A 16 `mu F` capacitor is charged to a 20 Volt potential. Battery is then disconnected and inductor of inductance `40 mH ` is connected across the capacitor, So that LC oscillations are step-up. Maximum current in the coil is

Options

A. 0.4 A
B. `2 A`
C. 0.8 A
D. 0.2 A

Answer

To solve the problem, we need to find the maximum current in the inductor when it is connected across a charged capacitor. Let's break down the steps:

### Step 1: Calculate the maximum charge on the capacitor.
The maximum charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C \) is the capacitance (in farads),
- \( V \) is the voltage (in volts).

Given:
- \( C = 16 \, \mu F = 16 \times 10^{-6} \, F \)
- \( V = 20 \, V \)

Calculating \( Q \):
\[ Q = 16 \times 10^{-6} \, F \times 20 \, V = 320 \times 10^{-6} \, C = 320 \, \mu C \]

### Step 2: Use energy conservation to relate the energy in the capacitor to the energy in the inductor.
Initially, the energy stored in the capacitor \( U_C \) is given by:
\[ U_C = \frac{Q^2}{2C} \]
When the capacitor is fully discharged into the inductor, all this energy is transferred to the inductor as magnetic energy \( U_L \):
\[ U_L = \frac{1}{2} L I^2 \]
where:
- \( L \) is the inductance (in henries),
- \( I \) is the maximum current (in amperes).

Setting \( U_C = U_L \):
\[ \frac{Q^2}{2C} = \frac{1}{2} L I^2 \]

### Step 3: Solve for the maximum current \( I \).
From the equation, we can eliminate \( \frac{1}{2} \) on both sides:
\[ \frac{Q^2}{C} = L I^2 \]
Rearranging for \( I \):
\[ I^2 = \frac{Q^2}{LC} \]
\[ I = \frac{Q}{\sqrt{LC}} \]

### Step 4: Substitute the values for \( Q \), \( L \), and \( C \).
Given:
- \( Q = 320 \, \mu C = 320 \times 10^{-6} \, C \)
- \( L = 40 \, mH = 40 \times 10^{-3} \, H \)
- \( C = 16 \, \mu F = 16 \times 10^{-6} \, F \)

Substituting these values into the equation:
\[ I = \frac{320 \times 10^{-6}}{\sqrt{(40 \times 10^{-3})(16 \times 10^{-6})}} \]

### Step 5: Calculate the denominator.
Calculating \( LC \):
\[ LC = (40 \times 10^{-3}) \times (16 \times 10^{-6}) = 640 \times 10^{-9} = 6.4 \times 10^{-7} \]
Now, taking the square root:
\[ \sqrt{LC} = \sqrt{6.4 \times 10^{-7}} = 8 \times 10^{-4} \]

### Step 6: Calculate the current \( I \).
Now substituting back:
\[ I = \frac{320 \times 10^{-6}}{8 \times 10^{-4}} = 0.4 \, A \]

### Final Answer:
The maximum current in the coil is \( 0.4 \, A \).

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