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
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
Yes, you are correct! The energy stored in a capacitor is given by the formula: Energy = 1/2 * C * V^2, where C is the capacitance of the capacitor and V is the potential difference (voltage) across the plates of the capacitor.
(a) To calculate the energy used to produce the flash, we can substitute the given values into the formula:
C = 760 µF = 760 * 10^(-6) F
V = 315 V
Plugging these values into the formula, we have:
Energy = 1/2 * (760 * 10^(-6) F) * (315 V)^2
= 1/2 * 760 * 10^(-6) F * (315 V)^2
= 1/2 * 760 * 10^(-6) F * 99225 V^2
= (760 * 10^(-6) * 99225) J
≈ 75.528 J
Therefore, the energy used to produce the flash in this unit is approximately 75.528 joules.
(b) To find the effective power or "wattage" of the flash, we can use the formula for power:
Power = Energy / Time
The given time for the flash is 5.0 * 10^(-3) s. Plugging in the values:
Power = 75.528 J / (5.0 * 10^(-3) s)
= 75.528 J / 0.005 s
= 15105.6 W
The effective power or "wattage" of the flash is approximately 15105.6 watts.
Yes, you are correct. The energy stored in a capacitor is given by the formula:
Energy = (1/2) * C * V^2
where C is the capacitance and V is the potential difference between the plates of the capacitor.
To calculate the energy used to produce the flash in this unit, use the given values:
Capacitance (C) = 760 µF = 760 * 10^(-6) F
Potential difference (V) = 315 V
(a) Energy = (1/2) * C * V^2
= (1/2) * (760 * 10^(-6)) * (315)^2
= 1/2 * 760 * 10^(-6) * 99225
= 37.722 J (Joules)
So, the energy used to produce the flash in this unit is 37.722 Joules.
To calculate the effective power or "wattage" of the flash, divide the energy by the duration of the flash:
Duration of flash (t) = 5.0 * 10^(-3) s
(b) Power = Energy / Time
= 37.722 J / 5.0 * 10^(-3) s
= 7544.4 W (Watts)
So, the effective power or "wattage" of the flash is 7544.4 Watts.