The electronic flash on a camera uses a capacitor to store electrical energy from a battery. This energy is then dumped into a flash tube to produce light for photography. The amount of energy needed for the flash depends on the light power output and duration of the flash. Not all electrical energy is converted to light - some becomes thermal energy. Capacitors recharge between flashes using resistors, and the resistance can be calculated from the capacitor's capacitance and recharge time constant.

1. Physics Presentation:
Electronic Flash
Group 10:
MIRAFLORES, Trixie
DATOC, Hillary

2. Electronic Flash
— The electronic flash on a camera uses a capacitor.
A battery charges up the electronic flash capacitor
for a short time ( t) until it reaches the required
energy for a flash. The capacitor dumps this full
charge/energy (Uflash) into the flash tube in order to
produce an average light power output (Pave).
— This equation gives us the required energy for a
flash:
—
— Uflash = Pave * t

3. Electronic Flash
— This amount of energy is related to the energy stored in
a capacitor. Not all electrical energy in a capacitor can
be converted to light. The rest of the converted product
is made up of thermal energy. Every conversion of
electrical energy to light has its own output. The
relation between the electrical energy and the light
power output is:
—
— Uflash = (e/ 100) * Ucapacitor , where e = percent
efficiency in producing light power output, Uflash = light
power output, Ucapacitor = electrical energy stored in the
capacitor

4. Electronic Flash
— Capacitors used by electronic flashes usually
recharge after every flash by using resistors. The
value of resistance can be measured using the
equation:
—
— R = T/C, where R = resistance, T = time
constant (time between every capacitor recharge),
C = capacitance. The unit of R is ohm.

5. Electronic Flash

6. Examples
— 1. A camera can produce a flash with an average light
power output of 54.7 watts. If it requires 48.1 J to
produce a flash, how long will it take to reach the
required amount of energy?
—
— Uflash = Pave * t
—
— 48.1 = 54.7 ( t)
—
— t = 0.879 s

7. Examples
— 2. A capacitor used by an electric flash recharges after
every 1.2 seconds. Given that its resistor has a value of
5335 ohms, what is its capacitance?
—
— R = T/C
—
— 5335 ohms = 1.2 s /C
—
— C = 6402 F

8. Examples
— 3. An electronic flash attachment for a camera has a
light power output of 250 watts. If the camera has a
percent efficiency of 93%, how much electrical energy is
stored in its capacitor?
—
— Uflash = (e/ 100) * Ucapacitor
—
— 250 watts = (93/100) * Ucapacitor
—
— Ucapacitor = 268.8