Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 x 10⁻⁹ F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as: Vo =√2 x 230 = 325.22 V

Ans (a).

Current flowing in the circuit is given by the relation,

I₀=  V₀/√[(R² + (ωL- 1/ωC)²]

Where, Io = maximum at resonance

At resonance, we have

ω_R– 1/ω_RC =0

Where, ω_R  = Resonance angular frequency

Where , ω_R = Resonance angular frequency

Therefore ω_{R = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻⁹ ) = 4166.67 rad/s}

Resonant frequency , ν_R = ω_R/2π = 4166.67 /(2 x 3.14) = 663.48 Hz

And ,maximum current,   (I₀)_Max  =   V₀/R = 325.22/23 = 14.14 A

Ans (b).

Maximum average power absorbed by the circuit is given as:

 (P_av)_Max= 1/2 x  (I₀)²_MaxR

= 1/2 x (14.14)² x 23 = 2299.3 W

Hence, resonant frequency is 663.48 Hz.

Ans (c).

The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half,

=  ω_R ± Δω

= 2π (ν_R ± Δν)

Where,

Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s

Hence, change in frequency,

Δν = 1/2π    x    Δω  =95.83 /2π = 15.26 Hz

Therefore

ν_R +Δν  = 663.48 + 15.26 = 678.74 Hz

and ν_R – Δν  = 663.48 – 15.26 =648.22 Hz

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:

I’ = 1/√ 2 x (I₀)_Max

= 14.14/√ 2 = 10A

Ans (d).

Q-factor of the given circuit can be obtained using the relation,

Q = ω_R L/R  = 4166.67 x 0.12 /23 = 21.74

Hence, the Q-factor of the given circuit is 21.74.