24.45   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 1/675 s with an average light power output of 2.70 x 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 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?HINTS: Part (a) is about Physics 1. Energy = Power x Time, so U = 95% of 2.7E+05 x (1/675). Then one can use U = CV2 / 2 for part (b).

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Asked Sep 25, 2019
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24.45   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 1/675 s with an average light power output of 2.70 x 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 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

HINTS: Part (a) is about Physics 1. Energy = Power x Time, so U = 95% of 2.7E+05 x (1/675). Then one can use U = CV2 / 2 for part (b).

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Expert Answer

Step 1

a)

 

The relation between Energy and Power is given as,

E Pxt
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E Pxt

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Step 2

Here E is the energy stored in the flash, P is the average light power output and t is the time for

which the flash lasts.

 

Substituting the 2.7 x 105 W for P, 1/675 for t and solve for E as,

 

1
E (2.7x10 W
675
400 J
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1 E (2.7x10 W 675 400 J

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Step 3

The effective electrical energy U   that is c...

95
x 400 J
100
Е
'eff
- 380 J
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95 x 400 J 100 Е 'eff - 380 J

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