Electronics - Operational Amplifiers - Discussion

I'm also getting 4.8 khz please explain.

Please explain! I am not getting correct answer.

My answer coming 1.8khz

F = 1/2*pi*R*C

F = 4.8Khz.

Nisha could you please tell me how can we take that the value of Pi =1.34.

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

Take load resistance RL = R = 10K.

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

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.

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.

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