Question

A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 W is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit?

Answer 1

Given: The inductance of inductor is 0.12 H, the capacitance of the capacitor is 480 nF, the resistance of resistor is 23 Ω and the applied potential difference is 230 V.

The peak voltage is given as,

V 0 =V 2

Where, the applied potential difference is V.

By substituting the given values in the above equation, we get

V 0 =230 2  V =325.22 V

a)

The angular frequency at resonance is given as,

ω= 1 LC

Where, the inductance is L, the capacitance is C and the resonance angular frequency is ω.

By substituting the given values in the above equation, we get

ω= 1 0.12×480× 10 −9 = 1 5.76× 10 −8 = 10 4 2.4 =4166.67  rad/s

The resonance frequency is given as,

v= ω 2π

By substituting the given values in the above equation, we get

v= 4166.67 2( 3.14 ) =663.48 Hz

The maximum current flowing through the circuit is given as,

I 0 = V 0 R

Where, the resistance of the circuit is R.

By substituting the given values in the above equation, we get

I 0 = 325.22 23 =14.14 A

Thus, the maximum current of the circuit is 14.14 A.

b)

The maximum average power absorbed by the circuit is given as,

P m = 1 2 ( I 0 ) 2 R

By substituting the given values in the above equation, we get

P m = 1 2 ( 14.14 ) 2 ×23 =2299.3 W

Thus, the maximum average power absorbed by the circuit is 2299.3 W.

c)

Since, the power transferred to the circuit is half the power at resonant frequency.

The change in angular frequency given as,

Δω= R 2L

By substituting the given values in the above equation, we get

Δω= 23 2×( 0.12 ) =95.83  rad/s

The change in frequency is given as,

Δv= 1 2π Δω

By substituting the given values in the above equation, we get

Δv= 95.83 2π =15.26 Hz

The values of new frequencies are given as,

v 1 =v+Δv(1)

v 2 =v−Δv(2)

By substituting the given values in the above equations (1) and (2), we get

v 1 =663.48+15.26 =678.74 Hz

v 2 =663.48−15.26 =648.22 Hz

The current amplitude at these two frequencies is given as,

I ′ = 1 2 ( I 0 )

By substituting the given values in the above equation, we get

I ′ = 14.14 2 =10 A

Thus, the frequencies are 648.22 Hz, 678.74 Hz and the current amplitude is 10 A.

d)

The quality factor of the circuit is given as,

Q= ωL R

By substituting the given values in the above equation, we get

Q= 4166.67×0.12 23 =21.74

Thus, the Q-factor of the given circuit is 21.74.