A 0.05 H inductor is connected to a fully charged 80 `muF` capacitor. The maximum current in the inductor is found to be 2 A. Calculate the voltage across capacitor while it was getting charged.

Answer

To solve the problem, we need to calculate the voltage across the capacitor while it was getting charged. We will use the energy conservation principle, where the energy stored in the capacitor is equal to the energy stored in the inductor at maximum current.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (L) = 0.05 H
- Capacitance (C) = 80 µF = 80 × 10^(-6) F
- Maximum current (I) = 2 A

2. **Write the formula for energy stored in the capacitor:**
The energy (E) stored in the capacitor is given by the formula:
\[ E_{capacitor} = \frac{1}{2} C V^2 \]
where V is the voltage across the capacitor.

3. **Write the formula for energy stored in the inductor:**
The energy stored in the inductor is given by:
\[ E_{inductor} = \frac{1}{2} L I^2 \]

4. **Set the energies equal to each other:**
Since the energy in the capacitor is converted to energy in the inductor, we have:
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I^2 \]

5. **Cancel the \(\frac{1}{2}\) from both sides:**
\[ C V^2 = L I^2 \]

6. **Rearrange the equation to solve for V:**
\[ V^2 = \frac{L I^2}{C} \]
\[ V = \sqrt{\frac{L I^2}{C}} \]

7. **Substitute the known values into the equation:**
\[ V = \sqrt{\frac{0.05 \, \text{H} \cdot (2 \, \text{A})^2}{80 \times 10^{-6} \, \text{F}}} \]

8. **Calculate the values:**
\[ V = \sqrt{\frac{0.05 \cdot 4}{80 \times 10^{-6}}} \]
\[ V = \sqrt{\frac{0.2}{80 \times 10^{-6}}} \]
\[ V = \sqrt{2500} = 50 \, \text{V} \]

9. **Conclusion:**
The voltage across the capacitor while it was getting charged is **50 volts**.