Question

Two series resonant filters are shown below. Let the 3 dB bandwidth of filter 1 be B1 and that of filter 2 be B2. The value of \(\frac{B_1}{B_2}\) is ____________ (a) 0.25 (b) 1 (c) 0.5 (d) 0.75 I got this question by my school principal while I was bunking the class. This interesting question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory Select the correct answer from above options network theory questions and answers, network theory questions pdf, network theory question bank, network theory gate questions and answers pdf, mcq on network theory pdf, gate network theory questions and solutions, network theory mcq Test , ies network theory questions, Network theory Questions for GATE EC Exam, Network Theory MCQ (Multiple Choice Questions)

Answer 1

The correct option is (a) 0.25 To elaborate: For series resonant circuit, 3dB bandwidth is \(\frac{R}{L}\) B1 = \(\frac{R}{L_1}\) B2 = \(\frac{R}{L_2} = \frac{4R}{L_1}\) Hence, \(\frac{B_1}{B_2}\) = 0.25.