A resonance circuit having inductance and resistance 2- 10-4 H and 6-28 - respectively oscillates at 10 MHz frequency- The value of quality factor of this resonator is - -3-14-

Given : R = 6.28 Ω L = 2× 10^ 4 H f=10 MHz We know that quality factor (Q) is given by nbsp; Q=ω LR=2π f LR ⇒ Q=2π× 10× 10^6× 2× 10^ 46.28 ⇒ Q= nbsp;200 ...

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Standard XII

Physics

Impedance Triangle

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 quality factor of this resonator is $[π = 3.14]$

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Solution

Given :
$R = 6.28 Ω$
$L$ = $2 \times 10^{- 4} H$
$f = 10 MHz$

We know that quality factor $(Q)$ is given by

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

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

$\Rightarrow Q = 2000$

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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 quality factor of this resonator is [π=3.14]

Given : R=6.28 Ω L = 2×10−4 H f=10 MHz We know that quality factor (Q) is given by Q=ωLR=2πfLR ⇒Q=2π×10×106×2×10−46.28 ⇒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 quality factor of this resonator is $[π = 3.14]$

Given :
$R = 6.28 Ω$
$L$ = $2 \times 10^{- 4} H$
$f = 10 MHz$

We know that quality factor $(Q)$ is given by

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

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

$\Rightarrow Q = 2000$