Question

For the series circuit given below, the value of the cut-off frequencies are ____________ (a) 78.235 kHz; 16.215 kHz (b) 13.135 kHz; 81.531 kHz (c) 16.716 kHz; 15.124 kHz (d) 50.561 kHz; 18.686 kHz The question was asked by my college director while I was bunking the class. This is a very interesting question from Problems of Series Resonance Involving Quality Factor 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

Right option is (c) 16.716 kHz; 15.124 kHz Best explanation: Resonant Frequency, FR = \(\frac{1}{2π\sqrt{LC}} \) = \(\frac{1}{6.28\sqrt{(5 × 10^{-3})(0.02 × 10^{-6})}}\) = \(\frac{1}{6.28\sqrt{10^{-10}}}\) = \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz. Inductive Reactance, XL = 2πfL = (6.28) (15.92 × 10^3)(5 × 10^-6) = 0.5 kΩ ∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10 ∴ ∆F = \(\frac{f}{Q} = \frac{15.9 kHz}{10}\) = 1.592 kHz ∴ Bandwidth = \(\frac{∆F}{2}\) = ±796 Hz. Therefore, f2 = f + \(\frac{∆F}{2}\) = 16.716 kHz and f1 = f – \(\frac{∆F}{2}\) = 15.124 kHz.