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Draw LC circuit and calcilate impedence associated with it

Asked by dawarsonali783 | 15 May, 2021, 17:34: PM

Expert Answer

Figure shows the series and parallel circuit of Inductor and capacitor .

Impedance Z_s 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 Z_L is impedance of inductor, Z_C is impedance of capacitor and

ω is the angular frequency of applied alternating voltage in the circuit

Hence from eqn.(1) , we get impedance Z_s 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

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Impedance Z_p 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 Z_p 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 Thiyagarajan K | 15 May, 2021, 21:28: PM

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