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GATE Electrical 2012 | Question: 32

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The circuit shown is a

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