The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 700 碌F capacitor is 360 V.
(a) Determine the energy that is used to produce the flash in this unit.
(b) Assuming that the flash lasts for 5.0E-3 s, find the effective power or ';wattage'; of the flash.The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the f
Here,
capacitance, C = 700 uF
or, C = 700 * 10 ^ - 6 F
potential difference, V = 360 V
Energy, E = ?
time, t = 5 * 10 ^ -3 s
Power, P = ?
(a)
We know,
E = 1/2 CV^2
or, E = 1/2 * 700 * 10 ^ -6 * (360)^2
so, E = 45.36 Joule (ans.)
(b)
We know,
P = E/t
or, P = 45.36/(5*10^-3)
so, P = 9072 Watt (ans.)
Hope this was helpful.
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