A photographic flash unit consists of a xenon-filled tube. It gives a flash of average power 200 W for 0.04s. The flash is due to discharge of a fully charged capacitor of `40muF`. The voltage to which it is charged before a flash is given by the unit is

Options

A. 1500V
B. 2000V
C. 2500V
D. 3000V

Explanation

To find the voltage to which the capacitor is charged in the photographic flash unit, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Energy Stored in the Capacitor**:
The energy (U) stored in the capacitor can be calculated using the formula:
\[ U = P \times t \]
where \( P \) is the power and \( t \) is the time duration. Given that \( P = 200 \, \text{W} \) and \( t = 0.04 \, \text{s} \):
\[ U = 200 \, \text{W} \times 0.04 \, \text{s} = 8 \, \text{J} \]

2. **Use the Energy Formula for a Capacitor**:
The energy stored in a capacitor is also given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage. We know \( U = 8 \, \text{J} \) and \( C = 40 \, \mu\text{F} = 40 \times 10^{-6} \, \text{F} \).

3. **Substitute the Known Values**:
Substitute the values into the energy formula:
\[ 8 = \frac{1}{2} \times (40 \times 10^{-6}) \times V^2 \]

4. **Rearrange the Equation**:
Rearranging the equation to solve for \( V^2 \):
\[ V^2 = \frac{8 \times 2}{40 \times 10^{-6}} = \frac{16}{40 \times 10^{-6}} = \frac{16 \times 10^6}{40} \]
Simplifying this gives:
\[ V^2 = 0.4 \times 10^6 = 4 \times 10^5 \]

5. **Take the Square Root**:
Now take the square root to find \( V \):
\[ V = \sqrt{4 \times 10^5} = \sqrt{4} \times \sqrt{10^5} = 2 \times 10^{2.5} = 2 \times 100 \times \sqrt{10} \approx 632.45 \, \text{V} \]

### Final Answer:
The voltage to which the capacitor is charged before a flash is approximately:
\[ V \approx 632.45 \, \text{V} \]

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