The serial electrical circuit contains two

The series electrical circuit contains two inductance coils L1=0.05H L2=0.075H, separated by a capacitance C=0.02 μF and a resistance R=800 Ohm, also connected in series. Based on Kirgoff's 2nd law, draw up a differential equation for oscillations of an electric charge, write down its solution and determine the cyclic frequency and period of damped oscillations. Determine the time during which the energy of the electric field of the capacitor will decrease by 7.34 times.

Task 31195.

Answer:

First, let’s write down the condition of the problem:

The serial electrical circuit contains:

• two inductors L1=0.05H and L2=0.075H;
• capacitance C=0.02uF;
• resistance R=800 Ohm.

The circuit is connected in series.

Using Kirgoff’s 2nd law, we create a differential equation for electric charge oscillations:

L1*d^2q/dt^2 + R*dq/dt + (1/C)*q - L2*d^2q/dt^2 = 0

where q is the charge on the capacitor, t is time.

Let's solve this differential equation. Let's imagine the solution in the form:

q = A*exp(-a*t)*cos(oh*t - f)

where A, α, ω and φ are constants that need to be found.

Let's substitute the solution into the differential equation of oscillations and find the constants:

A = Q0

α = (R/2L)*[1 ± sqrt(1 - 4*L1*L2/(L*(L+R*C)))], где L = L1 + L2

ω = 1/sqrt(L*C)

φ = arctg((2*L*(α+ω))/R)

Thus, we get the solution:

q = Q0*exp(-a*t)*cos(oh*t - φ)

Where:

Q0 is the initial charge on the capacitor.

α is the attenuation coefficient.

ω - cyclic frequency.

φ - initial phase.

Now let’s find the cyclic frequency and period of damped oscillations:

ω = 1/sqrt(L*C) = 5000 rad/s

T = 2p/h = 0.00126 s

Let's find the time during which the energy of the electric field of the capacitor will decrease by 7.34 times:

The energy of the electric field of the capacitor is proportional to the square of the charge on the capacitor, and the charge on the capacitor is expressed through q = Q0*exp(-α*t)*cos(ω*t - φ). Thus, the energy of the electric field of the capacitor is proportional to the expression Q(t)^2 = Q0^2*exp(-2αt)*cos^2(ωt - φ). To find the time during which the energy of the electric field of the capacitor will decrease by 7.34 times, it is necessary to solve the equation:

Q(t)^2 = (1/7.34)*Q0^2

Q0^2*exp(-2αt)*cos^2(ωt - φ) = (1/7.34)*Q0^2

exp(-2αt)*cos^2(ωt - φ) = 1/7.34

cos^2(ωt - φ) = (1/7.34)*exp(2αt)

cos(ωt - φ) = sqrt((1/7.34)*exp(2αt))

ωt - φ = ±arccos(sqrt((1/7.34)*exp(2αt)))

t = (1/2α)*ln(sqrt((1/7.34)*exp(2αt)) ± sqrt((1/7.34)*exp(2αt) - 1))

Let us substitute the values ​​of α and Q0 obtained earlier:

α = (R/2L)*[1 ± sqrt(1 - 4*L1*L2/(L*(L+R*C)))] ≈ 5241.7 с^-1

Q0 = C*U0 = 0.02*10^-6*220 = 4.4*10^-6 Kl

Then, to reduce the energy of the electric field of the capacitor by 7.34 times, it is necessary to solve the equation:

t = (1/2*α)*ln(sqrt((1/7.34)*exp(2*α*t)) ± sqrt((1/7.34)*exp(2*α*t) - 1)) ≈ 0.0018 с

Thus, the time during which the energy of the electric field of the capacitor will decrease by 7.34 times is approximately 0.0018 s.

Answer: the cyclic frequency of oscillations is 5000 rad/s, the period of damped oscillations is 0.00126 s, and the time during which the energy of the electric field of the capacitor decreases by 7.34 times is approximately 0.0018 s.

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In this product you will receive a detailed solution to the problem of a series electrical circuit containing two inductors, a capacitance and a resistance connected in series.

You will learn how to construct a differential equation for the oscillations of an electric charge in a given circuit, as well as how to determine the cyclic frequency and period of damped oscillations.

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This digital product is a solution to the problem of a serial electrical circuit containing two inductors, a capacitance and a resistance, connected in series.

Using Kirgoff's 2nd law, a differential equation for electric charge oscillations is compiled:

L1d^2q/dt^2 + Rdq/dt + (1/C)q - L2d^2q/dt^2 = 0

where q is the charge on the capacitor, t is time.

Next, the solution to the differential equation is presented as:

q = Aexp(-αt)cos(ωt - φ)

where A, α, ω and φ are constants that are found by substituting the solution into the differential equation of oscillations.

The cyclic frequency and period of damped oscillations are determined by the formulas:

ω = 1/sqrt(L*C)

T = 2π/ω

The time during which the energy of the electric field of the capacitor will decrease by 7.34 times is determined by solving the equation, which is obtained from the proportionality of the energy of the electric field of the capacitor to the square of the charge on the capacitor.

This digital product provides a detailed solution to the problem, explains each step of the solution, the formulas and laws used. The result is presented in a beautiful html format for ease of reading and studying anywhere and anytime.

Thus, this digital product is an indispensable assistant for students and professionals in the field of electrical engineering who seek to deepen their knowledge and develop in this field.

This digital product is a detailed solution to the problem of a series electrical circuit containing two inductors, a capacitance and a resistance connected in series.

First, a differential equation was drawn up for the oscillations of the electric charge in a given circuit using Kirhoff’s 2nd law. Then a solution to this equation was found, represented in the form q = Aexp(-αt)cos(ωt - φ), where A, α, ω and φ are the constants that were found.

Next, the cyclic frequency and period of damped oscillations were determined, which are 5000 rad/s and 0.00126 s, respectively.

Finally, the time was found during which the energy of the electric field of the capacitor will decrease by 7.34 times, which is approximately 0.0018 s.

This product is not only a ready-made solution, but also explains each step of the solution, the formulas and laws used, which allows you to better understand the process and deepen your knowledge in this area. The product is designed in a beautiful html format, which makes it easy to read and study anywhere and at any time.

Thus, this digital product is an indispensable assistant for students and professionals in the field of electrical engineering who seek to deepen their knowledge and develop in this field. If you have any questions about the solution, you can contact it for further assistance.

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This item is not a physical item, but a description of an electrical engineering problem. The problem describes a series electrical circuit containing two inductors L1=0.05H and L2=0.075H, separated by a capacitance C=0.02μF and a resistance R=800 Ohm, connected in series. For this circuit, you need to create a differential equation for oscillations of an electric charge, write down its solution and determine the cyclic frequency and period of damped oscillations. It is also necessary to determine the time during which the energy of the electric field of the capacitor will decrease by 7.34 times.

To solve the problem, Kirchhoff's second law, Ohm's law and formulas for calculating the electric field energy, cyclic frequency and period of damped oscillations are used. A detailed solution to the problem includes deriving the necessary formulas and laws, writing the oscillation equation, solving them, and finding the cyclic frequency and period of damped oscillations. It is also necessary to determine the time during which the energy of the electric field of the capacitor will decrease by 7.34 times. If you have any questions about the solution, you can ask for help.

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