Question

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

Answer 1

The voltage and the current relationships in circuit analysis

$$i={dq \over dt}$$

voltage drop across the inductor

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

voltage drop across the resistor

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

voltage drop across the capacitor

$$V_C={q \over C}$$

Kirchhoff's Second Law

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

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

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

$${d^2q\over dt^2}+1000{dq \over dt}+1000000q=12$$

The characteristic equation is

$$r^2+1000r+1000000=0$$

$$D=(1000)^2-4(1000000)=-3(1000000)$$

$$r={-1000\pm i1000\sqrt{3} \over 2}$$

$$q_c=e^{-500t}(c_1\cos{(500 \sqrt{3}t)}+c_2\sin{(500 \sqrt{3}t)})$$

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

$$0+0+1000000A=12$$

$$A=1.2\times10^{-5}$$

$$Q(t)=1.2\times10^{-5}$$

$$q(t)=e^{-500t}(c_1\cos{(500 \sqrt{3}t)}+c_2\sin{(500 \sqrt{3}t)})+1.2\times10^{-5}$$

The initial charge of the capacitor is zero

$$q(0)=e^{-500(0)}(c_1\cos{(500 \sqrt{3}(0))}+c_2\sin{(500 \sqrt{3}(0))})+1.2\times10^{-5}=0$$

$$c_1=-1.2\times10^{-5}$$

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

$$q'(t)=-500e^{-500t}(-1.2\times10^{-5})\cos{(500 \sqrt{3}t)}-$$ $$-500e^{-500t}c_2\sin{(500 \sqrt{3}t)}+$$ $$+1.2\times10^{-5}(500 \sqrt{3})e^{-500t}\sin{(500 \sqrt{3}t)}+$$ $$+c_2500 \sqrt{3}e^{-500t}\cos{(500 \sqrt{3}t)}$$

$$q'(0)=-500(-1.2\times10^{-5})-0+0+c_2500 \sqrt{3}=0$$

$$c_2=-{\sqrt{3} \over 1.2}\times10^5$$

$$q(t)=e^{-500t}\big(-1.2\times10^{-5}\cos{(500 \sqrt{3}t)}-{\sqrt{3} \over 1.2}\times10^5 \sin{(500 \sqrt{3}t)}\big)+$$ $$+1.2\times10^{-5}$$

$Coulomb: [C]$

$$q(1)=e^{-500(1)}\big(-1.2\times10^{-5}\cos{(500 \sqrt{3}(1))}-{\sqrt{3} \over 1.2}\times10^5 \sin{(500 \sqrt{3}(1))}\big)+$$ $$+1.2\times10^{-5}\approx1.2\times10^{-5}(C)$$

Find the steady state charge.

$t\to\infin: q_p(t)\to1.2\times10^{-5} \ C$