A series circuit has a capacitor of 025 times 106 F and an i

A series circuit has a capacitor of 0.25 times 10^-6 F and an inductor of 1 H. If the initial charge on the capacitor is 10^-6 C and there is no initial current, find the charge Q on the capacitor at any time t. Find the resistance r so that the circuit described in part(a) would be critically damped.

Solution

the differential equation for the current in an LC circuit is given by:
i\'\'(t) + i(t)/(L*C) = 0
where i(t) is the current in the circuit at time t
L is the inductance of the inductor.
C is the capacitance of the capacitor
It is a second-order, linear, homogeneous, ordinary differential equation with constant coefficients.
so let us assume the solution is of the form i(t) = exp(k*t)

In this case, we get the characteristic equation:
k^2 + 1/(L*C) = 0
k = +/- i/sqrt(L*C)
where i = sqrt(-1) (imaginary number)

The roots of the characteristic equation are imaginary, so the solution can be written as:
i(t) = a*sin(t/sqrt(L*C)) + b*cos(t/sqrt(L*C))
where a and b are the constants of integration.
Now it is given that i(0) = 0, so therefore b = 0.

the relation ship between voltage accros a capacitor and a charge across a capacitor is given by:
V_C = Q/C

By Kirchhoff\'s law for voltages in a circuit, the sum of the voltage across the capacitor plus the voltage across the inductor must be zero, so -V_C = V_L = -Q/C

The voltage across the inductor is related to the rate that the current in the inductor is changing by:
V_L = L*di/dt = L*i\'
So:
-Q/C = L*i\'
i\' = -Q/(L*C)
From our solution for i, we have that:
i(t) = a*sin(t/sqrt(L*C))
i\'(t) = (a/sqrt(L*C))*cos(t/sqrt(L*C))
At t = 0, i\'(0) = (a/sqrt(L*C))
So:
a/sqrt(L*C) = -Qo/(L*C)
where Qo is the initial charge on the capacitor.
a = -Qo/sqrt(L*C)
The complete solution is then given by:
i(t) = -(Qo/sqrt(L*C))*sin(t/sqrt(L*C))

Here, L = 1H and C = 2.5*10^-7 F, so:

i(t) = (-2*10^-3 A)*sin(2000*t/sec)
The current through the capacitor is given by: i(t) = dQ(t)/dt
i(t) dt = dQ(t)
where Q(t) is the charge on the capacitor at time t.
So, in general, in an LC circuit:
dQ = (-Qo/sqrt(L*C))*sin(t/sqrt(L*C)) dt
Q(t) = Qo*cos(t/sqrt(L*C)) +c
where c is the constant of integration. We already know that at t = 0, Q(0) = Qo, so c = 0.
Hence,
Q(t) = (10^-6 coulombs)*cos(2000*t/sec)
where t is any time