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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) 0.75I 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

Answer»

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.



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