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Find the bandwidth of the circuit shown below.(a) 1(b) 2(c) 3(d) 4I got this question during an interview.The doubt is from Parallel Resonance topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

Right choice is (C) 3

For explanation: The bandwidth of the circuit is BW = fr/Q. we OBTAINED fr = 7.12 HZ and Q = 2.24. On substituting the given VALUES in the equation we get the bandwidth = 7.12/2.24 = 3.178Hz.



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