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) Hz`
B. `(10^5)/(2pi) Hz`
C. `2pi xx 10^5 Hz`
D. `2 pi xx 10^6 Hz`

Answer

To find the frequency at which the impedance of the LCR circuit is minimum, we need to determine the resonance frequency. The resonance frequency for an LCR circuit is given by the formula:

\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

Where:
- \( L \) is the inductance in henries (H)
- \( C \) is the capacitance in farads (F)

### Step 1: Identify the values of L and C
Given:
- \( R = 50 \, \Omega \) (not needed for resonance frequency)
- \( L = 1 \, \text{mH} = 1 \times 10^{-3} \, \text{H} \)
- \( C = 0.1 \, \mu\text{F} = 0.1 \times 10^{-6} \, \text{F} = 1 \times 10^{-7} \, \text{F} \)

### Step 2: Substitute the values into the resonance frequency formula
Now, substitute the values of \( L \) and \( C \) into the resonance frequency formula:

\[ f_0 = \frac{1}{2\pi\sqrt{(1 \times 10^{-3})(1 \times 10^{-7})}} \]

### Step 3: Calculate the product of L and C
Calculate \( LC \):

\[ LC = (1 \times 10^{-3}) \times (1 \times 10^{-7}) = 1 \times 10^{-10} \]

### Step 4: Calculate the square root of LC
Now, calculate the square root:

\[ \sqrt{LC} = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \]

### Step 5: Substitute back into the frequency formula
Now substitute this back into the frequency formula:

\[ f_0 = \frac{1}{2\pi(1 \times 10^{-5})} \]

### Step 6: Calculate the frequency
Now calculate \( f_0 \):

\[ f_0 = \frac{1}{2\pi \times 10^{-5}} = \frac{10^5}{2\pi} \, \text{Hz} \]

### Final Answer
Thus, the frequency at which the impedance of the circuit will be minimum is:

\[ f_0 = \frac{10^5}{2\pi} \, \text{Hz} \]