An `LCR` circuit contains `R=50 Omega, L=1 mH` and `C=0.1 muF`. The impedence of the circuit will be minimum for a frequency of

Options

A. `(10^(5))/(2pi)s^(-1)`
B. `(10^(6))/(2pi)s^(-1)`
C. `2pixx10^(5) s^(-1)`
D. `2pixx10^(6) s^(-1)`

Explanation

To find the frequency at which the impedance of the LCR circuit is minimum, we need to determine the resonant frequency. The resonant frequency occurs when the inductive reactance (XL) is equal to the capacitive reactance (XC).

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance \( R = 50 \, \Omega \)
- Inductance \( L = 1 \, \text{mH} = 1 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 0.1 \, \mu\text{F} = 0.1 \times 10^{-6} \, \text{F} \)

2. **Write the formulas for inductive and capacitive reactance:**
- Inductive reactance:
\[ X_L = \omega L \]
- Capacitive reactance:
\[ X_C = \frac{1}{\omega C} \]

3. **Set the inductive reactance equal to the capacitive reactance for resonance:**
\[ X_L = X_C \implies \omega L = \frac{1}{\omega C} \]

4. **Rearranging the equation:**
\[ \omega^2 = \frac{1}{LC} \]

5. **Substituting the values of L and C:**
\[ \omega^2 = \frac{1}{(1 \times 10^{-3})(0.1 \times 10^{-6})} \]

6. **Calculate the product of L and C:**
\[ LC = 1 \times 10^{-3} \times 0.1 \times 10^{-6} = 1 \times 10^{-10} \]

7. **Calculate \( \omega^2 \):**
\[ \omega^2 = \frac{1}{1 \times 10^{-10}} = 10^{10} \]

8. **Taking the square root to find \( \omega \):**
\[ \omega = \sqrt{10^{10}} = 10^5 \, \text{rad/s} \]

9. **Convert \( \omega \) to frequency \( f \):**
\[ f = \frac{\omega}{2\pi} = \frac{10^5}{2\pi} \]

10. **Calculate the frequency:**
\[ f \approx \frac{10^5}{6.2832} \approx 15915.5 \, \text{Hz} \]

### Final Answer:
The impedance of the circuit will be minimum at a frequency of approximately **15915.5 Hz**.