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 will follow these steps:

### Step 1: Identify the given values
- Inductance, \( L = 0.05 \, \text{H} \)
- Capacitance, \( C = 80 \, \mu\text{F} = 80 \times 10^{-6} \, \text{F} \)
- Maximum voltage across the capacitor, \( V_{max} = 200 \, \text{V} \)

### Step 2: Calculate the angular frequency \( \omega_0 \)
The formula for the angular frequency of an LC circuit is given by:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
Substituting the values of \( L \) and \( C \):
\[ \omega_0 = \frac{1}{\sqrt{0.05 \times 80 \times 10^{-6}}} \]

### Step 3: Calculate \( LC \)
First, calculate \( LC \):
\[ LC = 0.05 \times 80 \times 10^{-6} = 4 \times 10^{-6} \, \text{H}\cdot\text{F} \]

### Step 4: Calculate \( \sqrt{LC} \)
Now, calculate \( \sqrt{LC} \):
\[ \sqrt{LC} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \]

### Step 5: Calculate \( \omega_0 \)
Now, substitute back to find \( \omega_0 \):
\[ \omega_0 = \frac{1}{2 \times 10^{-3}} = 500 \, \text{rad/s} \]

### Step 6: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega_0 L \]
Substituting the values:
\[ X_L = 500 \times 0.05 = 25 \, \Omega \]

### Step 7: Calculate the maximum current \( I_{max} \)
The maximum current can be calculated using the formula:
\[ I_{max} = \frac{V_{max}}{X_L} \]
Substituting the values:
\[ I_{max} = \frac{200}{25} = 8 \, \text{A} \]

### Final Answer
The maximum current in the circuit is \( 8 \, \text{A} \).

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