The frequencies at which the current amplitude in an LCR series circuit becomes (1/√2) times its maximum value, are 212 rad s -1 and 232 rad s -1. The value of resistance in the circuit is R =5 Ω. The self inductance in the circuit is mH.

Question

The frequencies at which the current amplitude in an LCR series circuit becomes (1/√2) times its maximum value, are 212 rad s -1 and 232 rad s -1. The value of resistance in the circuit is R =5 Ω. The self inductance in the circuit is mH. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Band width =232-212=( R / L ) ∴ L =(5/20)=250 mH

Q. The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{2}$ times its maximum value, are $212 r a d s^{- 1}$ and $232 r a d s^{- 1}$ . The value of resistance in the circuit is $R = 5Ω$ . The self inductance in the circuit is____ $m H$ .

Band width $= 232 - 212 = \frac{R}{L}$
$∴ L = \frac{5}{20} = 250 m H$

Q. The frequencies at which the current amplitude in an LCR series circuit becomes 2​1​ times its maximum value, are 212rads−1 and 232rads−1. The value of resistance in the circuit is R=5Ω. The self inductance in the circuit is____ mH.

Band width =232−212=LR​ ∴L=205​=250mH

Q. The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{2}$ times its maximum value, are $212 r a d s^{- 1}$ and $232 r a d s^{- 1}$ . The value of resistance in the circuit is $R = 5Ω$ . The self inductance in the circuit is____ $m H$ .

Band width $= 232 - 212 = \frac{R}{L}$
$∴ L = \frac{5}{20} = 250 m H$