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A series LCR circuit containing a resist...

A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the voltage across resistance and inductance are 60V and 40 V, respectively,
At what frequency, the current in the circuit lags the voltage bu `45^(@)` ?

A

`16xx10^(5)"rad "s^(-1)`

B

`8xx10^(5)"rad "s^(-1)`

C

`4xx10^(5)"rad "s^(-1)`

D

`2xx10^(5)"rad "s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
At resonance as X = 0,
`I = (epsilon_(0))/(R )=(60)/(120)=(1)/(2)A=0.5 A`
As `V_(L)=IX_(L)=I omega L rArr L = (V_(L))/(I omega)`
`therefore L = (40)/((0.5)xx4xx10^(5))=0.2 mH`
Also, `omega_(0)=(1)/(sqrt(LC)) therefore C=(1)/(L omega_(0)^(2))`
`=(1)/(0.2xx10^(-3)xx(4xx10^(5))^(2))=(1)/(32)mu F`
Now in case of series LCR circuit.
`therefore tan phi = (X_(L)-X_(C ))/(R )`
So current will lag the applied voltage by `45^(@)` if,
`tan 45^(@)=(omega L-(1)/(omega C))/(R )` ....(i)
or `R = omega L -(1)/(omega C) " " (because tan 45^(@)=1)`
or `120=omegaxx2xx10^(4)-(1)/(omega(1//32)xx10^(-6))` (using (i))
`rArr omega^(2)-6xx10^(5)omega-16xx10^(10)=0`
`therefore omega = (6xx10^(5)pm sqrt((6xx10^(5))^(2)+64xx10^(10)))/(2)`
`therefore omega = (6xx10^(5)+10xx10^(5))/(2)=8xx10^(5)"rad "s^(-1)`
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