Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for 1/675 s with an average light power output of 2.7❝105 W. (a) If the conversion of electrical 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? (b) The capacitor has a potential difference between its plates of 125V when the stored energy equals the value calculated in part (a). What is the capacitance?

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priya678 priya678 · Answer 1 · 2020-03-26T07:03:57+01:00

Answer:

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time [tex]t = \frac{1}{675}[/tex]

Average power [tex]P = 2.7 \times 10^{5}[/tex] W

From power equation,

[tex]P= \frac{E}{t}[/tex]

So energy in one light is given by,

[tex]E = Pt[/tex]

[tex]E = 2.7 \times 10^{5} \times \frac{1}{675} = 400[/tex] J

Since efficiency is 95 % so we can write, energy stored in one flash,

[tex]E_{tot} = \frac{400}{0.95} = 421[/tex] J

(B)

From the formula of energy stored in capacitor,

[tex]E = \frac{1}{2}C V^{2}[/tex]

Where [tex]E = E_{tot}[/tex] and [tex]V = 125[/tex] V

[tex]C = \frac{2E}{V^{2} }[/tex]

[tex]C = \frac{2 \times 421}{15625}[/tex]

[tex]C = 0.0537 F[/tex]