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In R-L-C series circuit, we have same cu...

In `R-L-C` series circuit, we have same current at angular frequencies `omega_(1)` and `omega_(2)`. The resonant frequency of circuit is

A

`(omega_(1)^(2))/(omega_(2))`

B

`(omega_(2)^(2))/(omega_(1))`

C

`sqrt(omega_(1) omega_(2))`

D

`omega_(1) + omega_(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`omega_(1) L - (1)/(omega_(1) C) = omega_(2) L - (1)/(omega_(2) C)`
Solving `omega_(1) omega_(2) = (1)/(sqrt(LC)) = omega_(0)^(2)`
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Knowledge Check

  • The resonant frequency of an L - C circuit is

    A
    ` ( 1 )/( 2pi sqrt ( L C)) `
    B
    ` ( 1 ) /( 2 pi ) sqrt (( L ) / ( C)) `
    C
    ` ( 1 )/( 2 pi ) sqrt ( ( L ) /( C)) `
    D
    ` ( 1 )/( 2pi ) sqrt (( C) / ( L)) `

  • For an LCR series circuit with an aac source of angular frequency omega .

    A
    circular will be capacitive if `omegagt(1)/(sqrt(LC))`
    B
    circular will be inductive if `omega=(1)/(sqrt(LC))`
    C
    Power factor of circuit will be unity if capacitive reactance equals inductive reactance
    D
    circular will be leading voltage if `omegagt(1)/(sqrt(LC))`

  • In a series LCR circuit, impedance Z is same at two frequencies f_1 and f_2 . Therefore, the resonant frequency of this circuit is

    A
    `(f_1+f_2)/2`
    B
    `(2f_1f_2)/(f_1+f_2)`
    C
    `sqrt(f_1^2+f_2^2)/2`
    D
    `sqrt(f_1f_2)`

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