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GATE ECE 2012 | Question: 41

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The circuit shown is a 

  1. low pass filter with $f_{3\:dB}=\frac{1}{(R_1+R_2)C}\: rad/s$
  2. high pass filter with $f_{3\:dB}=\frac{1}{R_1C}\: rad/s$
  3. low pass filter with $f_{3\:dB}=\frac{1}{R_1C}\: rad/s$
  4. high pass filter with $f_{3\:dB}=\frac{1}{(R_1+R_2)C}\: rad/s$
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