Yes, you're correct!
For part (a), you can use the formula: energy = 1/2 * C * V^2, where C is the capacitance (760 µF) and V is the potential difference (315 V).
Energy = (1/2) * (760 * 10^(-6) F) * (315 V)^2
Now, calculate the energy:
Energy ≈ 0.035596 J
For part (b), you can find the effective power or "wattage" of the flash using the formula: power = energy/time, where time is given as 5.0 * 10^(-3) s.
Power = 0.035596 J / (5.0 * 10^(-3) s)
Now, calculate the power:
Power ≈ 7.1192 W
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 760 µF capacitor is 315 V.
(a) Determine the energy that is used to produce the flash in this unit.
(b) Assuming that the flash lasts for 5.0*10-3 s, find the effective power or "wattage" of the flash.
energy = 1/2 C v^2, right?
power= energy/time
1 answer
1 answer