Question

Find the charge on the capacitor in an LRC-series circuit when L = 1 2 h, R = 10 Ω, C = 0.01 f, E(t) = 50 V, q(0) = 1 C, and i(0) = 0 A. q(t) = C What is the charge on the capacitor after a long time? C

Answer 1

Answer:

Explanation:

Consider

L = 1 2 h, R = 10 Ω, C = 0.01 f, E(t) = 50 V, q(0) = 1 C, and i(0) = 0 A. q(t) = C

$Lq''+Rq'+\frac{q}{C} =E(t)$

$\frac{1}{2} q''+10q'+100q=50\\\\q''+20q'+200q=100\\\\(D^2+20q+200)q=100$

The auxilliary equation of the differential equation is as follows

$m^2+20m+200=0\\\\m=-10 \pm10i\\\\q_c=e^{-10t}(A \cos (10t)+B \sin (10t)\\\\q_p=\frac{1}{(D^2+20D+200)} 100 \\\\=\frac{1}{(D^2+20D+200)} 100e^{0t}\\\\\frac{100}{(0+0+200)} \\\\=\frac{1}{2}$

Hence, the general solution is as follow

$q(t)=q_c+q_p\\\\q(t)=e^{-10t}(A \cos (10t)+B \sin (10t)+\frac{1}{2} \\\\q'(t)=e^{-10t}(-10A \sin (10t)+10B \cos (10t))-10e^{-10t}(A \cos (10t)+B \sin (10t)$

$q(0)=1$ ⇒ $A+\frac{1}{2} = 1$⇒ $A=\frac{1}{2}$

$i(0)=q'(0)=0$ ⇒ $10B - 10A=0$ ⇒ $B=A=\frac{1}{2}$

Hence,

$q(t)=e^(-10t)(\frac{1}{2} \cos (10t) + \frac{1}{2} \sin (10t)+\frac{1}{2}$

Therefore ,the charge on the capacitor is 1/2

Answer 2

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→∞