Consider an ideal op-amp filter shown in the figure below

$Then, the cut-off frequency f_{c} is equal to$

21.85 kHz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

17.55 kHz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15.9 kHz

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

19.65 kHz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 15.9 kHz
$f_{c} = \frac{1}{2 π R_{2} C}$

$= \frac{1}{2 π \times 10 \times 10^{3} \times 10^{- 9}} = 15.9 kHz$

Suggest Corrections

Similar questions

Both the resistance R_{1} and R_{2} are constructed with a tolerance of 5%, then the range in which the gain of the amplifier can approximately vary is

then the value of \frac{V_{out}}{V_{{in}_{2}} - V_{{in}_{1}}} is equal to

The value of output voltage V_{0} is equal to

V_{0}

Join BYJU'S Learning Program