Question

A series RLC circuit has a quality factor $Q$ of 1000 at a center frequency of $10^{6}$ rad/s. The possible values of $R$, $L$ and $C$ are

• A. $R=1\,\Omega,\; L=1\,\mu\mathrm{H}\; \text{and}\; C=1\,\mu\mathrm{F}$
• B. $R=0.1\,\Omega,\; L=1\,\mu\mathrm{H}\; \text{and}\; C=1\,\mu\mathrm{F}$
• C. $R=0.01\,\Omega,\; L=1\,\mu\mathrm{H}\; \text{and}\; C=1\,\mu\mathrm{F}$
• D. $R=0.001\,\Omega,\; L=1\,\mu\mathrm{H}\; \text{and}\; C=1\,\mu\mathrm{F}$

Hint:

For a {series} RLC at resonance: $Q=\omega_0 L/R=1/(\omega_0 R C)$. If $L$ and $C$ already satisfy $\omega_0$, then $Q$ immediately fixes $R$.

Answer 1

The Correct Option is D

Step 1 (Resonant frequency check): For a series RLC, $\omega_0=\dfrac{1}{\sqrt{LC}}$. With $L=1\,\mu\mathrm{H}=10^{-6}\,$H and $C=1\,\mu\mathrm{F}=10^{-6}\,$F, $\sqrt{LC}=\sqrt{10^{-12}}=10^{-6}$, hence $\omega_0=10^{6}\,$rad/s, matching the given value.
Step 2 (Quality factor at resonance): For a series RLC, $Q=\dfrac{\omega_0 L}{R}=\dfrac{1}{\omega_0 R C}$. With $\omega_0 L = 10^{6}. 10^{-6}=1$, we get $Q=\dfrac{1}{R}$.
Step 3 (Solve for $R$): Given $Q=1000$, $R=\dfrac{1}{Q}=0.001\,\Omega$. The only option with $R=0.001\,\Omega$ (and the same $L,C$ that satisfy $\omega_0$) is (D).
\[ \boxed{R=0.001\,\Omega,\; L=1\,\mu\mathrm{H},\; C=1\,\mu\mathrm{F}} \]