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A series LCR circuit with L= 0.12H, C=48...

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

Text Solution

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Inductance, L=0.12H
Capacitance, `C=480nF=480xx10^(-9)F`
Resistance, R=23`Omega`
Supply voltage `V=230V`
Peak voltage is given as:
`V_(0)=sqrt(2)xx230=325.22V`
Current flowing in the circuit is given by the relation,
Where,
`I_(0)=` maximum at resonance
At resonance, we have
`omega_(R)L-(I)/(omega_(g)C)=0`
where,
`omega_(R)=` Resonance angular frequency
`thereforeomega_(R)=(1)/sqrt(LC)`
`=(1)/(sqrt(0.12xx480xx10^(-9)))=4166.67` rad/s
`therefore` Resonant frequency, `V_(R)=(omega_(R))/(2pi)=(416 6 .6 7)/(2xx3.14)=663.48Hz`
And, maximum current `(I _(0))_("max") =(V_(0))/(R)=(325.22)/(23)=14.14A`
(b) Maximum average power absorbed by the circuit is given as:
`(P_(as))_("ma x")=(1)/(2)(I_(0))_("Max")^(2)R`
`=(1)/(2)xx(14.14)^(2)xx23= 2 299.3W`
Hence, resonant frequ ency `(V_(R))` is 663.48Hz.
(c) The power tranferred to the circuit is h alf the powr at resonant frequency.
Freqeuncies at which power transferred is half, `=omega_(R)+-Deltaomega`
`=2pi(V_(R)+-Deltav)`
where,
`Deltaomega=(R)/(2L)`
`=(23)/(2xx0.12)=95.83` rad/s
Hence, change in frequency, `Delta v=(1)/(2pi)Deltaomega=(95.83)/(2pi)=15.26Hz`
`thereforeV_(R)+DeltaV=663.48+15.26=678.74Hz`
And, ` v_(R)-DeltaV=663.48 -15.26= 648.22Hz`
Hence, at 648.22Hz and 678.74Hz frequencies, the power transferred is half
At these frequencies, current amplitude can be given as:
`I'(1)/sqrt(2)xx(I_(0))_("Max")`
`(14.14)/sqrt(2)=10A`
(d) Q-factor of the gven circuit can be obtained using the relation ` Q=(omega_(R)L)/(R)`
`=(4166.67xx0.12)/(23)=21.74`
Hence, the Q-factor of the given circuit is 21.74.
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