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 kHzThis 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

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

1.	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 kHzThis 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» 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.

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