A capasitor of capacitance `1 (mu)F` is charged to a potential of 1 V, it is connected in parallel to an inductor of inductance `10^(-3)H`. The maximum current that will flow in the circuit has the value

Options

A. (A) `sqrt(1000)mA`
B. (B) `1mA`
C. (C) `1(mu)A`
D. (D) `1000 mA`

Explanation

To solve the problem, we need to find the maximum current that will flow in a circuit consisting of a capacitor and an inductor connected in parallel. Here are the steps to find the solution:

### Step 1: Identify the Given Values
- Capacitance \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- Voltage \( V = 1 \, V \)
- Inductance \( L = 10^{-3} \, H \)

### Step 2: Calculate the Charge on the Capacitor
The charge \( Q \) stored in the capacitor can be calculated using the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q = (1 \times 10^{-6} \, F) \times (1 \, V) = 1 \times 10^{-6} \, C \]

### Step 3: Find the Angular Frequency \( \omega \)
The angular frequency \( \omega \) for an LC circuit is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]
Substituting the values of \( L \) and \( C \):
\[ \omega = \frac{1}{\sqrt{(10^{-3} \, H)(1 \times 10^{-6} \, F)}} = \frac{1}{\sqrt{10^{-9}}} = 10^3 \, rad/s \]

### Step 4: Relate Maximum Current to Charge and Angular Frequency
The maximum current \( I_0 \) in the circuit can be expressed in terms of the maximum charge \( Q_0 \) and angular frequency \( \omega \):
\[ I_0 = \frac{Q_0}{\sqrt{LC}} \]
Substituting \( Q_0 = 1 \times 10^{-6} \, C \) and \( \sqrt{LC} = \sqrt{10^{-9}} \):
\[ I_0 = \frac{1 \times 10^{-6}}{10^{-3}} = 10^3 \, A \]

### Step 5: Convert to Milliamperes
Since \( 1 \, A = 1000 \, mA \):
\[ I_0 = 10^3 \, A = 1000 \, mA \]

### Final Answer
The maximum current that will flow in the circuit is:
\[ \boxed{1000 \, mA} \]

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