Question

For the circuit given below, the cut-off frequency of the filter is ________________

(a) 3225.8 Hz

(b) 7226 Hz

(c) 3225.8 kHz

(d) 7226 kHz

This question was posed to me by my school teacher while I was bunking the class.

This interesting question is from Advanced Problems on Filters topic in chapter Filters and Attenuators of Network Theory

Answer 1

Right answer is (c) 3225.8 kHz

Easy explanation: We know that, F = \(\frac{1}{2π×(R_1×R_2)×(C_1×C_2)}\)

Where, R1 = 4700 Ω, R2 = 4700 Ω, C1 = 0.047 × 10^-6 F, C2 = 0.047×10^-6 F

∴ F = \(\frac{1}{2π×(4700×4700)×(C_1×C_2)}\)

= \(\frac{1}{2π×48796.81×10^{-12}}\)

= \(\frac{10^6}{2π×0.04879681}\)

= \(\frac{10^6}{0.30654156042}\) = 3225.8 kHz.