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 
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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