Capacitance of the capacitor, C = 100 μF = 100 x 10^-6 F = 10⁻⁴ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Ans (a).

Frequency of oscillations, ν= 60 Hz

Angular frequency, ω = 2π ν = 2π x 60 rad/s =120π rad/s

For a RC circuit, we have the relation for impedance as: Z = √(R² +1/ω²c²)

Peak voltage, V₀ = V√2 = 110√2. Maximum current is given as:

I₀= V₀ /Z  =  V₀/√(R² +1/ω²c²)

=110 √2 / √(R² +1/ω²c²)

=110 √2 / √(40² +1/(120π)²(10⁻⁴)²)

=3.24A

Ans (b).

In a capacitor circuit, the voltage lags behind the current by a phase angle of φ. This angle is given by the relation:

tan φ =(1/ωC)/R = 1/ωCR

=1/(120 π10⁻⁴x 40)

= 0.6635

Therefore, φ =tan⁻¹ (0.6635) = 33.56º

= 33.56π/180 rad

Therefore ,Time lag = φ/ω = 33.56 π/(180 x 120 π) = 1.55 x 10⁻³ s

=1.55 ms

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.

Inductance of the inductor, L = 20 mH = 20 x 10⁻³ H

Capacitance of the capacitor, C = 50 pF = 50 x 10⁻⁶ F

Initial charge on the capacitor, Q = 10 mC = 10 x 10^-3 C

Ans (a).

Total energy stored initially in the circuit is given as:

E = 1/2 x Q²/C =  (10 x 10^-3)²/ (2 x 50 x 10⁻⁶ ) = 1 J

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.

Ans (b).

Natural frequency of the circuit is given by the relation,

ν =  1/[2π√(LC)] =  1/(2π√(20 x 10⁻³ x 50 x 10⁻⁶) = 10³/2π = 159.24 Hz

Natural angular frequency, ω_r = 1/√(LC)

= 1/ √(20 x 10⁻³ x 50 x 10⁻⁶) = 1/ √ 10⁻⁶= 10³ rad/s

Hence, the natural frequency of the circuit is 10³ rad/s.

Ans (c).

(i) For time period (T = 1/ν =1/_159.24 = 6.28 ms), total charge on the capacitor at time t,

Q’ = Qcos2πt/T

For energy stored is electrical, we can write Q’ = Q.

Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =0,T/2,T,3T/2

(ii) Magnetic energy is the maximum when electrical energy, Q’ is equal to 0.

Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time, t = T/4 ,3T/4,5T/4—-

Ans (d).

Q¹ = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = 1/2(maximum energy)

=>1/2  x (Q¹)² /C =1/2 x (1/2C  x  Q²) =  Q²/4C

Q¹ =Q/√2

But Q¹ = Q cos(2π/T) .t

=>  Q/√2= Q cos(2π/T) .t

=> cos(2π/T) .t= 1/√2 = cos (2n+1)π/4; where n= 0,1,2,…

t= (2n+1)T/8

Hence, total energy is equally shared between the inductor and the capacity at time,

t =T/8,3T/8,5T/8,…

Ans (e).

If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.

The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency, ν1 = 800 kHz = 800 x 10³ Hz

Upper tuning frequency, ν2 = 1200 kHz = 1200 x 10³ Hz

Effective inductance of circuit L = 200μH = 200 x 10⁻⁶ H

Capacitance of variable capacitor for ν1 is given as_:

C₁ = 1/(ω₁)²_L

Where, ω₁ = Angular frequency for capacitor C₁= 2πν1= 2π x  800 x 10³ rad/s

Therefore,

C₁ = 1/(2π x 800 x 10³ )² x 200 x 10⁻⁶  = 1.9809 x 10⁻¹⁰ F = 198 pF

Capacitance of variable capacitor for ν2 is given as ;

C2 = 1/(ω₂)²_L

Where, ω₂ = Angular frequency for capacitor C2 = 2πν2 = 2π x 1200 x 10³ rad/s

Therefore ,

C2 = 1/(2π x 1200 x 10³ )² x 200 x 10⁻⁶  = 0.8804  x 10⁻¹⁰ F = 88 pF

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

Capacitance of the capacitor, C = 30μF = 30 x 10⁻⁶F

Inductance of the inductor, L = 27 mH = 27 x 10⁻³ H

Charge on the capacitor, Q = 6 mC = 6 x 10⁻³ C

Total energy stored in the capacitor can be calculated as:

E = 1/2   x Q²/C  = 1/2 x (6 x 10⁻³)²/(30 x 10⁻⁶) = 6/10 = 0.6 J

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

Inductance, L = 2.0 H

Capacitance, C = 32 μF = 32 x 10^-6 F

Resistance, R = 10Ω

Resonant frequency is given by the relation,

ω_{r= 1/ √ (LC) = 1/ √ (2 x 32 x 10^-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s}

Now ,Q-value of the circuit is given as :

Q = 1/R_√ (L/C) = (1/10) x _{√[2/(32 x 10^-6 ) ]}

= 1/(10 x 4 x 10⁻³ ) = 25

Hence, the Q-Value of this circuit is 25.