Question

Если в колебательном контуре частота колебаний составляет 10 МГц,то их период равен 10 мкс,верно утверждение?

Answer 1

T=1/ частоту
10^-6 = 1 / 10^6

Answer 2

Answer:

The statement is not correct. If the frequency of oscillation in a circuit is 10 MHz, then the period of the oscillation is not 10 µs. The correct period can be calculated using the formula:

Period (T) = 1 / Frequency (f)

So, in this case, the period would be:

Period (T) = 1 / 10,000,000 Hz = 0.1 µs

This means that the period of oscillation for a 10 MHz frequency is 0.1 µs, not 10 µs.

Explanation:

The period of oscillation in a circuit is the time taken for one complete cycle of the oscillation. It is inversely proportional to the frequency of the oscillation. As the frequency increases, the period decreases, and vice versa. The relationship between period and frequency is given by the formula: Period (T) = 1 / Frequency (f).

Example:

For a frequency of 10 MHz: Period (T) = 1 / 10,000,000 Hz = 0.1 µs

Therefore, the correct period for a 10 MHz frequency is 0.1 µs, not 10 µs.