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CBSE Class 12-science Answered

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Asked by jain.pradeep | 16 Feb, 2020, 08:17: PM
answered-by-expert 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
 
Error converting from MathML to accessible text. ...................................(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
--------------------------------------------------------------
 
At resosnance we have XC = XL or  1/( ωC )  = Lω   or  LC = 1/ω2
 
-----------------------------------------------
 
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
-----------------------------------------------------------------
 
 
Answered by Thiyagarajan K | 17 Feb, 2020, 09:56: PM

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