Question

a simple series circuit has an inductor of 1 henry a capacitor of 10^-6 farads and a resistor of 1000 ohms.the initial charge on the capacitor is zero.if a 12 volt battery is connected to the circuit and the circuit is closed at t=0 find the charge on the capacitor 1 second later and the steady state charge

Answer 1

Kirchhoffs Second Law: Impressed Voltage $E(t)$ on a closed loop = Sum of voltage drop in the loop: 

$$L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)$$

where $i=\frac{dq}{dt},\, V_L=L\frac{di}{dt},\, V_R=Ri,\,V_C=\frac{q}{C}$

We have $L=1\,H,\,C=10^{-6}\,F,\,R=1000\,\text{Ohm}$

and $E(t)=12\,V,\, q(0)=0, \, q'(0)=0$

Solution: (see [1])

$q(t)=4\cdot 10^{-6} \cdot e^{-500t}(3e^{500t}-\sqrt{3} \sin (500 \sqrt 3 t)-3\cos (500\sqrt 3 t))$

$q(1)\approx 12\cdot10^{-6}$ C (see [2])

$\lim\limits_{t\to \infty} q(t)=12 \cdot 10^{-6}$ C (see [3])

(here C is Coulomb)

Appendix. (Theoretical solution of differential equation)

The characteristic equation is $\lambda^2+10^3\lambda+10^6=0$

$\lambda=-500\pm i\,500\sqrt 3$

Hence, the complementary solution is $q_c(t)=c_1e^{-500t}\sin(500\sqrt 3 t)+c_2e^{-500t}\cos(500\sqrt 3 t)$

Find the partial solution using method of undetermined coefficient

$q_p(t)=A, \,q_p'(t)=0, \,q_p''(t)=0$ substitute in original equation:

$10^6A=12 \rightsquigarrow A=12\cdot10^{-6} \rightsquigarrow q_p(t)=12\cdot10^{-6}$

So, the general solution is $q(t)=q_c+q_p$

$q(0)=0 \rightsquigarrow c_2=-12\cdot10^{-6}$

$q'(0)=0 \rightsquigarrow -500c_2+500 \sqrt{3} c_1=0$

$c_1=-4\cdot \sqrt{3} \cdot10^{-6}$

Finally, $q(t)=-4\cdot 10^{-6} e^{-500t}(\sqrt 3 \sin (500\sqrt 3 t) +3\cos(500\sqrt 3 t))+12\cdot10^{-6}$

It is equal to solution given by [1]

Moreover,

$\sqrt 3 \sin (500\sqrt 3 t) +3\cos(500\sqrt 3 t)=2\sqrt{3}\cos(500\sqrt3 t+\pi/6)$

$q(t)=-8\sqrt{3}\cdot10^{-6}e^{-500t}\cos(500\sqrt{3}t+\pi/6)+12\cdot10^{-6}$

Obviously, firs term $\ll1$ for $t=1$ and $\to0$ when $t\to\infty$.

[1] https://www.wolframalpha.com/input/?i=q%27%27%28t%29%2B1000+q%27%28t%29%2B10%5E6+q%28t%29%3D12%2C+q%280%29%3D0%2C+q%27%280%29%3D0

[2] https://www.wolframalpha.com/input/?i=-%28-3+e%5E500+%2B+3+cos%28500+sqrt%283%29%29+%2B+sqrt%283%29+sin%28500+sqrt%283%29%29%29%2F%28250000+e%5E500%29&assumption=%22ClashPrefs%22+-%3E+%7B%22Math%22%7D

[3] https://www.wolframalpha.com/input/?i=%28e%5E%28-500+t%29+%283+e%5E%28500+t%29+-+3+cos%28500+sqrt%283%29+t%29+-+sqrt%283%29+sin%28500+sqrt%283%29+t%29%29%29%2F250000%2C+t+to+infinity