An AC circuit contains a resistance R, an inductance L and a capacitance C connected in series across an alternator of constant voltage and variable frequency. At resonant frequency, it is found that the inductive reactance, the capacitive reactance and the resistance are equal and the current in the circuit is `i_(0)`. Find the current in the circuit at a frequency twice that of the resonant frequency.

Explanation

At Resonance `R = omega_(0)L=(1)/(omega_(0)C)`
At `f=2f_(0),omega_(0)L=2pi fL=2pi(2f_(0))L`
`= 2 omega_(0)L=2R`
`f = 2f_(0)`,
`(i)/(omega C)=(1)/(2pi fC)=(11)/(2pi(2f_(0))C)=(1)/(2).(1)/(omega_(0)C)=(R )/(2)`
current (i) `= (E_(0))/(Z)`
`i = (i_(0)R)/(sqrt(R^(2)+(omega L-(I)/(omega C))^(2)))`
`i=(i_(0)R)/(sqrt(R^(2)+(2R-(R )/(2))^(2))) " " (because E_(0)=i_(0)R)`
`= (i_(0)R)/(sqrt(R^(2)+(9R^(2))/(4)))=(2i_(0)R)/(R sqrt(13))`
`i = (2i_(0))/(sqrt(13))`