Question

Помогите пожалуйстаЗадача по физики:Определите емкостное сопротивление батареи из двух конденсаторов емкостью С1=С2=2мкФ, соединенных параллельно в цепь переменного тока с частотой V=50Гц. Ответ: 796 Ом.

Answer 1

Определим сначала общую емкость при паралельном соединении C1+C2=4мкФ
Сопротивление цепи тогда будет
Xc=1/wC=10(в шестой степени)/314x4=796,1 Ом

Answer 2

To determine the capacitive resistance of a battery consisting of two capacitors connected in parallel in an AC circuit, we need to use the formula for capacitive reactance. Capacitive reactance (Xc) is given by the formula:

Xc = 1 / (2πfC)

Where: - Xc is the capacitive reactance in ohms (Ω) - π is a mathematical constant approximately equal to 3.14159 - f is the frequency of the AC current in hertz (Hz) - C is the capacitance of the capacitor in farads (F)

In this case, we have two capacitors with the same capacitance, C1 = C2 = 2μF, connected in parallel in an AC circuit with a frequency of f = 50Hz.

Let's calculate the capacitive reactance using the given values:

Xc = 1 / (2π * 50 * 2 * 10^-6)

Calculating this expression gives us a value of approximately 796Ω.

Therefore, the capacitive resistance of the battery consisting of two capacitors with a capacitance of 2μF each, connected in parallel in an AC circuit with a frequency of 50Hz, is approximately 796Ω.

Please note that the answer provided is based on the information given and the formula for capacitive reactance.