MENU
Question -

A series LCR circuitwith = 0.12 H, = 480 nF, =23 Ω is connected to a 230 V variable frequency supply.

(a) What isthe source frequency for which current amplitude is maximum. Obtain thismaximum value.

(b) What isthe source frequency for which average power absorbed by the circuit is maximum.Obtain the value of this maximum power.

(c) Forwhich frequencies of the source is the power transferred to the circuit halfthe power at resonant frequency? What is the current amplitude at thesefrequencies?

(d) What isthe Q-factor of the given circuit?



Answer -

Inductance, L =0.12 H

Capacitance, C =480 nF = 480 × 10−9 F

Resistance, R =23 Ω

Supply voltage, V =230 V

Peak voltage is given as:

V0 = = 325.22 V

(a) Current flowing inthe circuit is given by the relation, 

Where,

I0 =maximum at resonance

At resonance, we have

Where,

ωResonanceangular frequency

Resonantfrequency, 

And, maximum current 

(b) Maximumaverage power absorbed by the circuit is given as:

Hence, resonant frequency () is 

(c) Thepower transferred to the circuit is half the power at resonant frequency.

Frequenciesat which power transferred is half, = 

Where,

Hence, change in frequency, 

And 

Hence, at 648.22 Hz and 678.74 Hzfrequencies, the power transferred is half.

At these frequencies, currentamplitude can be given as:

(d) Q-factorof the given circuit can be obtained using the relation, 

Hence, the Q-factor of the givencircuit is 21.74.

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×