CBSE Class 12-science Answered
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Asked by jain.pradeep | 16 Feb, 2020, 08:17: PM
Expert Answer
If alternating voltage v = vm sin(ωt) is applied to LCR circuit, then current i in the circuit is given by
i = im sin(ωt + φ) ...................(1)
where vm and im are peak voltage and peak current respectively. ω = 2πf is angular frequency
Peak current im is given by, im = vm / z ......................(2) ,
where z is the impedence in the circuit which is given by
...................................(3)
where R is the resistance in the circuit, XC = 1/( ω C ) is capacitive reactance and XL = Lω is the inductive reactance.
Phase angle φ is given as, tanφ = ( XC - XL ) / R .........................(4)
Hence plot of current i as a function of ω is obtained using eqn.(1), (2), (3) and (4) using the known values of vm ,
Capacitance C , inductance L and resistance R
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At resosnance we have XC = XL or 1/( ωC ) = Lω or LC = 1/ω2
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At resosnace, XC = XL . Hence impedence z is minimum and we get maximum current.
This maximum current is decided by Resistance R in the circuir. If Resistance is less we get more current at resosnance.
Hence for fine tuning, circuit with resistance R1 is more suitable because R1 < R2
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Answered by Thiyagarajan K | 17 Feb, 2020, 09:56: PM
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