Electronic flash units for camera contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for 1675s with an average light power output of 2.70 x 105 W . (a) If the conversion of electric energy to light is 95% efficient ( the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash

420J

Explanation:

Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.

Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:

Power = Energy / time

Substituting:

2.7 * 10⁵ W = Energy / (1/675)

Energy = 2.7 * 10⁵ W * 1/675 = 400J

Therefore the energy emitted as light is 400J.

Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:

Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400

Energy(capacitor) = 420J

gets less

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