Question

In a series RLC circuit with \( R = 10\ \Omega \), \( L = 0.1\text{ H} \), \( C = 100\ \mu\text{F} \), the resonant frequency, Q-factor and bandwidth respectively are:

• A. \( 50\text{ Hz}, 3.14, 15.9\text{ Hz} \)
• B. \( 50\text{ Hz}, 2.15, 9\text{ Hz} \)
• C. \( 40\text{ Hz}, 2.15, 9\text{ Hz} \)
• D. \( 50\text{ Hz}, 3.14, 16\text{ Hz} \)

Hint:

Bandwidth calculation can save time! Calculate \(BW = \frac{R}{2\pi L}\) first. Here, \(\frac{10}{2\pi(0.1)} = \frac{50}{\pi} \approx 15.9\text{ Hz}\). Checking the options reveals that only Option (A) features \(15.9\text{ Hz}\) for the bandwidth value.

Answer 1

The Correct Option is A

Concept: In a series RLC resonant network, several frequency-domain specifications define its selectiveness and energy storage capability:
• Resonant Frequency in Hertz (\(f_0\)): The frequency at which inductive reactance equals capacitive reactance, causing the net reactive component to vanish: \[ f_0 = \frac{1}{2\pi\sqrt{LC}} \end{itemize} \item Quality Factor (\(Q\)): A dimensionless value that measures the sharpness of resonance, formulated as: \[ Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} \]
• Bandwidth (\(BW\)): The range of frequencies over which the power remains above half of its peak value, related to resonance by: \[ BW = \frac{f_0}{Q} = \frac{R}{2\pi L} \]

Step 1: Calculating the resonant frequency \(f_0\).
Given components: \(R = 10\ \Omega\), \(L = 0.1\text{ H}\), and \(C = 100\ \mu\text{F} = 100 \times 10^{-6}\text{ F} = 10^{-4}\text{ F}\). First, find the product under the radical: \[ LC = 0.1 \times 10^{-4} = 10^{-5}\text{ s}^2 \] \[ \sqrt{LC} = \sqrt{10^{-5}} = \sqrt{10 \times 10^{-6}} = \sqrt{10} \times 10^{-3} \approx 3.162 \times 10^{-3} \] Now substitute this back to compute the frequency in Hz: \[ f_0 = \frac{1}{2 \pi \sqrt{10 \times 10^{-6}}} = \frac{10^3}{2 \pi \sqrt{10}} \approx \frac{1000}{2 \times 3.1416 \times 3.1623} = \frac{1000}{19.869} \approx 50.33\text{ Hz} \approx 50\text{ Hz} \]

Step 2: Calculating the Quality Factor \(Q\).
We use the expression relating the circuit components directly to evaluate \(Q\): \[ Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{0.1}{10^{-4}}} = \frac{1}{10}\sqrt{1000} = \frac{\sqrt{1000}}{10} = \frac{31.622}{10} \approx 3.16 \] Reviewing our options, \(3.14\) serves as the closest numerical representation. Let us recalculate using \(\omega_0 = 2\pi f_0 \approx 2 \times 3.1416 \times 50.33 = 316.22\text{ rad/s}\): \[ Q = \frac{\omega_0 L}{R} = \frac{316.22 \times 0.1}{10} = \frac{31.622}{10} \approx 3.16 \]

Step 3: Calculating the Bandwidth \(BW\).
Using the fundamental configuration formula: \[ BW = \frac{R}{2\pi L} = \frac{10}{2 \times \pi \times 0.1} = \frac{10}{0.2 \pi} = \frac{50}{\pi} \approx \frac{50}{3.14159} \approx 15.915\text{ Hz} \] Thus, compiling our values together, we get \(f_0 \approx 50\text{ Hz}\), \(Q \approx 3.14\), and \(BW \approx 15.9\text{ Hz}\), which matches Option (A).