Answer to Question #315343 in Differential Equations for sandip

Question

Consider an electric circuit with an inductance of 0.05 henry, a resistance of 20 ohms, a condenser of capacitance of 100 micro farads and an E = 100 volts. Find I and Q given the initial conditions Q = 0, I = 0 at t = 0

Answer 1

Due to Kirchhoff's law:

$U_L+U_R+U_C=E$ (1)

where

$U_L=L\frac{di}{dt}$ ; $U_R=iR$ ; $U_C=\frac1C\int idt$ .

Let's differentiate (1):

$L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac 1C i=0$

$\frac{d^2i}{dt^2}+\frac RL\frac{di}{dt}+\frac {1}{LC} i=0$

Characteristic equation:

$\lambda^2+\frac RL\lambda+\frac {1}{LC}=0$

$\frac RL=\frac {20}{0.05}=400$; $\frac {1}{LC}=\frac {1}{0.05\cdot 100\cdot 10^{-6}}=2\cdot10^5$.

$\lambda^2+400\lambda+2\cdot 10^5=0$

$\lambda=-200\pm400j$ , where $j=\sqrt{-1}$ .

$i(t)=e^{-200t}(C_1 \cos {400t}+C_2 \sin {400t})$.

$i|_{t=0}=C_1=0;$

$L\frac{di}{dt}|_{t=0}+Ri|_{t=0}+U_C|_{t=0}=E$ (2)

$U_C|_{t=0}=\frac 1C Q|_{t=0}=0$ ; (3)

$Ri|_{t=0}=0$; (4)

$\frac{di}{dt}=C_2e^{-200t}(-200\cdot\sin{400t}+400\cdot\cos{400t})$;

$\frac{di}{dt}|_{t=0}=400\cdot C_2$. (5)

Substitute (3), (4) and (5) into (2):

$L\cdot400\cdot C_2=E$;

$C_2=\frac{E}{400\cdot L}=\frac{100}{400\cdot 0.05}=5$

$i(t)=5\cdot e^{-200t}\cdot \sin{400t}.$

$Q(t)=\int5\cdot e^{-200t}\cdot \sin{400t} dt=$

$-\frac{1}{200}e^{-200t}(\sin{400t}+2\cos{400t})+C_3$;

$Q|_{t=0}=-\frac{1}{200}\cdot2+C_3=0$;

$C_3=\frac{1}{100}$

$Q(t)=\frac{1}{200}[2-e^{-200t}(\sin{400t}+2\cos{400t})]$ .

Answer:

$i(t)=5\cdot e^{-200t}\cdot \sin{400t}$ ;

$Q(t)=\frac{1}{200}[2-e^{-200t}(\sin{400t}+2\cos{400t})]$ .