Question

В импульсной фотовспышке лампа питается от конденсатора емкостью 800 мкф заряженного до напряжения 300В найти энергию вспышки и среднюю мощность если продолжительность разрядки 2,4мс

Answer 1

Ровно 500
1) 800+300 и разделить 2.4=500
2)800-300 =500

Answer 2

Calculation of Flash Energy and Average Power

To calculate the flash energy and average power of an impulse flash lamp, we need to know the capacitance of the capacitor, the voltage it is charged to, and the discharge duration.

Given: - Capacitance (C) = 800 μF - Voltage (V) = 300 V - Discharge duration (t) = 2.4 ms

Flash Energy Calculation

The flash energy (E) can be calculated using the formula:

E = 0.5 * C * V^2

Substituting the given values:

E = 0.5 * 800 μF * (300 V)^2

To calculate the flash energy, we need to convert the capacitance from microfarads (μF) to farads (F) and the voltage from volts (V) to joules (J).

1 μF = 1e-6 F 1 V = 1 J

Converting the units:

E = 0.5 * 800e-6 F * (300 V)^2 E = 0.5 * 800e-6 F * (300 J)^2 E = 0.5 * 800e-6 F * 90000 J E = 0.036 J

Therefore, the flash energy is 0.036 joules.

Average Power Calculation

The average power (P) can be calculated using the formula:

P = E / t

Substituting the given values:

P = 0.036 J / 2.4 ms

To calculate the average power, we need to convert the discharge duration from milliseconds (ms) to seconds (s).

1 ms = 1e-3 s

Converting the units:

P = 0.036 J / 2.4e-3 s P = 0.036 J / 0.0024 s P = 15 W

Therefore, the average power is 15 watts.

Please note that the above calculations are based on the given values and assumptions.