevents excessive loading on the filters output while its low output impedance prevents the filters cutoff frequency point from being affected by changes in the impedance of the load.

Take load resistance RL = R = 10K.

Because the formula 1/(2*pi*RL*c).

I think 4.8khz is correct answer. Because of LPF cutoff frequency formula is 1/2*pi*R*c.