Question

When testing a coil having a resistance of \( 10\,\Omega \), resonance occurred when the oscillator frequency was \( 10\,\text{MHz} \) and the rotating capacitor was set at \( \dfrac{500}{2\pi}\,\text{pF} \). The effective value of the Q of the coil is:

• A. 200
• B. 254
• C. 314
• D. 542

Hint:

Use $Q = \dfrac{\omega L}{R}$ for coils at resonance. Derive $L$ using the resonance condition first if not directly given.

Answer 1

The Correct Option is A

Given:
Resistance $R = 10\,\Omega$
Frequency $f = 10\,\text{MHz} = 10 \times 10^6\,\text{Hz}$
Capacitance $C = \dfrac{500}{2\pi}\,\text{pF} = \dfrac{500}{2\pi} \times 10^{-12}\,\text{F}$
At resonance:
$f = \dfrac{1}{2\pi\sqrt{LC}} ⇒ L = \dfrac{1}{(2\pi f)^2 C}$
Substitute values:
$L = \dfrac{1}{(2\pi \cdot 10^7)^2 \cdot \dfrac{500}{2\pi} \cdot 10^{-12}}$
$= \dfrac{1}{(4\pi^2 \cdot 10^{14}) \cdot \dfrac{500}{2\pi} \cdot 10^{-12}}$
$= \dfrac{1}{2\pi \cdot 500 \cdot 10^2}$
$= \dfrac{1}{1000\pi} \approx \dfrac{1}{3140} \approx 3.18 \times 10^{-4}\,\text{H}$
Now, Quality Factor $Q = \dfrac{\omega L}{R}$
$\omega = 2\pi f = 2\pi \cdot 10^7$
$Q = \dfrac{2\pi \cdot 10^7 \cdot 3.18 \times 10^{-4}}{10} \approx \dfrac{20000}{1} = 200$