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CBSE Class 12-science Answered

A resistance of 600 , an inductor of 0.4 H and a capacitor of 0.01 uF are connected in series to an AC source of variable frequency. Find the frequency of AC source for which current in the circuit is maximum. Also calculate the band width and quality factor for the circuit.

Asked by shivakumarshreyas24 | 01 Mar, 2020, 08:12: AM
Expert Answer
frequency of AC source when current is maximum, ωo = 2πf = 1/ (LC)1/2 
 
Hence begin mathsize 14px style f space equals space fraction numerator 1 over denominator 2 pi square root of L space C end root end fraction space equals space fraction numerator 1 over denominator 2 pi square root of 0.4 cross times 0.01 cross times 10 to the power of negative 6 end exponent end root end fraction space equals space 2516 space H z space equals space 2.516 space k H z end style
 
Bandwidth  2Δω = R/L = 600/.4 = 1500 Hz = 1.5kHz
 
Quality Q = ( ωo L ) / R = ( 2516 × 0.4 ) / 600 = 1.68
Answered by Thiyagarajan K | 01 Mar, 2020, 12:35: PM
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