Class 12 Physics Alternating Current Resonance

Resonance

• Resonance is defined as the tendency of the system to oscillate at greater amplitude at some frequencies than at others.
• It is common among the systems that have a tendency to oscillate at a particular frequency and that frequency is known as natural frequency.
• It is common among the systems which have the tendency to oscillate at a particular frequency.

Examples:-

• A pendulum oscillates at its natural frequency. If a push is given  to  pendulum its  amplitude increases. This frequency with which pendulum oscillates with a greater amplitude is known as the resonance frequency. • Swing. A child when sitting on a swing, he swings at his natural frequency. But if someone gives a push to the swing from the behind at the same frequency with which the swing was swinging earlier. Then the amplitude increases ,this is known as resonance and frequency is known as resonant frequency.

Problem:- Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

Inductance, L = 2.0 H

Capacitance, C = 32 μF = 32 × 10−6 F

Resistance, R = 10 Ω

Resonant frequency is given by the relation,

ωr =(1√LC)

=(1/√2 × 32 × 10−6)

=1/(8 × 10−3) = 125 rad/s.

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

Q = (1/R)√(L/C)

=(1/10)√(2)/(32 × 10−6)

=1/(10 × 4 × 10−3) = 25

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

Resonance of Series LCR circuit

• At resonant frequency: Amplitude is  maximum.
• In LCR circuit, current amplitude is given as:- Im  = (Vm/Z).
• => Im  = (Vm/√( R2 + (XC - XL)2)
• At resonance ,Im = max => Z = minimum when (XC - XL) =0 => XC = XL
• =>(1/ωC) = ωL => ω =(1/√LC).
• This value is known as resonant frequency ω0 =(1/√LC) .
• From the graph we can see that the value of Im increases with the value of ω, it reaches a maximum value which is ω0 and then again it decreases.

Important Note: -

• Resonance is exhibited by a circuit only if both L and C are present in the circuit.
• Only then the voltages across L and C cancel each other (as both being out of phase) and the current amplitude is (Vm/R), the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC The above graph shows the variation of im with w for two cases:

(i) R = 100 W, (ii) R = 200 W, L = 1.00 mH.

Problem:- A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:- At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.

Resistance, R = 20 Ω

Inductance, L = 1.5 H

Capacitance, C = 35 μF = 30 × 10−6 F

AC supply voltage to the LCR circuit, V = 200 V

Impedance of the circuit is given by the relation,

Z = √R2 + (XL − XC)2

At resonance, XL = XC

∴ Z = R = 20 Ω

Current in the circuit can be calculated as:

I = (V/Z)

=(200/20) = 10 A

Hence, the average power transferred to the circuit in one complete cycle:

VI = 200 × 10 = 2000 W.

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