Class 12 Physics Alternating Current | Applications of Resonance |

**Applications of Resonance**

Resonance circuits have variety of applications. They are as following:-

Tuning circuit of radio or TV set:-

- Inside radio there is a circuit known as tuner circuit. This tuner circuit is LCR circuit.
- Every radio has an antenna which receives signals from multiple stations.
- When we are tuning the knob of the radio to connect to particular station we are changing the capacitance of the capacitor in the circuit.
- As capacitance is changing the resistance also changes and when the natural frequency matches with the resonant frequency then the amplitude will attain the maximum value.
- As a result we will be able to hear song.
- When the amplitude is minimum we won’t be able to hear any song and when amplitude is near to maximum value we will be able to hear the song but the clarity won’t be very clear.

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** Problem:**- A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

** Answer**:-

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

Lower tuning frequency, ν_{1} = 800 kHz = 800 × 10^{3} Hz

Upper tuning frequency, ν_{2} = 1200 kHz = 1200 × 10^{3} Hz

Effective inductance of circuit L = 200 μH = 200 × 10^{−6} H

Capacitance of variable capacitor for ν_{1} is given as:

C_{1} = (1/ω_{1}2L)

Where,

ω_{1} = Angular frequency for capacitor C_{1}

= 2πν_{1}

= 2π × 800 × 10^{3} rad/s

∴ C_{1} = (1/ (2π × 800 × 10^{3})^{2} × 200 × 10^{−6})

= 1.9809 × 10^{−10} F = 198 pF

Capacitance of variable capacitor for ν_{2} is given as:

C_{2} = (1/ω_{2}2L)

Where,

ω_{2} = Angular frequency for capacitor C_{2}

= 2πν_{2}

= 2π × 1200 × 10^{3} rad/s

∴ C_{2} = (1/(2π × 1200 × 10^{3})^{2} × 200 × 10^{−6}

= 0.8804 × 10^{−10} F = 88 pF

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

** Problem**:-

The given Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

** Answer**:-

Inductance of the inductor, L = 5.0 H

Capacitance of the capacitor, C = 80 μH = 80 × 10^{−6} F

Resistance of the resistor, R = 40 Ω

Potential of the variable voltage source, V = 230 V

(a) Resonance angular frequency is given as:

ω_{r} = 1/√LC

=1/√(5 × 80 × 10^{−6})

= (10^{3}/20) = 50 rad/s

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

(b) Impedance of the circuit is given by the relation:

Z = √R^{2} + (X_{L} − X_{C})^{2}

At resonance, X_{L} = X_{C} ⇒ Z = R = 40 Ω

Amplitude of the current at the resonating frequency is given as:

I_{o} = (V_{o}/Z)

Where,

V_{o} = Peak voltage = √2 V

∴ I_{o} = √2 V/(Z)

= (√2 × 230)/40 = 8.13 A

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

(c) RMS potential drop across the inductor,

(V_{L})_{rms} = I × ω_{r}L

Where,

I_{rms} = (I_{o} /√2)

= (√2 V)/(√2 Z)

= (230/40) = (23/4)A

∴ (V_{L})_{RMS} = (23/4) x(50x5)

= 1437.5 V

Potential drop across the capacitor:

∴ (V_{C})_{RMS} = I × (1/ω_{r}C)

=(23/4)×(1/50 × 80 × 10^{−6}) = 1437.5 V

Potential drop across the resistor:

(V_{R})_{RMS} = IR = (23/4) × 40 = 230 V

Potential drop across the LC combination:

V_{LC} = I(X_{L} − X_{C})

At resonance, X_{L} = X_{C} ⇒ V_{LC }= 0

Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.

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