Question

simple series circuit inductor 1 henry capacitor of 10 power minus 6 Faradas and a resistor of 1000 ohms . the initial charge on a 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

The voltage and the current relationships in circuit analysis

$i= \frac{dq}{dt} ​$

voltage drop across the inductor

$V_L =L\frac{di}{dt}=L\frac{d^2q}{dt^2}​$

voltage drop across the resistor

$V_R=Ri=R\frac{dq}{dt}$

voltage drop across the capacitor

$V _C=\frac{q}{C}$

Kirchhoff's Second Law

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

Given that $L=1$H, $R=1000$ohm, $C=10^{-6}$ F, $E(t)=12$ V

$1\frac{d^2q}{dt^2}+1000\frac{dq}{dt}+\frac{1}{10^{-6}}q=12\\ \frac{d^2q}{dt^2}+1000\frac{dq}{dt}+1000000q=12\\$

The characteristic equation is

$r^2+1000r+1000000=0\\ D=(1000)^2-4\cdot1000000=-3000000\\ r_1=\frac{-1000+i1000\sqrt3}{2}=-500+500\sqrt3i\\ r_2=\frac{-1000-i1000\sqrt3}{2}=-500-500\sqrt3i\\ q_C=e^{-500t}(c_1\cos(500\sqrt3t)+c_2\sin(500\sqrt3t))$

Find the particular solution of the non-homogeneous differential equation using the method of the undetermined coefficients.

Assume that $Q(t)=A$ is a solution of the non-homogeneous differential equation where $E(t)=12$

Differentiate the assumption with respect to $t$

$Q'=0, Q''=0$

Substitute in the original equation

$0+0+1000000A=12\\ A=1.2\cdot10^{-5}\\ Q(t)=1.2\cdot10^{-5}\\ q(t)= e^{-500t}(c_1\cos(500\sqrt3t)+c_2\sin(500\sqrt3t))+\\ +1.2\cdot10^{-5}$

The initial charge of the capacitor is zero

$q(0)= e^{-500\cdot 0}(c_1\cos(500\sqrt3\cdot 0)+\\ +c_2\sin(500\sqrt3\cdot 0))+1.2\cdot10^{-5}=0\\ c_1=-1.2\cdot10^{-5}$

The circuit is closed at $t=0: i(0)=q'(0)=0\\$

$q'(t)= -500e^{-500t}(-1.2\cdot10^{-5})(\cos(500\sqrt3t)-\\ -500e^{-500t}(c_2\sin(500\sqrt3t))+\\ + 1.2\cdot10^{-5}(-500\sqrt3)e^{-500t}(\sin(500\sqrt3t)+\\ +c_2\cdot500\sqrt3e^{-500t}\cos(500\sqrt3t))\\ q'(0)=-500(-1.2\cdot10^{-5})+c_2\cdot500\sqrt3=0\\ c_2=-0.4\sqrt3\cdot10^{-5}\\ q(t)=e^{-500t}(-1.2\cdot10^{-5}\cos(500\sqrt3t)-\\ -0.4\sqrt3\cdot10^{-5}\sin(500\sqrt3t))+1.2\cdot10^{-5}$

Coulomb:

$q(1)=e^{-500\cdot(1)}(-1.2\cdot10^{-5}\cos(500\sqrt3\cdot(1))-\\ -0.4\sqrt3\cdot10^{-5}\sin(500\sqrt3\cdot(1)))+1.2\cdot10^{-5}\approx\\ \approx1.2\cdot10^{-5}(C)$

Find the steady state charge.

$t\to\infty:q_{p}(t)\to1.2\cdot10^{-5}C$