Respuesta :
Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.
The problem can be modeled through a linear equation, in the form:
[tex]10^5 Q +300Q'+0.2Q''=0[/tex]
With the initial conditions as,
[tex]Q(0) = 10^{-6}[/tex]
[tex]Q'(0)= 0[/tex]
Where Q(t) is the charge.
The general solution of a linear equation is given as:
[tex]y(x) = c_1e^{-ax}+c_2e^{-bx}[/tex]
Applying this definiton in our differential equation we have that
[tex]Q(t) = C_1e^{at}+C_2e^{bt}[/tex]
To find b and a we use the first equation and find the roots:
[tex]r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}[/tex]
[tex]r_{a,b} = {-1000,-500}[/tex]
Then we have
[tex]Q(t) = C_1e^{-1000t}+C_2e^{-500t}[/tex]
To find the values of the Constant we apply the initial conditions, then
[tex]Q(0)= 10^{-6} = C_1+C_2[/tex]
And for the derivate:
[tex]Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}[/tex]
[tex]0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}[/tex]
[tex]0 = -1000C_1-500C_2[/tex]
We have a system of 2x2:
[tex](1) 10^{-6} = C_1+C_2[/tex]
[tex](2) 0 = -1000C_1-500C_2[/tex]
Solving we have:
[tex]C_1 = -10^{-6}[/tex]
[tex]C_2 = 2*10^{-6}[/tex]
The we can replace at the equation and we have that the Charge at any moment is given by,
[tex]Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}[/tex]
If we obtain the derivate we find also the Current, then
[tex]I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}[/tex]
The charge Q on the capacitor at any time t, for a series circuit of capacitor, is,
[tex]q(t)=(-10^{-6})e^{-(1000)t}+(2\times10^{-6})e^{-(500)t}[/tex]
What is electric potential energy of a capacitor?
The electric potential energy of a capacitor is the energy stored by it.
The initial charge on the capacitor is 10−6 C and there is no initial current.
A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. Let the charge on these components, by the differential equation form, is q, q' and q" respectively. Therefore,
[tex]10^5q+300q'+0.2q"=0\\0.2q"+300q'+10^5q=0[/tex]
On solving this equation, the roots of the equation, we get are (-1000, -500). The general solution of the above linear equation is,
[tex]q(t)=C_1e^{-at}+C_2e^{-bt}[/tex]
Put the values of a and b as,
[tex]q(t)=C_1e^{-(1000)t}+C_2e^{-(500)t}[/tex]
Now, the initial charge on the capacitor is 10−6 C at the time zero. Therefore, the equation, become as,
[tex]q(0)=C_1e^{(-1000)(0)}+C_2e^{(-500)(0)}\\10^{-6}=C_1(1)+C_2(1)\\C_1+C_2=10^{-6}\pi[/tex].....1
Differentiate the above equation, with respect to t,
[tex]q'(t)=-1000C_1e^{(-1000)t}-500C_2e^{(-500)t}[/tex]
Put the initial values again,
[tex]0=-1000C_1e^{(-1000)0}-500C_2e^{(-500)0}\\-1000C_1-500C_2=0[/tex].....2
Solving the equation 1 and 2, we get,
[tex]C_1=-10^{-6}\\C_2=2\times10^{-6}[/tex]
Put these values of constant, to find the charge as,
[tex]q(t)=(-10^{-6})e^{-(1000)t}+(2\times10^{-6})e^{-(500)t}[/tex]
Thus, the charge Q on the capacitor at any time t, for a series circuit of capacitor is,
[tex]q(t)=(-10^{-6})e^{-(1000)t}+(2\times10^{-6})e^{-(500)t}[/tex]
Learn more about the capacitor here;
https://brainly.com/question/14542122
