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Question: In a series LCR circuit having L = 30 mH, R = 8\(\Omega\)and the resonant frequency is 50 Hz. The qu...

In a series LCR circuit having L = 30 mH, R = 8Ω\Omegaand the resonant frequency is 50 Hz. The quality factor of the circuit is

A

0.118

B

11.8

C

118

D

1.18

Answer

1.18

Explanation

Solution

: Here, , L = 30 mH

=30×103H,R=8Ω,υr=50Hz= 30 \times 10^{- 3}H,R = 8\Omega,\upsilon_{r} = 50Hz

As , ωr=2πυr\omega_{r} = 2\pi\upsilon_{r}

=2×3.14×50=314Hz= 2 \times 3.14 \times 50 = 314Hz

\therefore Quality factor,

Q=ωrLR=314×30×1038=1.18Q = \frac{\omega_{r}L}{R} = \frac{314 \times 30 \times 10^{- 3}}{8} = 1.18