Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for a time interval of 1.50×10^−3 s with an average light power output of 2.90×10^5 W a) If the conversion of electrical energy to light has an efficiency of 90.0 % (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash?
b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part A. What is the capacitance?

(a) E = 483.33 J

(b) C = 0.062 F = 62 mF

Explanation:

(a)

First we need to calculate the energy output for one flash. For that purpose we have the formula:

E₀ = Pt

where,

E₀ = Energy output for one flash = ?

P = Light Output Power for one flash=  2.9 x 10⁵ W

t = time interval for one flash = 1.5 x 10⁻³ s

Therefore,

E₀ = (2.9 x 10⁵ W)(1.5 x 10⁻³ s)

E₀ = 435 J

Now, for energy to be stored in capacitor, we use the following formula:

Efficiency = E₀/E

where,

E = Energy required to be stored in capacitor for one flash = ?

Efficiency = 90% = 0.9

Therefore,

0.9 = 435 J/E

E = 435 J/0.9

E = 483.33 J

(b)

The energy stored in the capacitor is given by the formula:

E = (1/2)(CV²)

where,

C = Capacitance = ?

V = Voltage = 125 V

Therefore,

483.33 J = (1/2)(C)(125 V)²

C = (483.33 J)(2)/(125 V)²

C = 0.062 F = 62 mF

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