Answer:
Explanation:
Given that,
In an LRC circuit
L = 1/2h
R = 10 Ω,
C = 0.01 f
E(t) = 50 V,
q(0) = 1 C, and
i(0) = 0 A.
q(t) = C
We can to fine the charge after a long time, let say t→∞
The Kirchoff second law for the system is
L•dq²/dt + R•dq/dt + q/C = E(t)
Divide through by L
dq²/dt + R/L •dq/dt + q/LC = E(t)/L
Now inserting the values of R, L, C and E
dq²/dt+10/½ •dq/dt +q/½×0.01=50/½
dq²/dt + 20•dq/dt + 200q = 100
Let solve the differential equation
First : homogenous solution
Using D operator
D² + 20D +200 = 0
Solving the quadratic equation using formula method
D = (-b±√b²-4ac)/2a
D = (-20±√20²-4×1×200) /2
D = (-20±√400-800)/2
D = (-20±20•i)/2
D = -10±10•i
So we have a complex solution
Then, the complementary solution is
q(t) = e(-10t)[ Acos10t + BSin10t]
A and B are constant
Let find the particular solution using the method of undetermine coefficient
Let assume particular solution of
q(t) = C, I.e q(t) Is a constant
So, inserting this into the equation below
dq²/dt + 20•dq/dt + 200q = 100
200q = 100
q = 100/200
q = ½
Then, the particular solution is ½
So, the total solution is the sum of particular solution and complementary solution
q = e(-10t)[ Acos10t + BSin10t] + ½
Using the initial conditions
q(0) = 1
1 = e(0) [ACos0 + BSin0] +½
1 = A+½
A = ½
Also i(0) = 0
I(t) = q'(t)
Then,
q'(t) = -10•e(-10t)[ Acos10t + BSin10t] + e(-10t)[ -10Asin10t + 10BCos10t]
0 = -10e(0) [ ACos0 + BSin0] + e(0)[-10ASin0 +10BCos0]
0 = -10(A) + 10B
A=B=½
So the general equation becomes
q(t) = e(-10t)[ ½cos10t + ½Sin10t] + ½
So, as t→∞, the aspect of e(-10t) become zero
So the charge stabilizes at q = ½C after a long time
q = ½C as t→∞
