Question

ОЧЕНЬ СРОЧНО, ФИЗИКА ПОМОГИТЕ ПРОШУУУУ Батарея з трьох послідовно з’єднаних конденсаторів ємностями 15 пФ, 25 пФ і 35 пФ ввімкнена до джерела електричної енергії, від якого був наданий заряд 7нКл. Визначити загальну ємність, напругу та енергію кожного конденсатора.​

Answer 1

Ответ:

Объяснение:

Дано:

С1=15пФ=1,5*10⁻¹¹Ф

С2=25пФ=2,5*10⁻¹¹Ф

С3=35пФ=3,5*10⁻¹¹Ф

q=7нКл=7*10⁻⁹Кл

----------------

С-?; U1-3-?; W1-3-?

При последовательном соединении заряды на всех конденсаторах одинаковы:

q1=q2=q3=q

При последовательном соединении:

1/С=1/С1+1/С2+1/С3

С=С1*С2*С3/(С1*С2+С1*С3+С2*С3)=15*25*35/(15*25+15*35+25*35)=

7,4пФ=7,4*10⁻¹²Ф

Напряжения на конденсаторах равны:

U1=q/C1=7*10⁻⁹/1,5*10⁻¹¹=467В

U2=q/C2=7*10⁻⁹/2,5*10⁻¹¹=280В

U1=q/C3=7*10⁻⁹/3,5*10⁻¹¹=200В

Энергия конденсаторов равна:

W1=C1*U1²/2=1,5*10⁻¹¹*467²/2=1,6*10⁻⁶Дж

W2=C2*U2²/2=2,5*10⁻¹¹*280²/2=9,8*10⁻⁷Дж

W3=C3*U3²/2=3,5*10⁻¹¹*200²/2=7*10⁻⁷Дж

Answer 2

Calculation of Total Capacitance:

To find the total capacitance of the three capacitors connected in series, we can use the formula:

1/C_total = 1/C1 + 1/C2 + 1/C3

Given that the capacitances of the three capacitors are 15 pF, 25 pF, and 35 pF respectively, we can substitute these values into the formula:

1/C_total = 1/15 pF + 1/25 pF + 1/35 pF

Now, let's calculate the total capacitance:

1/C_total = (35 + 21 + 15) / (15 * 25 * 35) pF

C_total = 71 / (15 * 25 * 35) pF

C_total ≈ 0.027 pF

Therefore, the total capacitance of the three capacitors connected in series is approximately 0.027 pF.

Calculation of Voltage:

To find the voltage across each capacitor, we can use the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the voltage.

Given that the charge provided to the capacitors is 7 nC (nanocoulombs), we can calculate the voltage across each capacitor using the formula:

V1 = Q / C1

V2 = Q / C2

V3 = Q / C3

Substituting the values of charge and capacitance:

V1 = 7 nC / 15 pF

V2 = 7 nC / 25 pF

V3 = 7 nC / 35 pF

Now, let's calculate the voltage across each capacitor:

V1 ≈ 0.467 V

V2 ≈ 0.28 V

V3 ≈ 0.2 V

Therefore, the voltage across the first capacitor is approximately 0.467 V, the voltage across the second capacitor is approximately 0.28 V, and the voltage across the third capacitor is approximately 0.2 V.

Calculation of Energy:

To find the energy stored in each capacitor, we can use the formula:

E = (1/2) * C * V^2

where E is the energy, C is the capacitance, and V is the voltage.

Let's calculate the energy stored in each capacitor:

E1 = (1/2) * 15 pF * (0.467 V)^2

E2 = (1/2) * 25 pF * (0.28 V)^2

E3 = (1/2) * 35 pF * (0.2 V)^2

Now, let's calculate the energy stored in each capacitor:

E1 ≈ 0.051 nJ (nanojoules)

E2 ≈ 0.014 nJ

E3 ≈ 0.007 nJ

Therefore, the energy stored in the first capacitor is approximately 0.051 nJ, the energy stored in the second capacitor is approximately 0.014 nJ, and the energy stored in the third capacitor is approximately 0.007 nJ.

Please note that the calculations provided are based on the given values and formulas.