A series LCR circuit with `L = 0.12 H, C = 480 nF, R = 23 Omega` is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?

Explanation

As shown in above resonance curve, a and b are half power points. The half power width is given by formula,
`Delta omega - ( R )/( L )`
`:. Delta(2pi f ) = ( R )/( L )`
`:. 2 pi Delta f = ( R )/( L )`
`:. Delta f= ( R )/( 2 pi L )`
`:. ( Delta f)/( 2 ) = ( R )/( 4 pi L )`
`= ( 23)/( 4 xx 3.14 xx 0.12)`
`= 15.26 Hz`
Now, frequency at point a is,
`f_(1) = f_(0) - ( Delta f)/( 2)`
`= 663.5-15.26`
= 648.24 Hz
Similarly, frequency at point b is,
`f_(2) = f_(0) - (Delta f)/( 2)`
`= 663.5 + 15.26`
= 678.76 Hz
Current at points a and b are,
`((I_(rms))_("max"))/(sqrt(2))`
`= ( I_(m))/(sqrt(2))`
`= (14.14)/(1.414)`
`10A`