Question

Лампа- вспышка питается от батареи конденсаторов емкостью 500 мкФ, заряженного до напряжения 220 В. Определить среднюю мощность вспышки, если он длится 2,5 с.Лампа- спалах живиться від батареї конденсаторів ємністю 500 мкФ,зарядженої до напруги 220 В. Визначити середню потужність спалаху, якщо він триває 2,5 с.

Answer 1

Энергия заряженного конденсатора E=C*U^2/2
P=E/t
P=C*U^2/2*t=500*10^-6*484*10^2/2,5=9,68 Вт

Answer 2

The question is in Russian and Ukrainian languages, but I will answer in Russian. A flash lamp is powered by a capacitor battery with a capacity of 500 μF, charged to a voltage of 220 V. Determine the average power of the flash if it lasts 2.5 s.

To answer this question, we need to use the formula for the energy stored in a capacitor:

$$E = \frac{1}{2}CV^2$$

where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.

Using the given values, we can calculate the energy of the capacitor battery as:

$$E = \frac{1}{2}(500 \times 10^{-6})(220)^2$$

$$E = 12.1 \text{ J}$$

The average power of the flash is the energy divided by the time:

$$P = \frac{E}{t}$$

where P is the power in watts, and t is the time in seconds.

Using the given values, we can calculate the average power of the flash as:

$$P = \frac{12.1}{2.5}$$

$$P = 4.84 \text{ W}$$

Therefore, the average power of the flash is 4.84 W.