An oscillator circuit contains an inductor 0.05 H and a capacitor of capacity `80 mu F`. When the maximum voltage across the capacitor is 200 V, the maximum current (in amperes) in the circuit is

Options

A. 4
B. 8
C. 10
D. 16

Answer

To solve the problem, we need to find the maximum current in an LC oscillator circuit given the values of inductance, capacitance, and maximum voltage.

### Step-by-step Solution:

1. **Identify the given values:**
- Inductance (L) = 0.05 H
- Capacitance (C) = 80 µF = 80 x 10^(-6) F
- Maximum Voltage (V_max) = 200 V

2. **Use the formula for maximum current in an LC circuit:**
The maximum current (I_max) can be calculated using the formula:
\[ I_{\text{max}} = V_{\text{max}} \sqrt{\frac{C}{L}} \]

3. **Substitute the values into the formula:**
- Substitute V_max = 200 V, C = 80 x 10^(-6) F, and L = 0.05 H into the formula:
\[ I_{\text{max}} = 200 \sqrt{\frac{80 \times 10^{-6}}{0.05}} \]

4. **Calculate the fraction inside the square root:**
- First, calculate \( \frac{80 \times 10^{-6}}{0.05} \):
\[ \frac{80 \times 10^{-6}}{0.05} = 80 \times 10^{-6} \times 20 = 1.6 \times 10^{-3} \]

5. **Calculate the square root:**
- Now, take the square root:
\[ \sqrt{1.6 \times 10^{-3}} \approx 0.04 \]

6. **Calculate the maximum current:**
- Now multiply by V_max:
\[ I_{\text{max}} = 200 \times 0.04 = 8 \text{ A} \]

### Final Answer:
The maximum current in the circuit is **8 A**.

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