Answer :
Final answer:
To provide a flash with an efficiency of 93.0%, a capacitor must store approximately 107.5% of the energy of the flash, which is nearest to option (c) 107.7% of the energy of the flash.
Explanation:
If the conversion of electrical energy to light has an efficiency of 93.0%, to find out how much energy must be stored in the capacitor for one flash, we can use the formula relating energy output to efficiency. The efficiency (Eff) can be defined as the ratio of the useful energy output (Eout) to the total energy input (Einput), so Eff = Eout / Einput. This formula can be rearranged to calculate the total energy input required for a given efficiency and energy output: Einput = Eout / Eff.
In this case, if we have an energy output of 1 unit (the energy of the flash), and the efficiency is 93.0% or 0.930, the energy input required would be Einput = 1 unit / 0.930 = 1.0753 units. This means that the capacitor needs to store approximately 107.5% of the energy of the flash, which is most closely represented by option (c) 107.7% of the energy of the flash.
