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: Identify the circuit:
All four elements (source, R, L, C) sit in a single loop, so it is a series resonance circuit.

Step 2: Write the quality factor:
For a series LCR circuit $Q = \omega_0 L / R$. The given band width is $R/L$.

Step 3: Divide:
Dividing by $R/L$ is the same as multiplying by $L/R$. So
\[ \frac{Q}{R/L} = \frac{\omega_0 L}{R}\times\frac{L}{R} \]
and with $\omega_0 = 1/\sqrt{LC}$ we get $\sqrt{1/(LC)}\,L^2/R^2$.

Step 4: Why the others fail:
Each of the other three options carries a different power of $L/R$: zero power in (B), first power in (C), and a negative power in (D). Only the second power in (A) comes out of the division.

Final Answer:
Only option (A) has the correct $L^2/R^2$ factor. \[ \boxed{\sqrt{\frac{1}{LC}}\frac{L^2}{R^2}} \]