In an LCR circuit `10Omega` resistance, `0.5 mu`F capacitor and 8 H inductor are connected in series, their angular resonance frequency will be

Options

A. 800 rad/sec
B. 600 rad/sec
C. 500 rad/sec
D. 300 rad/sec

Answer

To find the angular resonance frequency of the given LCR circuit, we will follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 10 Ω
- Capacitance (C) = 0.5 µF = 0.5 × 10^(-6) F
- Inductance (L) = 8 H

### Step 2: Understand the resonance condition
At resonance in an LCR circuit, the inductive reactance (XL) is equal to the capacitive reactance (XC):
\[ X_L = X_C \]

### Step 3: Write the formulas for inductive and capacitive reactance
- Inductive Reactance:
\[ X_L = \omega L \]
- Capacitive Reactance:
\[ X_C = \frac{1}{\omega C} \]

### Step 4: Set the reactances equal to each other
At resonance:
\[ \omega L = \frac{1}{\omega C} \]

### Step 5: Rearrange the equation to find angular frequency (ω)
Multiplying both sides by ωC gives:
\[ \omega^2 = \frac{1}{LC} \]
Thus,
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 6: Substitute the values of L and C into the formula
Substituting the values:
\[ L = 8 \, \text{H} \]
\[ C = 0.5 \times 10^{-6} \, \text{F} \]
\[ \omega = \frac{1}{\sqrt{8 \times 0.5 \times 10^{-6}}} \]

### Step 7: Calculate the product of L and C
Calculating:
\[ LC = 8 \times 0.5 \times 10^{-6} = 4 \times 10^{-6} \]

### Step 8: Find the square root
Now, take the square root:
\[ \sqrt{LC} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \]

### Step 9: Calculate ω
Now calculate ω:
\[ \omega = \frac{1}{2 \times 10^{-3}} = 500 \, \text{rad/s} \]

### Final Answer
The angular resonance frequency is:
\[ \omega = 500 \, \text{rad/s} \]

---