(a) Given: L = 0.12 H

C = 480 nF

Converting into Farad, we get 480 × 10^-9F

R = 23Ω

_{Voltage = 230 V}

_{Peak voltage can be calculated as follows:}

_V0 = √2V

_{Substituting values we get}

_V0 = 230√2

_V0 = 325.22 V

_{Current flowing through the circuit is given by the following:}

I₀ =

At resonating frequency, ω_R L - 1/ω_RC = 0

ω_R = 1/√LC

= 1/ √0.12 (H) × 480 × 10^-9 (F) = 4166.67 rad/s

Resonant frequency can be calculated using the formula:

v_R = ω_R/2π

v_R = 4166.67(Hz)/2 × 3.14

v_R = 663.48 Hz

_{Maximum current is calculated as follows:}

_(I0)_max = V₀/R = 325.22(Hz)/23 = 14.14 A

_{(b) The maximum power absorbed by the circuit can be calculated as follows:}

_Pav = 1/2(I₀)²_maxR

_{Substituting the values, we get}

_Pav = 1/2 × (14.14)²(A) × 23 (Ω)

_Pav = 2299.33 W

_{Therefore, the resonating frequency is 663.48 Hz.}

_{(c) The power transferred is equal to the half of the power at the resonating frequency.}

_{Frequency at which power is half = ωR} �∆ω

_{Or = 2π (vR} �∆v)

_{Where ∆ω = R/2L}

_{Substituting the values we get}

_{∆ω = 23/2(Ω) × 0.12(H)}

_{∆ω = 95.83 rad/s}

_{Therefore, change in frequency is written as follows:}

_{∆v = 1/2π × ∆ω}

_{Substituting the values, we get}

_{∆v = 95.83(rad/s)/2π = 15.26 Hz}

_vR + ∆v is calculated as follows:

_{663.48(Hz) + 15.26(Hz) = 678.74 Hz}

_vR-∆v is calculated as follows:

_{663.48(Hz) – 15.26(Hz) = 648.22 Hz}

_{The current amplitude is calculated as follows:}

_{I’ = 1/√2 × (I0})_max

_{I’ = 14.14(A)/ √2}

_{I’ = 10 A}

_{(d) Q factor can be calculated as follows:}

_{Q = ωR}L/R

_{Q = (4166.67 rad/s) × 0.12H/23Ω}

_{⇒ Q = 21.74}