Question

Два конденсатора емкостью 0,2 мкФ и 0,1 мкФ включены последовательно в цепь переменного тока напряжением 220 В. Частота переменного тока равна 50 Гц. Найдите силу тока в цепи и падение напряжения на каждом конденсаторе. Написать что дано, найти и решение))

Answer 1

Дано:
C1 = 0,2 мкФ = 0,2*10⁻⁶ Ф
C2 = 0,1 мкф = 0,1*10⁻⁶ Ф
U = 220 B
ν = 50 Гц
__________________
 I - ?
 U1 - ?
 U2 - ?

Конденсаторы соединены ПОСЛЕДОВАТЕЛЬНО, значит их общая емкость равна:
С = С1*С2 / (С1+С2) = 0,2*0,1 / (0,2+0,1) ≈ 0,067 мкФ = 0,067*10⁻⁶Ф
Ёмкостное сопротивление:
Xc = 1/(2*π*v*C) = 1 / (2*3,14*50*0,067*10⁻⁶) ≈ 47 500 Ом

Сила тока : 
I = U / Xc = 220/47500 ≈ 0,0046 A = 4,6 мА
Заряд на обкладках:

q = C*U = 0,067*10⁻⁶*220 = 1,47*10⁻⁵ Кл

U1 = q / C1 = 1,47*10⁻⁵ / 0,2*10⁻⁶ = 73,5 В
U2 = 220 - 73,5 =146,5 В

Answer 2

Given:

Two capacitors with capacitances of 0.2 µF and 0.1 µF are connected in series in a circuit with an alternating current voltage of 220 V. The frequency of the alternating current is 50 Hz.

To Find:

1. The current in the circuit. 2. The voltage drop across each capacitor.

Solution:

The total impedance of capacitors in series can be calculated using the formula:

Z = 1 / (jωC)

Where: - Z = Impedance - j = Imaginary unit - ω = Angular frequency (2πf) - C = Capacitance

The total impedance of capacitors in series is the sum of the individual impedances.

The current in the circuit can be calculated using Ohm's law:

I = V / Z

Where: - I = Current - V = Voltage - Z = Impedance

The voltage drop across each capacitor can be calculated using the formula:

Vc = I * Zc

Where: - Vc = Voltage drop across the capacitor - I = Current - Zc = Impedance of the capacitor

Let's calculate the impedance of each capacitor and then find the current in the circuit and the voltage drop across each capacitor.

Impedance of Capacitors:

The impedance of each capacitor can be calculated using the formula Z = 1 / (jωC).

For the 0.2 µF capacitor: - Z1 = 1 / (j * 2πf * C1) - Z1 = 1 / (j * 2π * 50 * 0.2 * 10^-6)

For the 0.1 µF capacitor: - Z2 = 1 / (j * 2πf * C2) - Z2 = 1 / (j * 2π * 50 * 0.1 * 10^-6)

Current in the Circuit:

The total impedance of capacitors in series is the sum of the individual impedances. - Z_total = Z1 + Z2

Using Ohm's law, the current in the circuit can be calculated: - I = V / Z_total

Voltage Drop Across Each Capacitor:

The voltage drop across each capacitor can be calculated using the formula Vc = I * Zc.

Let's calculate the impedance of each capacitor, the current in the circuit, and the voltage drop across each capacitor using the given values and the formulas above.