Question

An ac circuit contains a resistance of 1 k$\Omega$, a capacitor of 0.1 $\mu$F and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately: ____.

• A. 13.5 kHz
• B. 15.9 kHz
• C. 10.1 kHz
• D. 20.7 kHz

Hint:

To speed up calculations involving $2\pi$, remember that $1/2\pi \approx 0.159$. This makes $0.159 \times 10^5$ immediately recognizable as 15.9 kHz.

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem comes from the "Alternating Current" (AC) chapter. It focuses on the phenomenon of resonance in a series LCR circuit. Resonance is a state where the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other out, leading to minimum impedance and maximum current.
Step 2: Key Formulas and Approach:
The condition for resonance is $X_L = X_C$, which leads to the formula for resonant frequency: \[ f_r = \frac{1}{2\pi \sqrt{LC}} \] Note that the resistance ($R$) does not affect the resonant frequency itself, only the sharpness (Q-factor) of the resonance.
Step 3: Detailed Explanation:

Identify given values:
$L = 1 \text{ mH} = 10^{-3} \text{ H}$.
$C = 0.1 \mu\text{F} = 10^{-7} \text{ F}$.
$R = 1 \text{ k}\Omega$ (Distractor for frequency calculation).

Calculate the product LC: \[ LC = 10^{-3} \times 10^{-7} = 10^{-10} \]
Calculate the square root: \[ \sqrt{LC} = \sqrt{10^{-10}} = 10^{-5} \]
Plug into frequency formula: \[ f_r = \frac{1}{2 \cdot \pi \cdot 10^{-5}} = \frac{10^5}{2\pi} \]
Compute: \[ f_r \approx \frac{100,000}{6.28} \approx 15,923 \text{ Hz} \]
Convert to kHz: $15,923 / 1000 \approx 15.9 \text{ kHz}$.
Step 4: Final Answer:
The resonance frequency is approximately 15.9 kHz.