0 votes 0 votes The circuit shown is a low pass filter with $\text{f}_{3\text{dB}}=\frac{1}{(R_1+R_2)C}\;\text{rad/s}$ high pass filter with $\text{f}_{3\text{dB}}=\frac{1}{R_1C}\;\text{rad/s}$ low pass filter with $\text{f}_{3\text{dB}}=\frac{1}{R_1C}\;\text{rad/s}$ high pass filter with $\text{f}_{3\text{dB}}=\frac{1}{(R_1+R_2)C}{rad/s}$ Others gate2012-in + – Milicevic3306 asked Mar 25, 2018 • edited Nov 20, 2020 by soujanyareddy13 Milicevic3306 7.9k points answer See all 0 reply