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Physics Question on LCR Circuit

Quality factor of LCR circuit having resistance R and inductance L at resonance frequency (ω)(\omega ) is given by:

A

(ωLR)2{{\left( \frac{\omega L}{R} \right)}^{2}}

B

(ωLR)1/2{{\left( \frac{\omega L}{R} \right)}^{1/2}}

C

RωL\frac{R}{\omega L}

D

ωLR\frac{\omega L}{R}

Answer

ωLR\frac{\omega L}{R}

Explanation

Solution

Here: Resistance =R,=R, inductance =L=L and resonance frequency =ω=\omega potential difference developed across the inductance =IωL=I\omega L and applied emf =IR=IR Hence quality factor Q=potential differenceapplied emf=IωLIR=ωLRQ=\frac{\text{potential}\,\text{difference}}{\text{applied}\,\text{emf}}=\frac{I\omega L}{IR}=\frac{\omega L}{R}