A series `AC` circuit consists of an inductor and a capacitor. The inductance and capacitance is respectively `1`henry and `25 muF` if the current is maximum in circuit then angular frequency will be

Options

A. 200
B. 100
C. 50
D. `200//2`

Answer

To solve the problem, we need to find the angular frequency (ω) at which the current in a series AC circuit consisting of an inductor and a capacitor is at its maximum. This occurs when the inductive reactance (XL) equals the capacitive reactance (XC), which is the condition for resonance.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Inductance (L) = 1 Henry
- Capacitance (C) = 25 microfarads = 25 × 10^(-6) F

2. **Understand the Condition for Maximum Current**:
- For maximum current in the circuit, the inductive reactance (XL) must equal the capacitive reactance (XC):
\[ X_L = X_C \]

3. **Express Reactances in Terms of Angular Frequency**:
- The inductive reactance is given by:
\[ X_L = \omega L \]
- The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]

4. **Set the Reactances Equal**:
- Setting the two expressions equal gives:
\[ \omega L = \frac{1}{\omega C} \]

5. **Rearranging the Equation**:
- Multiply both sides by ωC:
\[ \omega^2 L C = 1 \]
- Rearranging gives:
\[ \omega^2 = \frac{1}{LC} \]

6. **Substituting the Values**:
- Substitute L = 1 H and C = 25 × 10^(-6) F into the equation:
\[ \omega^2 = \frac{1}{1 \times 25 \times 10^{-6}} = \frac{1}{25 \times 10^{-6}} = \frac{10^6}{25} \]

7. **Calculating ω**:
- Taking the square root of both sides:
\[ \omega = \sqrt{\frac{10^6}{25}} = \frac{1000}{5} = 200 \text{ rad/s} \]

### Final Answer:
The angular frequency (ω) at which the current is maximum in the circuit is **200 rad/s**.

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