A photographic flash unit consists of a xenon-filled tube. It givers a flash of average power `2000 W` for `0.04 s. The flash is due to discharge of a fully charged capacitor of `40 muF` The valtage to which it is charged before a flash is given by The unit is .

Options

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

Explanation

To solve the problem, we need to determine the voltage to which the capacitor was charged before the flash. We can use the relationship between power, energy, and the capacitance of the capacitor.

### Step-by-Step Solution:

1. **Calculate the Energy Used in the Flash:**
The average power of the flash is given as \( P = 2000 \, \text{W} \) and the duration of the flash is \( t = 0.04 \, \text{s} \).
The energy \( E \) used during the flash can be calculated using the formula:
\[ E = P \times t \]
Substituting the values:
\[ E = 2000 \, \text{W} \times 0.04 \, \text{s} = 80 \, \text{J} \]

2. **Relate the Energy to the Capacitor's Charge:**
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. We know that the capacitance \( C = 40 \, \mu\text{F} = 40 \times 10^{-6} \, \text{F} \).

3. **Set the Two Energy Equations Equal:**
From the two expressions for energy, we have:
\[ 80 \, \text{J} = \frac{1}{2} (40 \times 10^{-6} \, \text{F}) V^2 \]

4. **Solve for \( V^2 \):**
Rearranging the equation to solve for \( V^2 \):
\[ 80 = 20 \times 10^{-6} V^2 \]
\[ V^2 = \frac{80}{20 \times 10^{-6}} = \frac{80}{20} \times 10^{6} = 4 \times 10^{6} \]

5. **Calculate \( V \):**
Taking the square root of both sides:
\[ V = \sqrt{4 \times 10^{6}} = 2 \times 10^{3} = 2000 \, \text{V} \]

### Final Answer:
The voltage to which the capacitor was charged before the flash is \( 2000 \, \text{V} \).

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