The amplitude of current in series L - C - R circuit connected to an AC of frequency `omega` is given by `i_(m)=(V_(m))/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))` where `X_(L) and X_(C )` are inductive and inductive and `V_(m)` is amplitude of voltage. Starting from this equation, show that sharpness of resonance in the circuit is equal to quality factor of the circuit.

Explanation

AC through L-C-R Circuit Suppose that an inductor (L), capacitor (C ) and resitor (R ) are connected in series to an AC source. I is the current passed through this circuit. As R,L,C are in series, therefore at any instant through the three elements AC has the same amplitude and phase. Let it be represented by `i = i_(m) sin omega t`.
However,voltage across each element berars a different phase relationship with the current.

`V_(L)=i_(m)X_(L)[V_(L)"is maximum voltage across L"]`
`V_(C)=i_(m)X_(C)[V_(C)"is maximum voltage across C"]`
`V_(R)=i_(m)R[V_(R)"is maximum voltage across R"]`
Inside the phasor diagram of each L,C and R are given. To form phasor diagram for L-C-R circuit combine all these phasor diagram.

SInce voltage `(V_(L))` is in upward direction and voltage `(V_(C))` in downward direction so net voltage upto point A is `V_(L) - V_(C) ("assuming" V_(L) gt V_(C))` and net maximum voltage is `E_(0)` From phasor diagram.
`OB=sqrt((OC)^(2)+(CB)^(2))=sqrt(V_(R)^(2)+(V_(L)-V_(C)^(2)))`
`V_(m)=sqrt((I_(m)R)^(2)+(i_(m)X_(L)-i_(m)X_(C))^(2))`
`V_(m)=isqrt(R^(2)+(X_(L)-X_(C))^(2))`
So, amplitude of current in series L-C-R circuit is
`i_(m)=(V_(m))/(sqrt(R^(2)+(X_(L)-X_(c))^(2)))`