Question

In the circuit shown, the ratio of the quality factor and the band width is
(Given: band width = R/L)

• A. \(\sqrt{\frac{1}{LC}}\frac{L^2}{R^2}\)
• B. \(\sqrt{\frac{1}{LC}}\)
• C. \(\sqrt{\frac{1}{LC}}\frac{L}{R}\)
• D. \(\sqrt{\frac{1}{LC}}\frac{R}{L}\)

Hint:

Use \(Q=\omega_0 L/R\) with \(\omega_0=1/\sqrt{LC}\), then divide by the given band width \(R/L\).

Answer 1

The Correct Option is A

Step 1: Read the circuit:
The figure shows a source E, a resistor R, an inductor L and a capacitor C all joined in one loop. So this is a series LCR circuit driven by an AC source.

Step 2: Recall the formulas:
At resonance the angular frequency is \(\omega_0 = \sqrt{\frac{1}{LC}}\). The quality factor of a series LCR circuit is \(Q = \frac{\omega_0 L}{R}\). The question tells us to take the band width as \(\Delta\omega = \frac{R}{L}\).

Step 3: Form the ratio:
We need \(Q\) divided by the band width.
\[ \frac{Q}{\Delta\omega} = \frac{\omega_0 L/R}{R/L} = \omega_0 \cdot \frac{L}{R}\cdot\frac{L}{R} = \omega_0\frac{L^2}{R^2} \]
Put \(\omega_0 = \sqrt{1/(LC)}\):
\[ \frac{Q}{\Delta\omega} = \sqrt{\frac{1}{LC}}\,\frac{L^2}{R^2} \]

Step 4: Check the other options:
Option (B) is just \(\omega_0\), which has lost the \(L^2/R^2\) factor. Option (C) has only \(L/R\), which is \(Q/\omega_0\), not \(Q/\Delta\omega\). Option (D) has \(R/L\), which is the band width itself. None of these matches the division we did.

Final Answer:
The ratio of quality factor to band width is \(\omega_0 L^2/R^2\), which is option (A). \[ \boxed{\sqrt{\frac{1}{LC}}\frac{L^2}{R^2}} \]