A capacitor of capacity `2muF` is changed to a potential different of `12V` . It is then connected across an inductor of inductance `0.6mH` What is the current in the circuit at a time when the potential difference across the capacitor is `6.0V` ?

Answer

To solve the problem step by step, we will use the principles of energy conservation in the LC circuit formed by the capacitor and the inductor.

### Step 1: Calculate the initial energy stored in the capacitor
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage.

Given:
- \( C = 2 \mu F = 2 \times 10^{-6} F \)
- \( V = 12 V \)

Substituting the values:
\[ E_{\text{initial}} = \frac{1}{2} \times (2 \times 10^{-6}) \times (12^2) \]
\[ E_{\text{initial}} = \frac{1}{2} \times (2 \times 10^{-6}) \times 144 \]
\[ E_{\text{initial}} = (1 \times 10^{-6}) \times 144 = 144 \times 10^{-6} J = 0.000144 J \]

### Step 2: Write the expression for final energy in the circuit
When the capacitor is connected to the inductor, the energy will be distributed between the capacitor and the inductor. The final energy in the circuit can be expressed as:
\[ E_{\text{final}} = \frac{1}{2} C V_1^2 + \frac{1}{2} L I^2 \]
where \( V_1 \) is the voltage across the capacitor at the moment of interest and \( I \) is the current.

Given:
- \( V_1 = 6 V \)
- \( L = 0.6 mH = 0.6 \times 10^{-3} H \)

Substituting the values:
\[ E_{\text{final}} = \frac{1}{2} \times (2 \times 10^{-6}) \times (6^2) + \frac{1}{2} \times (0.6 \times 10^{-3}) \times I^2 \]
\[ E_{\text{final}} = (1 \times 10^{-6}) \times 36 + (0.3 \times 10^{-3}) I^2 \]
\[ E_{\text{final}} = 36 \times 10^{-6} + 0.3 \times 10^{-3} I^2 \]

### Step 3: Apply conservation of energy
According to the conservation of energy:
\[ E_{\text{initial}} = E_{\text{final}} \]
Substituting the values:
\[ 144 \times 10^{-6} = 36 \times 10^{-6} + 0.3 \times 10^{-3} I^2 \]

### Step 4: Solve for \( I^2 \)
Rearranging the equation:
\[ 144 \times 10^{-6} - 36 \times 10^{-6} = 0.3 \times 10^{-3} I^2 \]
\[ 108 \times 10^{-6} = 0.3 \times 10^{-3} I^2 \]

Dividing both sides by \( 0.3 \times 10^{-3} \):
\[ I^2 = \frac{108 \times 10^{-6}}{0.3 \times 10^{-3}} = \frac{108}{0.3} \times 10^{-3} = 360 \times 10^{-3} \]

### Step 5: Calculate \( I \)
Taking the square root:
\[ I = \sqrt{360 \times 10^{-3}} = 6 \times 10^{-1} = 0.6 A \]

Thus, the current in the circuit when the potential difference across the capacitor is \( 6.0 V \) is \( 0.6 A \).

---