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Draw LC circuit and calcilate impedence associated with it

 

Asked by dawarsonali783 15th May 2021, 5:34 PM
Answered by Expert
Answer:
Figure shows the series and parallel circuit of Inductor and capacitor .
 
Impedance Zs in Series circuit is calculated as follows
 
begin mathsize 14px style Z subscript s space equals space Z subscript L space plus space Z subscript C space equals space j space left parenthesis L omega space right parenthesis space plus space fraction numerator 1 over denominator j space left parenthesis space C space omega space right parenthesis end fraction space equals space j space left parenthesis L omega space right parenthesis space plus space fraction numerator j over denominator j squared space left parenthesis space C space omega space right parenthesis end fraction space equals space j open parentheses L omega space minus space fraction numerator 1 over denominator C space omega end fraction close parentheses end style .............(1)
Where ZL is impedance of inductor,   ZC is impedance of capacitor and 
ω  is the angular frequency of applied alternating voltage in the circuit
 
Hence from eqn.(1) , we get impedance Zs of series circuit , begin mathsize 14px style Z subscript s space equals space L space omega space minus space fraction numerator 1 over denominator C space omega end fraction end style
--------------------------------------------
 
Impedance Zp in Parallel circuit is calculated as follows
 
begin mathsize 14px style Z subscript p space equals space fraction numerator Z subscript L space Z subscript C over denominator Z subscript L plus Z subscript C end fraction space equals space fraction numerator left parenthesis space j space L space omega space right parenthesis space open parentheses begin display style fraction numerator 1 over denominator j space C space omega end fraction end style close parentheses space over denominator left parenthesis space j space L space omega space right parenthesis space plus space open parentheses begin display style fraction numerator 1 over denominator j space C space omega end fraction end style close parentheses end fraction space equals space fraction numerator j space L space omega over denominator 1 space minus space omega squared space L C end fraction space equals space fraction numerator j space L space omega over denominator 1 space minus space open parentheses omega squared over omega subscript o squared close parentheses end fraction end style........................ (2)
Where begin mathsize 14px style omega subscript o space equals space fraction numerator 1 over denominator square root of L space C end root end fraction end style is the resonant frequency of LC circuit
Hence from eqn.(2) , we get impedance Zp of parallel circuit , begin mathsize 14px style Z subscript p space equals space fraction numerator space L space omega over denominator 1 space minus space omega squared space L C end fraction space equals space fraction numerator space L space omega over denominator 1 space minus space open parentheses omega squared over omega subscript o squared close parentheses end fraction end style


Answered by Expert 15th May 2021, 9:28 PM
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