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.

Inductance, L = 80 mH = 80 x 10⁻³ H,
Capacitance, C = 60 pF = 60 x 10⁻⁶ F
Supply voltage, V = 230 V,

Frequency, ν = 50 Hz,

Angular frequency, ω = 2πν = 100 π rad/s

Peak voltage, Vo= √2 = 230√2 V

Ans (a).

Maximum current is given as :

Io = Vo /(1/ωL -ωC)

= 230 √3 /[(100π x 80 x 10⁻³ )- 1/ (100π x 60 x 10⁻⁶ )]

= 230√2 / (8π -1000/6π) = -11.63 A

The negative sign appears because ωL <1/ωC

Amplitude of maximum current, |I₀| = 11.63 A

Hence, rms value of current =I = I₀/√2 =11.63 /√2  = -8.22 A

Ans (b).

Potential difference across the inductor,

V_L= I x ωL

= 8.22 x 100π x 80 x 10⁻³= 206.61 V

Potential difference across the capacitor,

V_c = I x 1/ωC

= 8.22 x 1/(100π x 60 x 10⁻⁶ )  = 436.3 V

Ans (c).

Average power consumed by the inductor is zero as actual voltage leads the current by π/2

Ans (d).

Average power consumed by the capacitor is zero as voltage lags current by π/2

Ans (e).

The total power absorbed (averaged over one cycle) is zero.

Capacitance of the capacitor, C = 100 μF = 100 x 10^-6 F =10⁻⁴F
Resistance of the resistor, R = 40Ω
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 x 10³ Hz
Angular Frequency, ω = 2π ν = 2 x π x 12 x 10³ Hz
= 24π x 10³ rad/s 
Peak voltage, V₀ = V√2 = 110√2 V
Maximum Current ,I₀=  V₀/√(R² +1/ω²c²)

=110 √2 /√[(40)² +1/(24π x 10³)²(10⁻⁴)²]

=110 √2/√[1600 + (10/24π)²] = 3.9 A

For an RC circuit, the voltage lags behind the current by a phase angle of φ given as:

tan φ = (1/ωC)/R = 1/ωCR = 1/(24π x 10³  x 10⁻⁴ x 40)

tan φ = 1/96π

Therefore , φ = 0.2º or 0.2 π/180 rad

Therefore ,Time lag = φ/ω  =  0.2 π/(180 x 24π x 10³)

=4.69 x 10⁻³s

= 0.04μs

Hence, φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor. In a dc circuit, after the steady state is achieved, ω = 0.

Hence, capacitor C amounts to an open circuit.