A resonance circuit having inductance and resistance 2- 10-4 H and 6-28 - respectively oscillates at 10 MHz frequency- The value of quantity factor of this resonator is

Given R=6.28 Ω ; f=10 MHz or 10× 10^6 Hz ; L=2× 10^ 4 H Quality factor Q=X_LR=LωR=2π LfR Substituting the data gives, nbsp;Q=2× 10^ 4× 2× 3.14× 10× 10^66. ...

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Standard XII

Physics

LC Oscillation

A resonance c...

Question

A resonance circuit having inductance and resistance $2 \times 10^{- 4} H$ and $6.28 Ω$ respectively oscillates at $10 MHz$ frequency. The value of quantity factor of this resonator is_______

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Solution

Given $R = 6.28 Ω; f = 10 MHz or 10 \times 10^{6} Hz; L = 2 \times 10^{- 4} H$

Quality factor $Q = \frac{X_{L}}{R} = \frac{L ω}{R} = \frac{2 π L f}{R}$

Substituting the data gives,

$Q = \frac{2 \times 10^{- 4} \times 2 \times 3.14 \times 10 \times 10^{6}}{6.28}$

$\Rightarrow Q = 2000$

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A resonance circuit having inductance and resistance 2×10−4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of quantity factor of this resonator is_______

Given R=6.28 Ω ; f=10 MHz or 10×106 Hz ; L=2×10−4 H Quality factor Q=XLR=LωR=2πLfR Substituting the data gives, Q=2×10−4×2×3.14×10×1066.28 ⇒Q=2000

A resonance circuit having inductance and resistance $2 \times 10^{- 4} H$ and $6.28 Ω$ respectively oscillates at $10 MHz$ frequency. The value of quantity factor of this resonator is_______

Given $R = 6.28 Ω; f = 10 MHz or 10 \times 10^{6} Hz; L = 2 \times 10^{- 4} H$

Quality factor $Q = \frac{X_{L}}{R} = \frac{L ω}{R} = \frac{2 π L f}{R}$

Substituting the data gives,

$Q = \frac{2 \times 10^{- 4} \times 2 \times 3.14 \times 10 \times 10^{6}}{6.28}$

$\Rightarrow Q = 2000$